Question 15 Marks
Solve the following system of equations by matrix method:
$3x + 4y + 7z = 14$
$2x - y + 3z = 4$
$x + 2y - 3z = 0$
AnswerThe given system of equations can be written in matrix form as follows:
$\begin{bmatrix}3&1\\ 3&-1\end{bmatrix}\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\begin{bmatrix}19\\ 23\end{bmatrix}$
$\text{AX}=\text{B}$
Here,
$\text{A}=\begin{bmatrix}3&1\\ 3&-1\end{bmatrix},\text{X}=\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}\text {and }\text{B}=\begin{bmatrix}19\\ 23\end{bmatrix}$
Now,
$\text{|A|}=\begin{vmatrix}3&1\\ 3&-1\end{vmatrix}$
$=-3-3$
$=-6\neq0$
So, the given system has a unique solution given by $X = A^{-1}B.$
Let $c_{ij}$ be the co-factors of the elements $a_{ij}$
$\text{C}_{11}{(-1)}^{1+1}{(-1)}=-1,\text{C}_{12}={(-1)}^{1+2}{(3)}=-3$
$\text{C}_{21}={(-1)}^{2+1}{(1)}=-1,\text{C}_{22}={(-1)}^{2+2}{(3)}=3$
$\text{adj}\ \text{A}=\begin{bmatrix}-1&-3\\ -1&3\end{bmatrix}^\text{T}$
$=\begin{bmatrix}-1&-1\\ -3&3\end{bmatrix}$
$\text{A}^{-1}=\frac{1}{\text{|A|}}\text{adj A}$
$=\frac{1}{-6}\begin{bmatrix}-1&-1\\ -3&3\end{bmatrix}$
$\text{X}=\text{A}^{-1}\text{B}$
$=\frac{1}{-6}\begin{bmatrix}-1&-1\\ -3&3\end{bmatrix}\begin{bmatrix}19\\ 23\end{bmatrix}$
$=\frac{1}{-6}\begin{bmatrix}-19-23\\ -57+69\end{bmatrix}$
$=\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\begin{bmatrix}\frac{-42}{-6}\\ \frac{12}{-6}\end{bmatrix}$
View full question & answer→Question 25 Marks
Two schools P and Q want to award their selected students on the values of Tolerance, Kindness and Leadership. The school P wants to award ₹ x each, ₹ y each and ₹ z each for the three respective values to 3, 2 and 1 students respectively with a total award money of ₹ 2,200. School Q wants to spend ₹ 3,100 to award its 4, 1 and 3 students on the respective values (by giving the same award money to the three values as school P). If the total amount of award for one prize on each values is ₹ 1,200, using matrices, find the award money for each value.
Apart from these three values, suggest one more value which should be considered for award.
Answerx, y and z be prize amount per student for Tolerence, Kindness and Leadership respectively.
As per the data in the question, we get
3x + 2y + z = 2200
4x + y + 3z = 3100
x + y + z = 1200
The above three simultaneous equations can be written in matrix form as
$\begin{bmatrix}3&2&1\\4&1&3\\1&1&1\end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}2200\\3100\\1200\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}3&2&1\\4&1&3\\1&1&1\end{bmatrix}^{-1}\begin{bmatrix}2200\\3100\\1200\end{bmatrix}\ \dots(1)$
$\text{A}=\begin{bmatrix}3&2&1\\4&1&3\\1&1&1\end{bmatrix}$
$|\text{A}|=3(-2)-2(1)+1(3)=-5$
$\text{cof }\text{A}=\begin{bmatrix}-2&-1&3\\-1&2&-1\\5&-5&-5\end{bmatrix}$
$\text{adj }\text{A}=(\text{cof }\text{A})^\text{T}=\begin{bmatrix}-2&-1&5\\-1&2&-5\\3&-1&-5\end{bmatrix}$
$\text{A}^{-1}=\frac{\text{adj }\text{A}}{|\text{A}|}=\frac{\begin{bmatrix}-2&-1&5\\-1&2&-5\\3&-1&-5\end{bmatrix}}{-5}$
From (1)
$\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\frac{\begin{bmatrix}-2&-1&5\\-1&2&-5\\3&-1&-5\end{bmatrix}}{-5}\begin{bmatrix}2200\\3100\\1200\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}-2&-1&5\\-1&2&-5\\3&-1&-5\end{bmatrix}\begin{bmatrix}-440\\-620\\-240\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}300\\400\\500\end{bmatrix}$
Excellence in extra-curricular activities should be another value considered for an award.
View full question & answer→Question 35 Marks
Find $A^{-1},$ If $\text{A}=\begin{bmatrix}1&2&5\\ 1&-1&-1\\ 2&3&-1\end{bmatrix}$. Hence solve the follwing system of linear equations:
$x + 2y +5z = 10, x- y - z = - 2, 2x + 3y - z = - 11$
Answer$\text{A}=\begin{bmatrix}1&2&5\\ 1&-1&-1\\ 2&3&-1\end{bmatrix}$ $\text{|A|}=1{(1+3)}-2{(-1+2)}+5{(5)}=4-2+25=27\neq0$ $\text{C}_{11}=4\\ \text{C}_{12}=-1\\ \text{C}_{13}=5$ $\text{C}_{31}=3\\ \text{C}_{32}=6\\ \text{C}_{33}=-3$ $\text{C}_{21}=17\\ \text{C}_{22}=-11\\ \text{C}_{23}=1$ $\text{A}^{-1}=\frac{1}{\text{|A|}}\times\text{adj A}=\frac{1}{27}\begin{bmatrix}4&17&3\\ -1&-11&6\\ 5&1&-3\end{bmatrix}$ Now, the given set of equations can be represented as: $\text{x}+2\text{y}+5\text{z}=10$ $\text{x}-\text{y}-\text{z}=-2$ $2\text{x}+3\text{y}-\text{z}=-11$ or $\begin{bmatrix}1&2&5\\ 1&-1&-1\\ 2&3&-1\end{bmatrix}\begin{bmatrix}\text{x}\\ \text{y}\\ \text{z}\end{bmatrix}=\begin{bmatrix}10\\ -2\\ -11\end{bmatrix}$ or $\text{X = A}^{-1}\times\text{B}$ $=\frac{1}{27}\begin{bmatrix}4&17&3\\ -1&-11&6\\ 5&1&-3\end{bmatrix}\begin{bmatrix}10\\ -2\\ -11\end{bmatrix}$ $=\frac{1}{27}\begin{bmatrix}40-34-33\\ -10+22-66\\ 50-2+33\end{bmatrix}=\frac{1}{27}\begin{bmatrix}-27\\ -54\\ 81\end{bmatrix}=\begin{bmatrix}-1\\ -2\\ 3\end{bmatrix}$Hence, $x = -1, y = -2, z = 3$
View full question & answer→Question 45 Marks
Show that the following system of linear equations is consistent and also find solution:
$x + y + z = 6$
$x + 2y + 3z = 14$
$x + 4y + 7z = 30$
AnswerThis system can be written as:
$\begin{bmatrix}1&1&1\\ 1&2&3\\ 1&4&7\end{bmatrix}\begin{bmatrix}\text{x}\\ \text{y}\\ \text{z}\end{bmatrix}=\begin{bmatrix}6\\ 14\\ 30\end{bmatrix}$
or $\text{AX = B}$
$\text{|A|}=1{(2)}-1{(4)}+1{(2)}$
$=2-4+2$
$=0$
So, A is singular, thus the given system has either no solutions or infinite solutions depending on as
$(\text{Adj A})\times\text{(B)}\neq0$ or $(\text{Adj A})\times\text{(B)}=0$
Let $C_{ij}$ be the co-factors of $a_{ij}$ in $A$
$\text{C}_{11}=2\\ \text{C}_{21}=-3\\ \text{C}_{31}=1$
$\text{C}_{12}=-4\\ \text{C}_{22}=6\\ \text{C}_{32}=-2$
$\text{C}_{13}=2\\ \text{C}_{23}=-3\\ \text{C}_{33}=1$
$\text{adj A}=\begin{bmatrix}2&-4&2\\ -3&6&-3\\ 1&-2&1\end{bmatrix}^\text{T}=\begin{bmatrix}2&-3&1\\ -4&6&-2\\ 2&-3&1\end{bmatrix}$
$(\text{adj A})\times\text{B}=\begin{bmatrix}2&-4&2\\ -3&6&-3\\ 1&-2&1\end{bmatrix}\begin{bmatrix}6\\ 14\\ 30\end{bmatrix}\begin{bmatrix}12-42+30\\ -24+84-60\\ 12-42+30\end{bmatrix}=\begin{bmatrix}0\\ 0\\ 0\end{bmatrix}$
So, $\text{AX = B}$ has infinite solutions.
Now, let $z = k$
So, $x + y = 6 - k$
$x + 2y = 14 - 3k$
which can be written as:
$\begin{bmatrix}1&1\\ 1&2\end{bmatrix}\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\begin{bmatrix}6-\text{k}\\ 14-\text{3k}\end{bmatrix}$
or $\text{AX = B}$
$\text{|A|}=1\neq0$
$\text{adj A}=\begin{bmatrix}2&-1\\-1&1\end{bmatrix}^\text{T}=\begin{bmatrix}2&-1\\-1&1\end{bmatrix}$
$\text{X = A}^{-1}\text{B}=\frac{1}{\text{|A|}}\text{adj A}\times\text{B}$
$\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\frac{1}{1}\begin{bmatrix}2&-1\\ -1&1\end{bmatrix}\begin{bmatrix}6-\text{k}\\ 14-3\text{k}\end{bmatrix}$
$=\begin{bmatrix}12-2\text{k}-14+3\text{k}\\ -6+\text{k}+14-3\text{k}\end{bmatrix}$
$=\begin{bmatrix}12-2\text{k}-14+3\text{k}\\ -6+\text{k}+14-3\text{k}\end{bmatrix}$
$\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\begin{bmatrix}-2+\text{k}\\ 8-2\text{k}\end{bmatrix}$
Hence, $x = k - 2$
$y = 8 - 2k$
$z = k$
View full question & answer→Question 55 Marks
If $\text{A}=\begin{bmatrix}1&-1&0\\ 2&3&4\\ 0&1&2\end{bmatrix}\text{and }\text{B}=\begin{bmatrix}2&2&-4\\ -4&2&-4\\ 2&-1&5\end{bmatrix}$ are two square matrices, find AB and hence solve the system of linear equations:
x - y = 3, 2x + 3y + 4z = 17, y + 2z = 7
AnswerHere,
$\text{A}=\begin{bmatrix}1&-1&0\\ 2&3&4\\ 0&1&2\end{bmatrix}\text{and}$$\text{B}=\begin{bmatrix}2&2&-4\\ -4&2&-4\\ 2&-1&5\end{bmatrix}$
Now,
$\text{AB}=\begin{bmatrix}1&-1&0\\ 2&3&4\\ 0&1&2\end{bmatrix}\begin{bmatrix}2&2&-4\\ -4&2&-4\\ 2&-1&5\end{bmatrix}$
$\Rightarrow\text{AB}=\begin{bmatrix}6&0&0\\ 0&6&0\\ 0&0&6\end{bmatrix}$
$\Rightarrow\text{AB}=6\begin{bmatrix}1&0&0\\ 0&1&0\\ 0&0&1\end{bmatrix}$
$\Rightarrow\text{AB}=6\text{I}_{3}$
$\Rightarrow\frac{1}{6}\text{AB}=\text{I}_{3}$
$\Rightarrow\big(\frac{1}{6}\text{B}\big)\text{A}=\text{I}_{3}\ (\because\text{AB}=\text{AB})$
$\Rightarrow\text{A}^{-1}=\frac{1}{6}\text{B}$
$\Rightarrow\text{A}^{-1}=\frac{1}{6}\begin{bmatrix}2&2&-4\\ -4&2&-4\\ 2&-1&5\end{bmatrix}$
$\text{X}=\text{A}^{-1}\text{B}$
$\text{X}=\frac{1}{6}\begin{bmatrix}2&2&-4\\ -4&2&-4\\ 2&-1&5\end{bmatrix}\begin{bmatrix}3\\ 17\\ 7\end{bmatrix}$
$\begin{bmatrix}\text{x}\\ \text{y}\\ \text{z}\end{bmatrix}=\frac{1}{6}\begin{bmatrix}6+34-28\\ -12+34-28\\ 6-17+35\end{bmatrix}$
$\begin{bmatrix}\text{x}\\ \text{y}\\ \text{z}\end{bmatrix}=\frac{1}{6}\begin{bmatrix}12\\ -6\\ 24\end{bmatrix}$
$\therefore$ x = 2, y = -1 and z = 4
View full question & answer→Question 65 Marks
Two schools A and B want to award their selected students on the values of sincerity, truthfulness and helpfulness. The school A wants to award ₹ x each, ₹ y each and ₹ z each for the three respective values to 3, 2 and 1 students respectively with a total award money of ₹ 1,600. School B wants to spend ₹ 2,300 to award its 4, 1 and 3 students on the respective values (by giving the same award money to the three values as before). If the total amount of award for one prize on each value is ₹ 900, using matrices, find the award money for each value. Apart from these three values, suggest one more value which should be considered for award.
AnswerLet the award money given for sincerity, truthfulness and helpfulness be ₹ x, ₹ y and ₹ z respectively.
Since, the total cash award is ₹ 900.
$\therefore$ x + y + z = 900 .....(1) Award money given by school A is ₹ 1,600.
$\therefore$ 3x + 2y + z = 1600 .....(2) Award money given by school B is ₹ 2,300.
$\therefore$ 4x + y + 3z = 2300 .....(3) The above system of equations can be written in matrix form CX = D as: $\begin{bmatrix}1&1&1\\3&2&1\\4&1&3\end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}900\\1600\\2300\end{bmatrix}$ Where, $\text{C}=\begin{bmatrix}1&1&1\\3&2&1\\4&1&3\end{bmatrix},\text{X}=\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}\text{and }\text{D}=\begin{bmatrix}900\\1600\\2300\end{bmatrix}$ Now, $|\text{C}|=\begin{vmatrix}1&1&1\\3&2&1\\4&1&3\end{vmatrix}$
$=1(6-1)-1(9-4)+1(3-8)$ $=5-5-5$
$=-5$Let $C_{ij}$ be the co-factors of elements $c_{ij}$ in $C = [c_{ij}]$. Then,
$\text{C}_{11}=(-1)^{1+1}\begin{vmatrix}2&1\\1&3\end{vmatrix}=5,$ $\text{C}_{12}=(-1)^{1+2}\begin{vmatrix}3&1\\4&3\end{vmatrix}=-5,$ $\text{C}_{13}=(-1)^{1+3}\begin{vmatrix}3&2\\4&1\end{vmatrix}=-5$ $\text{C}_{21}=(-1)^{2+1}\begin{vmatrix}1&1\\1&3\end{vmatrix}=-2,$ $\text{C}_{22}=(-1)^{2+2}\begin{vmatrix}1&1\\4&3\end{vmatrix}=-1,$ $\text{C}_{23}=(-1)^{2+3}\begin{vmatrix}1&1\\1&1\end{vmatrix}=3$ $\text{C}_{31}=(-1)^{3+1}\begin{vmatrix}1&1\\2&1\end{vmatrix}=-1,$ $\text{C}_{32}=(-1)^{3+2}\begin{vmatrix}1&1\\3&1\end{vmatrix}=2,$ $\text{C}_{33}=(-1)^{3+3}\begin{vmatrix}1&1\\3&2\end{vmatrix}=-1$ $\text{adj }\text{C}=\begin{bmatrix}5&-5&-5\\-2&-1&3\\-1&2&-1\end{bmatrix}^\text{T}$ $=\begin{bmatrix}5&-2&-1\\-5&-1&2\\-5&3&-1\end{bmatrix}$ $\Rightarrow\text{C}^{-1}=\frac{1}{|\text{C}|}\text{adj }\text{C}$
$=\frac{1}{-5}\begin{bmatrix}5&-2&-1\\-5&-1&2\\-5&3&-1\end{bmatrix}$ $\text{X}=\text{C}^{-1}\text{D}$ $\Rightarrow\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=-\frac{1}{5}\begin{bmatrix}4500-3200-2300\\-4500-4600+4600\\-4500+4800-2300\end{bmatrix}$ $\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=-\frac{1}{5}\begin{bmatrix}-1000\\-1500\\-2000\end{bmatrix}$ $\Rightarrow\text{x}=\frac{-1000}{-5},\text{y}=\frac{-1500}{-5}\ \text{and }\text{z}=\frac{-2000}{-5}$ $\therefore$ x = 200, y = 300 and z = 400.Hence, the award money for each value of sincerity, truthfulness and helpfulness is ₹ 200, ₹ 300 and ₹ 400.
One more value which should be considered for award hardwork.
View full question & answer→Question 75 Marks
Two institutions decided to award their employees for the three values of resourcefulness, competence and determination in the form of prices at the rate of Rs. x, y and z respectively per person. The first institution decided to award respectively 4, 3 and 2 employees with a total price money of Rs. 37000 and the second institution decided to award respectively 5, 3 and 4 employees with a total price money of Rs. 47000. If all the three prices per person together amount to Rs. 12000 then using matrix method find the value of x, y and z. What values are described in this equations?
AnswerLet x, y and z be the prize amount per person for Resourcefulness, Competence and Determination respectively.
As per the data in the question, we get
4x + 3y + 2z = 37000
5x + 3y + 4z = 47000
x + y + z = 12000
The above three simultaneous equations can be written in matrix form as
$\begin{bmatrix}4&3&2\\5&3&4\\1&1&1\end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}37000\\47000\\12000\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}4&3&2\\5&3&4\\1&1&1\end{bmatrix}^{-1}\begin{bmatrix}37000\\47000\\12000\end{bmatrix}\ \dots(1)$
$\text{A}=\begin{bmatrix}4&3&2\\5&3&4\\1&1&1\end{bmatrix}$
$|\text{A}|=4(-1)-3(1)+2(2)=-3$
$\text{cof }\text{A}=\begin{bmatrix}-1&-1&2\\-1&2&-1\\6&-6&-3\end{bmatrix}$
$\text{adj }\text{A}=(\text{cof }\text{A})^\text{T}=\begin{bmatrix}-1&-1&6\\-1&2&-6\\2&-1&-3\end{bmatrix}$
$\text{A}^{-1}=\frac{\text{adj }\text{A}}{|\text{A}|}=\frac{1}{-3}\begin{bmatrix}-1&-1&2\\-1&2&-1\\6&-6&-3\end{bmatrix}$
From (1)
$\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\frac{1}{-3}\begin{bmatrix}-1&-1&2\\-1&2&-1\\6&-6&-3\end{bmatrix}\begin{bmatrix}37000\\47000\\12000\end{bmatrix}$
$\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\frac{1}{-3}\begin{bmatrix}-12000\\-15000\\-9000\end{bmatrix}=\begin{bmatrix}4000\\5000\\3000\end{bmatrix}$
The values of x, y and z describe the amount of prizes per person of Resourcefulness, Competence and Determination.
View full question & answer→Question 85 Marks
If $\text{A}=\begin{bmatrix}2&-3&5\\ 3&2&-4\\ 1&1&-2\end{bmatrix}$, find $A^{-1}$ and hence solve the system of linear equations:
$2x - 3y + 5z = 11, 3x + 2y - 4z = -5, x + y + 2z = -3$
AnswerHere,
$\text{A}=\begin{bmatrix}2&-3&5\\ 3&2&-4\\ 1&1&-2\end{bmatrix}$
$\text{|A|}=\begin{bmatrix}2&-3&5\\ 3&2&-4\\ 1&1&-2\end{bmatrix}$
$=2{(-4+4)}+3{(-6+4)}+5{(3-2)}$
$=0-6+5$
$=-1$
Let $C_{ij}$ be the co-factors of the elements $a_{ij}$ in $A [a_{ij}].$ Then,
$\text{C}_{11}={(-1)}^{1+1}\begin{vmatrix}2&-4\\ 1&-2\end{vmatrix}=0,\\ \text{C}_{12}={(-1)}^{1+2}\begin{vmatrix}3&-3\\ 1&-2\end{vmatrix}=2,\\ \text{C}_{13}{(-1)}^{1+3}\begin{vmatrix}3&2\\ 1&1\end{vmatrix}=1$
$\text{C}_{21}={(-1)}^{2+1}\begin{vmatrix}-3&5\\ 1&-2\end{vmatrix}=-1,\\ \text{C}_{22}={(-1)}^{2+2}\begin{vmatrix}2&5\\ 1&-2\end{vmatrix}=-9,\\ \text{C}_{23}{(-1)}^{2+3}\begin{vmatrix}2&-3\\ 1&1\end{vmatrix}=-5$
$\text{C}_{31}={(-1)}^{3+1}\begin{vmatrix}-3&5\\ 2&-4\end{vmatrix}=2,\\ \text{C}_{32}={(-1)}^{3+2}\begin{vmatrix}2&5\\ 3&-4\end{vmatrix}=23,\\ \text{C}_{23}{(-1)}^{3+3}\begin{vmatrix}2&-3\\ 3&2\end{vmatrix}=13$
$\text{adj A}=\begin{bmatrix}0&2&1\\ -1&-9&-5\\ 2&23&13\end{bmatrix}^\text{T}$
$=\begin{bmatrix}0&-1&2\\ 2&-9&23\\ 1&-5&13\end{bmatrix}$
$\Rightarrow\text{A}^{-1}=\frac{1}{\text{|A|}}\text{adj A}$
$=\frac{1}{-1}\begin{bmatrix}0&-1&2\\ 2&-9&23\\ 1&-5&13\end{bmatrix}$
The given system of equations can be wriiten in matrix form as follows:
$\begin{bmatrix}2&-3&5\\ 3&2&-4\\ 1&1&-2\end{bmatrix}\begin{bmatrix}\text{x}\\ \text{y}\\ \text{z}\end{bmatrix}=\begin{bmatrix}11\\ -5\\ -3\end{bmatrix}$
$\text{X}=\text{A}^{-1}\text{B}$
$\Rightarrow\begin{bmatrix}\text{x}\\ \text{y}\\ \text{z}\end{bmatrix}=\frac{1}{-1}\begin{bmatrix}0&-1&2\\ 2&-9&23\\ 1&-5&13\end{bmatrix}\begin{bmatrix}11\\ -5\\ -3\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\ \text{y}\\ \text{z}\end{bmatrix}=\frac{1}{-1}\begin{bmatrix}0+5-6\\ 22+45-69\\ 11+25-39\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\ \text{y}\\ \text{z}\end{bmatrix}=\frac{1}{-1}\begin{bmatrix}-1\\ -2\\ -3\end{bmatrix}$
$\Rightarrow\text{x}=\frac{-1}{-1},\text{y}=\frac{-2}{-1}\text{ and }\text{z}=\frac{-3}{-1}$
$\therefore$ $x = 1, y = 2$ and $z = 3$
View full question & answer→Question 95 Marks
Solve the following system of equations by matrix method:
$5x +3y + z = 16$
$2x + y +3z = 19$
$x + 2y + 4z = 25$
AnswerHere,
$\text{A}=\begin{bmatrix}5&3&1\\ 2&1&3\\ 1&2&4\end{bmatrix}$
$\text{|A|}=\begin{vmatrix}5&3&1\\ 2&1&3\\ 1&2&4\end{vmatrix}$
$=5{(4-6)}-3{(8-3)}+1{(4-1)}$
$=-10-15+3$
$=-22$
Let $C_{ij}$ be the co-factors of the elements aij in $A [a_{ij}].$ Then,
$\text{C}_{11}={(-1)}^{1+1}\begin{vmatrix}1&2\\ 2&4\end{vmatrix}=-2,\\ \text{C}_{12}={(-1)}^{1+2}\begin{vmatrix}2&3\\ 1&4\end{vmatrix}=-5,\\ \text{C}_{13} ={(-1)}^{1+3}\begin{vmatrix}2&1\\ 1&2\end{vmatrix}=3$
$\text{C}_{21}={(-1)}^{2+1}\begin{vmatrix}3&1\\ 2&4\end{vmatrix}=-10,\\ \text{C}_{22}={(-1)}^{2+2}\begin{vmatrix}5&1\\ 1&4\end{vmatrix}=19,\\ \text{C}_{23} ={(-1)}^{2+3}\begin{vmatrix}5&3\\ 1&2\end{vmatrix}=-7$
$\text{C}_{31}={(-1)}^{3+1}\begin{vmatrix}3&1\\ 1&3\end{vmatrix}=-8,\\ \text{C}_{32}={(-1)}^{3+2}\begin{vmatrix}5&1\\ 2&3\end{vmatrix}=-13,\\ \text{C}_{33} ={(-1)}^{3+3}\begin{vmatrix}5&3\\ 2&1\end{vmatrix}=-1$
$\text{adj A}=\begin{bmatrix}-2&-5&3\\ -10&19&-7\\ 8&-13&-1\end{bmatrix}^\text{T}$
$=\begin{bmatrix}-2&-10&8\\ -5&19&-13\\ 3&-7&-1\end{bmatrix}$
$\Rightarrow\text{A}^{-1}=\frac{1}{\text{|A|}}\text{adj A}$
$=\frac{1}{-22}\begin{bmatrix}-2&-10&8\\ -5&19&-13\\ 3&-7&-1\end{bmatrix}$
$\text{X}=\text{A}^{-1}\text{B}$
$\Rightarrow\begin{bmatrix}\text{x}\\ \text{y}\\ \text{z}\end{bmatrix}=\frac{1}{-22}\begin{bmatrix}-2&-10&8\\ -5&19&-13\\ 3&-7&-1\end{bmatrix}\begin{bmatrix}16\\ 19\\ 25\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\ \text{y}\\ \text{z}\end{bmatrix}=\frac{1}{-22}\begin{bmatrix}-32-190+200\\ -80+361-325\\ 48-133-25\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\ \text{y}\\ \text{z}\end{bmatrix}=\frac{1}{-22}\begin{bmatrix}-22\\ -44\\ -110\end{bmatrix}$
$\Rightarrow\text{x}=\frac{-22}{-22},\text{y}=\frac{-44}{-22}$ and $\text{z}=\frac{-110}{-22}$
Hence, $x = 1, y = 2, z = 5$
View full question & answer→Question 105 Marks
Two factories decided to award their employees for three values of (a) adaptable to new techniques, (b) careful and alert in difficult situations and (c) keeping clam in tense situations, at the rate of ₹ x, ₹ y and ₹ z per person respectively. The first factory decided to honuor respectively 2, 4 and 3 employees with a total prize money of ₹ 29000. The second factory decided to honuor respectively 5, 2 and 3 employees with the prize money of ₹ 30500. If the three prizes per person together cost ₹ 9500, then
- Represent the above situation by matrix equation and form linear equation using matrix multiplication.
- Solve this equation by matrix method.
- Which values are reflected in the questions?
AnswerLet x, y and z be the prize amount per person for adaptibility, carefulness and calmness respectively.
as per the given data, we get
2x + 4y + 3z = 29000
5x + 2y + 3z = 30500
x + y + z = 9500
The above three simultaneous equations can be written in the matrix form as
$\begin{bmatrix}2&4&3\\5&2&3\\1&1&1\end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}29000\\30500\\9500\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}2&4&3\\5&2&3\\1&1&1\end{bmatrix}^{-1}\begin{bmatrix}29000\\30500\\9500\end{bmatrix}\ \dots(1)$
$\text{A}=\begin{bmatrix}2&4&3\\5&2&3\\1&1&1\end{bmatrix}$
$|\text{A}|=2(-1)-4(2)+3(3)=-1$
$\text{cof }\text{A}=\begin{bmatrix}-1&-2&3\\-1&-1&2\\6&9&-16\end{bmatrix}$
$\text{adj }\text{A}=(\text{cof }\text{A})^\text{T}=\begin{bmatrix}-1&-2&3\\-1&-1&2\\6&9&-16\end{bmatrix}$
$\text{A}^{-1}=\frac{\text{adj }\text{A}}{|\text{A}|}=\frac{=\begin{bmatrix}-1&-1&6\\-2&-1&9\\3&2&-16\end{bmatrix}}{-1}$$=\begin{bmatrix}1&1&-6\\2&1&-9\\-3&-2&16\end{bmatrix}$
From (1)
$\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}1&1&-6\\2&1&-9\\-3&-2&16\end{bmatrix}\begin{bmatrix}29000\\30500\\9500\end{bmatrix}\ \dots(1)$
$\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}2500\\3000\\4000\end{bmatrix}$
Keeping calm in a tense situation is more rewarding than carefulness, and carefulness is more rewarding than adaptability.
View full question & answer→Question 115 Marks
Show that the following system of linear equations is consistent and also find solutions:
$5x +3y + 7z = 4$
$3x + 26y + 2z = 9$
$7x + 2y + 10z = 5$
AnswerThis can be written as: $\begin{bmatrix}5&3&7\\ 3&26&2\\ 7&2&10\end{bmatrix}
\begin{bmatrix}\text{X}\\ \text{y}\\ \text{z}\end{bmatrix}
=\begin{bmatrix}4\\ 9\\ 5\end{bmatrix}$ Or $\text{A}\text{X}=\text{B}$
$\text{|A|}=5{(256)}-3{(16)}+7{(6-182)}$ $=0$ So, A is singular. Thus, the given system is either inconsistent or it is consistent with infinitely many solutions according as $\text{(Adj A)}\times\text{B}\neq0$ or $\text{(Adj A)}\times\text{B}=0$
Let $C_{ij}$ be the co-factors of $a_{ij}$ in A $\text{C}_{11}=256\\ \text{C}_{21}=-16\\ \text{C}_{31}=-176$ $\text{C}_{12}=-16\\ \text{C}_{22}=1\\ \text{C}_{32}=11$ $\text{C}_{13}=-176\\ \text{C}_{23}=11\\ \text{C}_{33}=121$
$\text{adj A}=\begin{bmatrix}256&-16&-176\\ -16&1&11\\ -176&11&121\end{bmatrix}^\text{T}=\begin{bmatrix}256&-16&-176\\ -16&1&11\\ -176&11&121\end{bmatrix}$ $\text{adj}\text{A}\times\text{B}=\begin{bmatrix}256&-16&-176\\ -16&1&11\\ -176&11&121\end{bmatrix}\begin{bmatrix}4\\ 9\\ 5\end{bmatrix}=\begin{bmatrix}0\\ 0\\ 0\end{bmatrix}$
Thus, AX = B has infinite many solutions. Now, let z = kThen, 5x + 3y = 4 - 7k
3x + 26y = 9 - 2k Which can be written as: $\begin{bmatrix}5&3\\ 3&26\end{bmatrix}\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\begin{bmatrix}4-7\text{k}\\ 9-2\text{k}\end{bmatrix}$
Or $\text{AX = B}$ $\text{|A|}=2$ $\text{adj A}=\begin{bmatrix}26&-3\\ -3&5\end{bmatrix}$ Now, $\text{x}=\text{A}^{-1}\text{B}=\frac{1}{\text{|A|}}\times\text{adj A}\times\text{B}$
$=\frac{1}{121}\begin{bmatrix}26&-3\\ -3&5\end{bmatrix}\begin{bmatrix}4-7\text{k}\\ 9-2\text{k}\end{bmatrix}$ $=\frac{1}{121}\begin{bmatrix}77-176\text{k}\\ 11\text{k}+33\end{bmatrix}$
$\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\begin{bmatrix}\frac{7-16\text{k}}{11}\\ \frac{\text{k}+3}{11}\end{bmatrix}$ There values of x, y, z satisfies the third eq. Hence, $\text{x}=\frac{7-16\text{k}}{11},\text{y}=\frac{\text{k}+3}{11},\text{z}=\text{k}$
View full question & answer→Question 125 Marks
The management committee of a residential colony decided to award some of its members (say x) for honesty, some (say y) for helping others and some others (say z) for supervising the workers to keep the colony neat and clean. The sum of all the awardees is 12. Three times the sum of awardees for cooperation and supervision added to two times the number of awardees for honesty is 33. If the sum of the number of awardees for honesty and supervision is twice the number of awardees for helping others, using matrix method, find the number of awardees of each category. Apart from these values, namely, honesty, cooperation and supervision, suggest one more value which the management of the colony must include for awards.
AnswerFrom the given data, we get
The following three equations:
x + y + z = 12
2x + 3y + 3z = 33
x - 2y + z = 0
This system of equations can be written in the matrix form as:
$\begin{bmatrix}1&1&1\\2&3&3\\1&-2&1\end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}12\\33\\0\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}1&1&1\\2&3&3\\1&-2&1\end{bmatrix}^1\begin{bmatrix}12\\33\\0\end{bmatrix}\ \dots(1)$
$\text{A}=\begin{bmatrix}1&1&1\\2&3&3\\1&-2&1\end{bmatrix}$
$|\text{A}|=1(9)-1(-1)+1(-7)=3$
$\text{cof}\cdot\text{A}=\begin{bmatrix}9&1&-7\\-3&0&3\\0&-1&1\end{bmatrix}$
$\text{adj }\text{A}=[\text{cof }\text{A}]^\text{T}=\begin{bmatrix}9&-3&0\\1&0&-1\\-7&3&1\end{bmatrix}$
$\text{A}^{-1}=\frac{\text{adj }\text{A}}{|\text{A}|}=\frac{1}{3}\begin{bmatrix}9&-3&0\\1&0&-1\\-7&3&1\end{bmatrix}$
$\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\frac{1}{3}\begin{bmatrix}9&-3&0\\1&0&-1\\-7&3&1\end{bmatrix}\begin{bmatrix}12\\33\\0\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}9&-3&0\\1&0&-1\\-7&3&1\end{bmatrix}\begin{bmatrix}4\\11\\0\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}36-33+0\\4+0+0\\-28+33+0\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}3\\4\\5\end{bmatrix}$
An award for organising different festivals in the colony can be included by the management.
View full question & answer→Question 135 Marks
Solve the following systems of homogeneous linear equations by matrix method:
$x + y - z = 0$
$x - 2y + z = 0$
$3x + 6y - 5z = 0$
Answer$x + y - z = 0 .....(1)$
$x - 2y + z = 0 .....(2)$
$3x + 6y - 5z = 0 .....(3)$
The given system of homogeneous equations can be written in matrix form as follows:
$\begin{bmatrix}1&1&-1\\1&-2&1\\3&6&-5\end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}$
$\text{AX}=\text{O}$
Here,
$\text{A}=\begin{bmatrix}1&1&-1\\1&-2&1\\3&6&-5\end{bmatrix},\text{X}=\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}\text{and }\text{O}=\begin{bmatrix}0\\0\\0\end{bmatrix}$
Now,
$|\text{A}|=\begin{vmatrix}1&1&-1\\1&-2&1\\3&6&-5\end{vmatrix}$
$=1(10-6)-1(-5-3)-1(6+6)$
$=4+8-12$
$=0$
So, the given system of homogeneous equations has non-trivial solution.
Substituting z = k in eq. (1) & eq. (2), we get
$x + y = k$ and $x - 2y = -k$
$\Rightarrow\begin{bmatrix}1&1\\1&-2\end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\end{bmatrix}=\begin{bmatrix}\text{k}\\-\text{k}\end{bmatrix}$
$\text{AX}=\text{B}$
Here,
$\text{A}=\begin{bmatrix}1&1\\1&-2\end{bmatrix},\text{X}=\begin{bmatrix}\text{x}\\\text{y}\end{bmatrix}\text{and }\text{B}=\begin{bmatrix}\text{k}\\-\text{k}\end{bmatrix}$
$|\text{A}|=\begin{vmatrix}1&1\\1&-2\end{vmatrix}=-3$
So, $A^{-1} $ exists.
$\text{adj }\text{A}=\begin{bmatrix}-2&-1\\-1&1\end{bmatrix}$
$\text{A}^{-1}=\frac{1}{|\text{A}|}\text{adj }\text{A}$
$\Rightarrow\text{A}^{-1}=\frac{1}{-3}\begin{bmatrix}-2&-1\\-1&1\end{bmatrix}$
$\text{X}=\text{A}^{-1}\text{B}$
$\Rightarrow\begin{bmatrix}\text{x}\\\text{y}\end{bmatrix}=\frac{1}{-3}\begin{bmatrix}-2&-1\\-1&1\end{bmatrix}\begin{bmatrix}\text{k}\\-\text{k}\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\\text{y}\end{bmatrix}=\frac{1}{-3}\begin{bmatrix}-2\text{k}+\text{k}\\-\text{k}-\text{k}\end{bmatrix}$
Thus, $\text{x}=\frac{\text{k}}{3},\text{y}=\frac{2\text{k}}{3}$ and $\text{z}=\text{k}$ (where k is any real number) satisfy the given system of equations.
View full question & answer→Question 145 Marks
Use product $\begin{bmatrix}1&-1&2\\0&2&-3\\3&-2&4\end{bmatrix}\begin{bmatrix}-2&0&1\\9&2&-3\\6&1&-2\end{bmatrix}$ to solve the system of equations $x + 3z = 9, -x + 2y - 2z = 4, 2x - 3y + 4z = -3.$
AnswerSuppose, $\text{A}=\begin{bmatrix}1&-1&2\\0&2&-3\\3&-2&4\end{bmatrix}\text{B}=\begin{bmatrix}-2&0&1\\9&2&-3\\6&1&-2\end{bmatrix}$
$\text{A}\times\text{B}=\begin{bmatrix}1&-1&2\\0&2&-3\\3&-2&4\end{bmatrix}\begin{bmatrix}-2&0&1\\9&2&-3\\6&1&-2\end{bmatrix}$
$=\begin{bmatrix}-2-9+12&0-2+2&1+3-4\\0+18-18&0+4-3&0-6+6\\-6-18+24&0-4+4&3+6-8\end{bmatrix}$
$=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
Since, $A \times B = I,$
$\therefore$ $B = A^{-1}.....(1)$
Now, the given system of equations is
$x + 3z = 9$
$-x + 2y - 2z = 4$
$2x - 3y + 4z = -3$
This can also be presented as:
$\begin{bmatrix}1&0&3\\-1&2&-2\\2&-3&4\end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}9\\4\\-3\end{bmatrix}$
Here, we can observe that $\begin{bmatrix}1&0&3\\-1&2&-2\\2&-3&4\end{bmatrix}=\text{A}^\text{T}$
So, $\text{A}^\text{T}=\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}9\\4\\-3\end{bmatrix}$
Multiply the above expression by $(A^T)^{-1}.$
$\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=(\text{A}^\text{T})^{-1}\begin{bmatrix}9\\4\\-3\end{bmatrix}$
$\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=(\text{A}^{-1})^\text{T}\begin{bmatrix}9\\4\\-3\end{bmatrix}$
$\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\text{B}^\text{T}\begin{bmatrix}9\\4\\-3\end{bmatrix}$ $[\text{Using}\ (1)]$
$\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}-2&0&1\\9&2&-3\\6&1&-2\end{bmatrix}^\text{T}\begin{bmatrix}9\\4\\-3\end{bmatrix}$
$=\begin{bmatrix}-2&9&6\\0&2&1\\1&-3&-2\end{bmatrix}\begin{bmatrix}9\\4\\-3\end{bmatrix}$
$=\begin{bmatrix}-18+36-18\\0+8-3\\9-12+6\end{bmatrix}$
$=\begin{bmatrix}0\\5\\3\end{bmatrix}$
Hence, $x = 0, y = 5$ and $z = 3.$
View full question & answer→Question 155 Marks
Solve the following system of equations by matrix method:$\frac{2}{\text{x}}+\frac{3}{\text{y}}+\frac{10}{\text{z}}=4,\frac{4}{\text{x}}-\frac{6}{\text{y}}+\frac{5}{\text{z}}=1,\frac{6}{\text{x}}+\frac{9}{\text{y}}-\frac{20}{\text{z}}=2:\text{x},\text{y},\text{z}\neq0$
AnswerLet
$\frac{1}{\text{x}}=\text{u},\frac{1}{\text{y}}=\text{v},\frac{1}{\text{z}}=\text{w}$
The above system can be written as:
$\begin{bmatrix}2&3&10\\ 4&-6&5\\ 6&9&-20\end{bmatrix}\begin{bmatrix}\text{u}\\ \text{v}\\ \text{w}\end{bmatrix}=\begin{bmatrix}4\\ 1\\ 2\end{bmatrix}$
Or AX = B
$\text{|A|}=2{(75)}-3{(-110)}+10{(72)}=1200\neq0$
So, the above system has a unique solution, given by
$\text{X}=\text{A}^{-1}\text{B}$
Let $C _{ ij }$ be the co-factors of $a _{ ij }$ in Al
$\text{C}_{11}=75\\ \text{C}_{21}=150\\ \text{C}_{31}=75$
$\text{C}_{12}=110\\ \text{C}_{22}=-100\\ \text{C}_{32}=30$
$\text{C}_{13}=72\\ \text{C}_{23}=0\\ \text{C}_{33}=-24$
$\text{adj A}=\begin{bmatrix}75&110&72\\ 150&-100&0\\ 75&30&-24\end{bmatrix}^\text{T}=\begin{bmatrix}75&150&75\\ 110&-100&30\\ 72&0&-24\end{bmatrix}$
Now,
$\text{X}=\text{A}^{-1}\text{B}=\frac{1}{\text{|A|}}\text{(Adj A)}\times\text{B}$
$=\frac{1}{1200}\begin{bmatrix}75&150&75\\ 110&-100&30\\ 72&0&-24\end{bmatrix}\begin{bmatrix}4\\ 1\\ 2\end{bmatrix}$
$\begin{bmatrix}\text{u}\\ \text{v}\\ \text{w}\end{bmatrix}=\frac{1}{1200}\begin{bmatrix}600\\ 400\\ 240\end{bmatrix}=\begin{bmatrix}\frac{1}{2}\\ \frac{1}{3}\\ \frac{1}{5}\end{bmatrix}$
Hence, $x = 2, y = 3, z = 5$
View full question & answer→Question 165 Marks
Solve the following systems of homogeneous linear equations by matrix method:
$2x - y + 2z = 0$
$5x + 3y - z = 0$
$x + 5y - 5z = 0$
Answer$2x - y + 2z = 0 5x + 3y - z = 0 x + 5y - 5z = 0$ $\begin{bmatrix}2&-1&2\\5&3&-1\\1&5&-5\end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}$ or, $\text{AX}=\text{O}$ $|\text{A}|=2(-10)+1(-24)+2(22)$$=-20-24+44$
$=0$
Hence, the system has infinite solutions. Let $\text{z}=\text{k}$$2\text{x}-\text{y}=-2\text{k}$
$5\text{x}+3\text{y}=\text{k}$
$\begin{bmatrix}2&-1\\5&3\end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\end{bmatrix}=\begin{bmatrix}-2\text{k}\\\text{k}\end{bmatrix}$
$\text{AX}=\text{B}$
$|\text{A}|=6+5=11\neq0$ So, $A^{-1}$ exists
Now, $\text{adj }\text{A}=\begin{bmatrix}3&-5\\1&2\end{bmatrix}'=\begin{bmatrix}3&1\\-5&2\end{bmatrix}$
$\text{x}=\text{A}^{-1}\text{B}=\frac{1}{|\text{A}|}(\text{adj }\text{A})\text{B}$$=\frac{1}{11}\begin{bmatrix}3&1\\-5&2\end{bmatrix}\begin{bmatrix}-2\text{k}\\\text{k}\end{bmatrix}=\begin{bmatrix}\frac{-5\text{k}}{11}\\\frac{12\text{k}}{11}\end{bmatrix}$
Hence, $\text{x}=\frac{-5\text{k}}{11},\ \text{y}=\frac{12\text{k}}{11},\ \text{z}=\text{k}$
View full question & answer→Question 175 Marks
Solve the following system of equations by matrix method.
$8x + 4y + 3z = 18$
$2x + y + z = 5$
$x + 2y + z = 5$
AnswerThe above system can be wrtten as:
$\begin{bmatrix}8&4&3\\ 2&1&1\\ 1&2&1\end{bmatrix}\begin{bmatrix}\text{x}\\ \text{y}\\ \text{z}\end{bmatrix}=\begin{bmatrix}18\\ 5\\ 5\end{bmatrix}$
Or AX = B
$\text{|A|}=8{(-1)}-4{(1)}+3{(3)}=-8-4+9=-3\neq0$
So, the above system has a unique solution, given by
$\text{X}=\text{A}^{-1}\text{B}$
Let $C_{ij}$ be the co-factors of $a_{ij}$ in A
$\text{C}_{11}=-1\\ \text{C}_{21}=2\\ \text{C}_{31}=1$
$\text{C}_{12}=-1\\ \text{C}_{22}=5\\ \text{C}_{32}=-2$
$\text{C}_{13}=3\\ \text{C}_{23}=-12\\ \text{C}_{33}=0$
$\text{adj A}=\begin{bmatrix}-1&-1&3\\ 2&5&-12\\ 1&-2&0\end{bmatrix}^\text{T}=\begin{bmatrix}-1&2&1\\ -1&5&-2\\ 3&-12&0\end{bmatrix}$
Now, $\text{X}=\text{A}^{-1}\text{B}=\frac{1}{\text{|A|}}\text{(Adj A)}\times\text{B}$
$=\frac{-1}{3}\begin{bmatrix}-1&2&1\\ -1&5&-2\\ 3&-12&0\end{bmatrix}\begin{bmatrix}18\\ 5\\ 5\end{bmatrix}$
$\begin{bmatrix}\text{x}\\ \text{y}\\ \text{z}\end{bmatrix}=\frac{-1}{3}\begin{bmatrix}-3\\ -3\\ -6\end{bmatrix}=\begin{bmatrix}1\\ 1\\ 2\end{bmatrix}$
Hence, x = 1, y = 1, z = 2
View full question & answer→Question 185 Marks
An amount of Rs. $10,000$ is put into three investments at the rate of $10,12$ and $15 \%$ per annum. The combined income is Rs. $1310$ and the combined income of first and second investment is Rs. $190$ short of the income from the third. Find the investment in each using matrix method.
AnswerLet the three investments are x, y, z
$\text{x}+\text{y}+\text{z}=10,000\ \dots(1)$
Also
$\frac{10}{100}\text{x}+\frac{12}{100}\text{y}+\frac{15}{100}\text{z}=1310$
$0.1\text{x}+0.12\text{y}+0.15\text{z}=1310\ \dots(2)$
Also
$\frac{10}{100}\text{x}+\frac{12}{100}\text{y}=\frac{15}{100}\text{z}-190$
$0.1\text{x}+0.12\text{y}+0.15\text{z}=-190\ \dots(3)$
The above system can be written as:
$\begin{bmatrix}1&1&1\\0.1&0.12&0.15\\0.1&0.12&-0.15\end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}10000\\1310\\-190\end{bmatrix}$
Or $\text{AX}=\text{B}$
$|\text{A}|=1(-0.036)-1(-0.03)+1(0)=-0.006\neq0$
So, the above system has a unique solution, given by
$X = A^{-1}B$
Let $C_{ij}$ be the co-factor of $a_{ij}$ in A
$\text{C}_{11}=-0.036,\text{C}_{12}=0.03,\text{C}_{13}=0$
$\text{C}_{21}=0.27,\text{C}_{22}=-0.25,\text{C}_{23}=-0.02$
$\text{C}_{31}=0.03,\text{C}_{32}=-0.05,\text{C}_{33}=0.02$
$\text{adj }\text{A}=\begin{bmatrix}-0.036&0.03&0\\0.27&-0.25&-0.02\\0.03&-0.05&0.02\end{bmatrix}^\text{T}=\begin{bmatrix}-0.036&0.27&0.03\\0.03&-0.25&-0.05\\0&-0.02&0.02\end{bmatrix}$
Now,
$\text{X}=\text{A}^{-1}\text{B}=\frac{1}{|\text{A}|}(\text{Adj}\ \text{A})\times\text{B}$
$=\frac{1}{-0.006}\begin{bmatrix}-0.036&0.27&0.03\\0.03&-0.25&-0.05\\0&-0.02&0.02\end{bmatrix}\begin{bmatrix}10000\\1310\\-190\end{bmatrix}$
$\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\frac{1}{-0.006}\begin{bmatrix}-12\\-18\\-30\end{bmatrix}=\begin{bmatrix}2000\\3000\\5000\end{bmatrix}$
Hence, x = Rs. 2000, y = Rs. 3000, z = Rs. 5000
View full question & answer→Question 195 Marks
A total amount of ₹ $7000$ is deposited in three different saving bank accounts with annual interest rates $5 \%, 8 \%$ and $8 \frac{1}{2} \%$ respectively. The total annual interest from these three accounts is ₹ $550$ . Equal amounts have been deposited in the $5 \%$ and $8 \%$ saving accounts. Find the amount deposited in each of the three accounts, with the help of matrices.
AnswerLet the amount deposited in each of the three accounts be Rs. x, Rs. x and Rs. y respecively.
Since, the total amount deposited is ₹ 7000.
$\text{x}+\text{x}+\text{y}=7000$
$2\text{x}+\text{y}=7000\ \dots(1)$
Total annual Interest is Rs. 550.
$\frac{5}{100}\text{x}+\frac{8}{100}\text{x}+\frac{17}{200}\text{y}=550$
$\Rightarrow26\text{x}+17\text{y}=110000$
The above system of equation can be written in a matrix form AX = B as:
$\begin{bmatrix}2&1\\26&17\end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\end{bmatrix}=\begin{bmatrix}7000\\110000\end{bmatrix}$
where, $\text{A}=\begin{bmatrix}2&1\\26&17\end{bmatrix},\text{X}=\begin{bmatrix}\text{x}\\\text{y}\end{bmatrix}\text{and }\text{B}=\begin{bmatrix}7000\\110000\end{bmatrix}$
Now,
$|\text{A}|=\begin{vmatrix}2&1\\26&17\end{vmatrix}$
$=34-26$
$=8$
Let $C _{ ij }$ be the co-factors of the elements $a _{ ij }$ in $A \left[ a _{ ij }\right]$. Then,
$\text{C}_{11}=(-1)^{1+1}17=17,\text{C}_{12}=(-1)^{1+2}26=-26$
$\text{C}_{21}=(-1)^{2+1}1=-1,\text{C}_{22}=(-1)^{2+2}2=2$
$\text{adj }\text{A}=\begin{bmatrix}17&-26\\-1&2\end{bmatrix}^\text{T}$
$=\begin{bmatrix}17&-1\\-26&2\end{bmatrix}$
$\Rightarrow\text{A}^{-1}=\frac{1}{|\text{A}|}\text{adj }\text{A}$
$=\frac{1}{8}\begin{bmatrix}17&-1\\-26&2\end{bmatrix}$
$\text{X}=\text{A}^{-1}\text{B}$
$\Rightarrow\begin{bmatrix}\text{x}\\\text{y}\end{bmatrix}=\frac{1}{8}\begin{bmatrix}17&-1\\-26&2\end{bmatrix}\begin{bmatrix}7000\\110000\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\\text{y}\end{bmatrix}=\frac{1}{8}\begin{bmatrix}119000-110000\\-182000+220000\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\\text{y}\end{bmatrix}=\frac{1}{8}\begin{bmatrix}9000\\38000\end{bmatrix}$
$\Rightarrow\text{x}=\frac{9000}{8}\text{and }\text{y}=\frac{38000}{8}$
$\therefore\ \text{x}=1125\text{ and }\text{y}=4750.$
View full question & answer→Question 205 Marks
Solve the following system of equations by matrix method:
$x + y + z = 6$
$x + 2z = 7$
$3x + y + z = 12$
AnswerHere,
$\text{A}=\begin{bmatrix}1&1&1\\ 1&0&2\\ 3&1&1\end{bmatrix}$
$\text{|A|}=\begin{vmatrix}1&1&1\\ 1&0&2\\ 3&1&1\end{vmatrix}$
$=1{(0-2)}-1{(1-6)}+1{(1-0)}$
$=-2+5+1$
$=4$
Let $C _{ ij }$ be the co-factors of the elements $a _{ ij }$ in $A \left[ a _{ ij }\right]$. Then,
$\text{C}_{11}={(-1)}^{1+1}\begin{vmatrix}0&2\\ 1&1\end{vmatrix}=-2,\\ \text{C}_{12}={(-1)}^{1+2}\begin{vmatrix}1&2\\ 3&1\end{vmatrix}=5,\\ \text{C}_{13}={(-1)}^{1+3}\begin{vmatrix}1&0\\ 3&1\end{vmatrix}=1$
$\text{C}_{21}={(-1)}^{2+1}\begin{vmatrix}1&1\\ 1&1\end{vmatrix}=0,\\ \text{C}_{22}={(-1)}^{2+2}\begin{vmatrix}1&1\\ 3&1\end{vmatrix}=-2,\\ \text{C}_{23}={(-1)}^{2+3}\begin{vmatrix}1&1\\ 3&1\end{vmatrix}=2$
$\text{C}_{31}={(-1)}^{3+1}\begin{vmatrix}1&1\\ 0&2\end{vmatrix}=2,\\ \text{C}_{32}={(-1)}^{3+2}\begin{vmatrix}1&1\\ 1&2\end{vmatrix}=-1,\\ \text{C}_{33}={(-1)}^{3+3}\begin{vmatrix}1&1\\ 1&0\end{vmatrix}=-1$
$=\begin{bmatrix}-2&0&2\\ 5&-2&-1\\ 1&2&-1\end{bmatrix}$
$\Rightarrow\text{A}^{-1}=\frac{1}{\text{|A|}}\text{adj A}$
$=\frac{1}{4}\begin{bmatrix}-2&0&2\\ 5&-2&-1\\ 1&2&-1\end{bmatrix}$
$\text{X}=\text{A}^{-1}\text{B}$
$\Rightarrow\begin{bmatrix}\text{x}\\ \text{y}\\ \text{z}\end{bmatrix}=\frac{1}{4}\begin{bmatrix}-2&0&2\\ 5&-2&-1\\ 1&2&-1\end{bmatrix}\begin{bmatrix}6\\ 7\\ 12\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\ \text{y}\\ \text{z}\end{bmatrix}=\frac{1}{4}\begin{bmatrix}-12+0+24\\ 30-14-12\\ 6-14-12\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\ \text{y}\\ \text{z}\end{bmatrix}=\frac{1}{4}\begin{bmatrix}12\\ 4\\ -20\end{bmatrix}$
$\Rightarrow\text{x}=\frac{12}{4},\text{y}=\frac{4}{4}\text{ and }\text{z}=\frac{-20}{4}$
$\therefore$ x = 3, y = 1 and z = -5
View full question & answer→Question 215 Marks
If $\text{A}=\begin{bmatrix}1&2&0\\-2&-1&-2\\0&-1&1\end{bmatrix}$, find $A^{-1}$, solve the system of linear equations $x - 2y = 10, 2x - y - z = 8, -2y + z = 7$
AnswerHere, $\text{A}=\begin{bmatrix}1&2&0\\-2&-1&-2\\0&-1&1\end{bmatrix}$ $|\text{A}| = 1(-1-2)+2(2)$
$= -3 + 4$
$= 1$
Let $C_{ij}$ be the cofactors of the elements $a_{ij}$ in $A = [a_{ij}]$. Then, $\text{C}_{11} = (-1)^{1+1}\begin{vmatrix}-1&-2\\-1&1\end{vmatrix}=-3,$$\text{C}_{12} = (-1)^{1+2}\begin{vmatrix}-2&-2\\0&1\end{vmatrix}=2,$$\text{C}_{13} = (-1)^{1+3}\begin{vmatrix}-2&-1\\0&-1\end{vmatrix}=2$ $\text{C}_{21} = (-1)^{2+1}\begin{vmatrix}2&0\\-1&1\end{vmatrix}=-2,$$\text{C}_{22} = (-1)^{2+2}\begin{vmatrix}1&0\\0&1\end{vmatrix}=1,$$\text{C}_{23} = (-1)^{2+3}\begin{vmatrix}1&2\\0&-1\end{vmatrix}=1,$ $\text{C}_{31} = (-1)^{3+1}\begin{vmatrix}2&0\\-1&-2\end{vmatrix}=-4,$$\text{C}_{32} = (-1)^{3+2}\begin{vmatrix}1&0\\-2&-2\end{vmatrix}=2,$$\text{C}_{33} = (-1)^{3+3}\begin{vmatrix}1&2\\-1&-2\end{vmatrix}=3$ $\therefore\ \text{adj }\text{A}=\begin{bmatrix}-3&2&2\\-2&1&1\\-4&2&3\end{bmatrix}^\text{T}$ $=\begin{bmatrix}-3&-2&-4\\2&1&2\\2&1&3\end{bmatrix}$ $\Rightarrow\text{A}^{-1}=\frac{1}{|\text{A}|}\text{adj }\text{A}$ $=\frac{1}{1}\begin{bmatrix}-3&-2&-4\\2&1&2\\2&1&3\end{bmatrix}$ $=\begin{bmatrix}-3&-2&-4\\2&1&2\\2&1&3\end{bmatrix}$ We know that, $(A^T)^{-1} = (A^{-1})^T$. Here, $C = A^T$ i. e. , $\text{C}=\begin{bmatrix}1&-2&0\\2&-1&-1\\0&-2&1\end{bmatrix}$ $\therefore\ \text{C}^{-1}=\begin{bmatrix}-3&2&2\\-2&1&1\\-4&2&3\end{bmatrix}$ or, CX = B where, $\text{C}=\begin{bmatrix}1&-2&0\\2&-1&-1\\0&-2&1\end{bmatrix}, \text{X}=\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}\text{and }\text{B}=\begin{bmatrix}10\\8\\7\end{bmatrix}$ Now, $\therefore$ $X = C^{-1}B$ $\Rightarrow\text{X}=\begin{bmatrix}-3&2&2\\-2&1&1\\-4&2&3\end{bmatrix}\begin{bmatrix}10\\8\\7\end{bmatrix}$ $\Rightarrow\text{X}=\begin{bmatrix}-30+16+14\\-20+8+7\\-40+16+21\end{bmatrix}$ $\Rightarrow\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}0\\-5\\-3\end{bmatrix}$ $\therefore$ x = 0, y = -5, and z = -3.
View full question & answer→Question 225 Marks
The sum of three numbers is $2$. If twice the second number is added to the sum of first and third, the sum is $1$. By adding second and third number to five times the first number, we get $6$. Find the three numbers by using matrices.
AnswerLet the three number be x, y and a
According to the question,
$x + y + 2$
$x + 2y + z = 1$
$5x + y + z = 6$
The given system of equations can be written in matrix form as follows:
$\begin{bmatrix}1&1&1\\1&2&1\\5&1&1\end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}2\\1\\6\end{bmatrix}$
$AX = B$
Here,
$\text{A}=\begin{bmatrix}1&1&1\\1&2&1\\5&1&1\end{bmatrix},\text{X}=\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}\text{and }\text{B}=\begin{bmatrix}2\\1\\6\end{bmatrix}$
$|\text{A}|=1(2-1)-1(1-5)+1(1-10)$
$=1+4-9$
$=-4$
Let $C _{ ij }$ be the co-factors of the elements $a _{ ij }$ in $A \left[ a _{ ij }\right]$. Then,
$\text{C}_{11}=(-1)^{1+1}\begin{vmatrix}2&1\\1&1\end{vmatrix}=1,$ $\text{C}_{12}=(-1)^{1+2}\begin{vmatrix}1&1\\5&1\end{vmatrix}=4,$ $\text{C}_{13}=(-1)^{1+3}\begin{vmatrix}1&2\\5&1\end{vmatrix}=-9$
$\text{C}_{21}=(-1)^{2+1}\begin{vmatrix}1&1\\1&1\end{vmatrix}=0,$ $\text{C}_{22}=(-1)^{2+2}\begin{vmatrix}1&1\\5&1\end{vmatrix}=-4,$ $\text{C}_{23}=(-1)^{2+3}\begin{vmatrix}1&1\\5&1\end{vmatrix}=4$
$\text{C}_{31}=(-1)^{3+1}\begin{vmatrix}1&1\\2&1\end{vmatrix}=-1,$ $\text{C}_{32}=(-1)^{3+2}\begin{vmatrix}1&1\\1&1\end{vmatrix}=0,$ $\text{C}_{33}=(-1)^{3+3}\begin{vmatrix}1&1\\1&2\end{vmatrix}=1$
$\text{adj }\text{A}=\begin{bmatrix}1&4&-9\\0&-4&4\\-1&0&1\end{bmatrix}^\text{T}$
$=\begin{bmatrix}1&0&-1\\4&-4&0\\-9&4&1\end{bmatrix}$
$\Rightarrow\text{A}^{-1}=\frac{1}{|\text{A}|}\text{adj }\text{A}$
$=\frac{1}{-4}\begin{bmatrix}1&0&-1\\4&-4&0\\-9&4&1\end{bmatrix}$
View full question & answer→Question 235 Marks
$\text{A}=\begin{bmatrix}3&-4&2\\ 2&3&5\\ 1&0&1\end{bmatrix}$, find $A^{-1}$ and hence solve the following system of equations:
$3x - 4y +2z = -1, 2x + 3y + 5z = 7, x + z = 2$
AnswerHere,
$\text{A}=\begin{bmatrix}3&-4&2\\ 2&3&5\\ 1&0&1\end{bmatrix}$
$\text{|A|}=3{(3-0)}+4{(2-5)}+2{(0-3)}$
$=9-12-6$
$=-9$
Let $C _{ ij }$ be the co-factors of the elements $a _{ ij }$ in $A \left[ a _{ ij }\right]$. Then,
$\text{C}_{11}={(-1)}^{1+1}\begin{vmatrix}3&5\\ 0&1\end{vmatrix}=3,\\ \text{C}_{12}={(-1)}^{1+2}\begin{vmatrix}2&5\\ 1&1\end{vmatrix}=3,\\ \text{C}-{13}={(-1)}^{1+3}\begin{vmatrix}2&3\\ 1&0\end{vmatrix}=-3$
$\text{C}_{21}={(-1)}^{2+1}\begin{vmatrix}-4&2\\ 0&1\end{vmatrix}=4,\\ \text{C}_{22}={(-1)}^{2+2}\begin{vmatrix}3&2\\ 1&1\end{vmatrix}=1,\\ \text{C}-{23}={(-1)}^{2+3}\begin{vmatrix}3&-4\\ 1&0\end{vmatrix}=-4$
$\text{C}_{31}={(-1)}^{3+1}\begin{vmatrix}-4&2\\ 3&5\end{vmatrix}=-26,\\ \text{C}_{32}={(-1)}^{3+2}\begin{vmatrix}3&2\\ 2&5\end{vmatrix}=-11,\\ \text{C}-{33}={(-1)}^{3+3}\begin{vmatrix}3&-4\\ 2&3\end{vmatrix}=17$
$\text{adj A}=\begin{bmatrix}3&3&-3\\ 4&1&-4\\ -26&-11&17\end{bmatrix}^\text{T}$
$=\begin{bmatrix}3&4&-26\\ 3&1&-11\\ -3&-4&17\end{bmatrix}$
$\Rightarrow\text{A}^{-1}=\frac{1}{|A|}\text{adj A}$
$=\frac{1}{-9}\begin{bmatrix}3&4&-26\\ 3&1&-11\\ -3&-4&17\end{bmatrix}$
$\text{AX = B}$
Here,
$\text{A}=\begin{bmatrix}3&-4&2\\ 2&3&5\\ 1&0&1\end{bmatrix},\text{X}=\begin{bmatrix}\text{x}\\ \text{y}\\ \text{z}\end{bmatrix}\text{and }\text{B}=\begin{bmatrix}-1\\ 7\\ 2\end{bmatrix}$
$\text{X = A}^{-1}\text{B}$
$\Rightarrow\text{X}=\frac{1}{-9}\begin{bmatrix}3&4&-26\\ 3&1&-11\\ -3&-4&17\end{bmatrix}\begin{bmatrix}-1\\ 7\\ 2\end{bmatrix}$
$\Rightarrow\text{X}=\frac{1}{-9}\begin{bmatrix}-3+28-25\\ -3+7-22\\ 3-28+34\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\ \text{y}\\ \text{z}\end{bmatrix}=\frac{1}{-9}\begin{bmatrix}-27\\ -18\\ 9\end{bmatrix}$
$\therefore$ x = 3, y = 2 and z = -1
View full question & answer→Question 245 Marks
If $\text{A}=\begin{pmatrix}2&3&1\\1&2&2\\-3&1&-1\end{pmatrix}$, find $A^{-1}$ and hence solve the system of equations $2x + y - 3z =13, 3x + 2y + z = 4, x + 2y - z = 8.$
AnswerWe have, $\text{A}=\begin{bmatrix}2&3&1\\1&2&2\\-3&1&-1\end{bmatrix}$
$\therefore\ |\text{A}|=\begin{vmatrix}2&3&1\\1&2&2\\-3&1&-1\end{vmatrix}$
$=2(-2-2)-3(1+6)+1(1+6)$
$=-8-15+7$
$=-16\neq0$
So, A is invertible.
Let $C _{ ij }$ be the co-factors of the elements $a _{ ij }$ in $A \left[ a _{ ij }\right]$. Then,
$\text{C}_{11}=(-1)^{1+1}\begin{vmatrix}2&2\\1&-1\end{vmatrix}=-2-2=-4$
$\text{C}_{12}=(-1)^{1+2}\begin{vmatrix}1&2\\-3&-1\end{vmatrix}=-1(-1+6)=-5$
$\text{C}_{13}=(-1)^{1+3}\begin{vmatrix}1&2\\-3&1\end{vmatrix}=1+6=7$
$\text{C}_{21}=(-1)^{2+1}\begin{vmatrix}3&1\\1&-1\end{vmatrix}=3+1=4$
$\text{C}_{22}=(-1)^{2+2}\begin{vmatrix}2&1\\-3&-1\end{vmatrix}=-2+3=1$
$\text{C}_{23}=(-1)^{2+3}\begin{vmatrix}2&3\\-3&1\end{vmatrix}=-1(2+9)=-11$
$\text{C}_{31}=(-1)^{3+1}\begin{vmatrix}3&1\\2&2\end{vmatrix}=6-2=4$
$\text{C}_{32}=(-1)^{3+2}\begin{vmatrix}2&1\\1&2\end{vmatrix}=-1(4-1)=-3$
$\text{C}_{33}=(-1)^{3+3}\begin{vmatrix}2&3\\1&2\end{vmatrix}=4-3=1$
$\therefore\ \text{Adj }\text{A}=\begin{bmatrix}-4&-5&7\\4&1&-11\\4&-3&1\end{bmatrix}^\text{T}=\begin{bmatrix}-4&4&4\\-5&1&-3\\7&-11&1\end{bmatrix}$
$\Rightarrow\text{A}^{-1}=\frac{\text{Adj }\text{A}}{|\text{A}|}=\frac{1}{-16}\begin{bmatrix}-4&4&4\\-5&1&-3\\7&-11&1\end{bmatrix}$
Now, the given system of equations is expressible as:
$\begin{bmatrix}2&1&-3\\3&2&1\\1&2&-1\end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}13\\4\\8\end{bmatrix}$
or $A^TX = B$, where $\text{X}=\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix},\text{B}=\begin{bmatrix}13\\4\\8\end{bmatrix}$
Now, $\big|\text{A}^\text{T}\big|=\big|\text{A}\big|=-16\neq0$
So, the given system of equations is consistent with a unique solution given by
$\text{X}=(\text{A}^\text{T})^{-1}\text{B}=(\text{A}^{-1})^\text{T}\text{B}$
$\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=-\frac{1}{16}\begin{bmatrix}-4&4&4\\-5&1&-3\\7&-11&1\end{bmatrix}^\text{T}\begin{bmatrix}13\\4\\8\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=-\frac{1}{16}\begin{bmatrix}-4&-5&7\\4&1&-11\\4&-3&1\end{bmatrix}\begin{bmatrix}13\\4\\8\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=-\frac{1}{16}\begin{bmatrix}-52-20+56\\52+4-88\\52-12+8\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=-\frac{1}{16}\begin{bmatrix}-16\\-32\\48\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}1\\2\\-3\end{bmatrix}$
Hence, x = 1, y = 2 and z = -3 is the required solution.
View full question & answer→Question 255 Marks
Solve the following system of equations by matrix method:
2x + y + z = 2
x + 3y − z = 5
3x + y − 2z = 6
Answer$\begin{bmatrix}2&1&1\\ 1&3&-1\\ 3&1&-2\end{bmatrix}\begin{bmatrix}\text{x}\\ \text{y}\\ \text{z}\end{bmatrix}=\begin{bmatrix}1\\ 5\\ 6\end{bmatrix}$
$\text{|A|}=2{(-5)}-1{(1)}+1{(-8)}$
$=-10-1-8=-19\neq0$
Hence, the unique solution, given by
$\text{X}=\text{A}^{-1}\text{B}$
$\text{C}_{11}=-5\\ \text{C}_{21}=3\\ \text{C}_{31}=-4$
$\text{C}_{12}=-1\\ \text{C}_{22}=-7\\ \text{C}_{32}=3$
$\text{C}_{13}=-8\\ \text{C}_{23}=1\\ \text{C}_{33}=5$
Next, $\text{X}=\text{A}^{-1}\times\text{B}=\frac{1}{\text{|A|}}\begin{bmatrix}-5&3&-4\\ -1&-7&3\\ -8&1&5\end{bmatrix}\begin{bmatrix}2\\ 5\\ 6\end{bmatrix}$
$=\frac{1}{-19}\begin{bmatrix}-10+15-24\\ -2-35+18\\ -16+5+30\end{bmatrix}$
$=\frac{1}{-19}\begin{bmatrix}-19\\ -19\\ 19\end{bmatrix}$
$=\frac{1}{-19}\begin{bmatrix}\text{x}\\ \text{y}\\ \text{z}\end{bmatrix}=\begin{bmatrix}1\\ 1\\ -1\end{bmatrix}$
Hence, x = 1, y = 1, z = -1
View full question & answer→Question 265 Marks
Solve the following system of equations by matrix method:
$x - y + 2z = 7$
$3x + 4y - 5z = -5$
$2x - y + 3z = 12$
AnswerThe above system can be written as:
$\begin{bmatrix}1&-1&2\\ 3&4&-5\\ 2&-1&3\end{bmatrix}\begin{bmatrix}\text{x}\\ \text{y}\\ \text{z}\end{bmatrix}=\begin{bmatrix}7\\ -5\\ 12\end{bmatrix}$
Or AX = B
$|\text{A}| = 1 (7) + 1 (19) + 2 (-11) = 4\neq0$
So, the above system has a unique solution, given by
$\text{X}=\text{A}^{-1}\text{B}$
Let $C_{ij}$_be the co-factors of $a_{ij}$_in A
$\text{C}_{11}=7\\ \text{C}_{21}=1\\ \text{C}_{31}=-3$
$\text{C}_{12}=-19\\ \text{C}_{22}=-1\\ \text{C}_{32}=11$
$\text{C}_{13}=-11\\ \text{C}_{23}=-1\\ \text{C}_{33}=7$
$\text{adj A}=\begin{bmatrix}7&-19&-11\\ 1&-1&-1\\ -3&11&7\end{bmatrix}^\text{T}=\begin{bmatrix}7&1&-3\\ -19&-1&11\\ -11&-1&7\end{bmatrix}$
Now,
$\text{X}=\text{A}^{-1}\text{B}=\frac{1}{\text{|A|}}\text{(Adj A)}\times\text{B}$
$=\frac{1}{4}\begin{bmatrix}7&1&-3\\ -19&-1&11\\ -11&-1&7\end{bmatrix}\begin{bmatrix}7\\ -5\\ 12\end{bmatrix}$
$\begin{bmatrix}\text{x}\\ \text{y}\\ \text{z}\end{bmatrix}=\frac{1}{4}\begin{bmatrix}8\\ 4\\ 12\end{bmatrix}=\begin{bmatrix}2\\ 1\\ 3\end{bmatrix}$
Hence, x = 2, y = 1, z = 3
View full question & answer→Question 275 Marks
Solve the following systems of homogeneous linear equations by matrix method:
$x + y + z = 0$
$x - y - 5z = 0$
$x + 2y + 4z = 0$
Answerx + y + z = 0 x - y - 5z = 0 x + 2y + 4z = 0 $|\text{A}|=\begin{bmatrix}1&1&1\\1&-1&-5\\1&2&4\end{bmatrix}$$=1(6)-1(9)+1(3)=9-9=0$
Hence, the system has infinite solutions.Let, $\text{z}=\text{k}$
$\text{x}+\text{y}=-\text{k}$
$\text{x}-\text{y}=5\text{k}$
$\begin{bmatrix}1&1\\1&-1\end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\end{bmatrix}=\begin{bmatrix}-\text{k}\\5\text{k}\end{bmatrix}$
or, $\text{AX}=\text{B}$
$|\text{A}|=-2\neq0$, hence $A^{-1}$ exists.
$\text{adj }\text{A}=\begin{bmatrix}-1&-1\\-1&1\end{bmatrix}$
So, $\text{x}=\text{A}^{-1}\text{B}=\frac{1}{|\text{A}|}(\text{adj }\text{A})\text{B}$$=\frac{1}{-2}\begin{bmatrix}-1&-1\\-1&1\end{bmatrix}\begin{bmatrix}-\text{k}\\5\text{k}\end{bmatrix}$
$\begin{bmatrix}\text{x}\\\text{y}\end{bmatrix}=\Big(\frac{1}{-2}\Big)\begin{bmatrix}\text{k}-5\text{k}\\\text{k}+5\text{k}\end{bmatrix}=\begin{bmatrix}2\text{k}\\-3\text{k}\end{bmatrix}$
x = 2k, y = -3k, z = k
View full question & answer→Question 285 Marks
Show that the following system of linear equation is inconsistent:
$3x − y − 2z = 2$
$2y − z = −1$
$3x − 5y = 3$
AnswerThe given system of equations can be written as follows:
$\text{AX = B}$
Here,
$\text{A}=\begin{bmatrix}3&-1&-2\\ 0&2&-1\\ 3&-5&0\end{bmatrix},\text{X}=\begin{bmatrix}\text{x}\\ \text{y}\\ \text{z}\end{bmatrix}\text{and }\text{B}=\begin{bmatrix}2\\ -1\\ 3\end{bmatrix}$
$\text{|A|}=\begin{vmatrix}3&-1&-2\\ 0&2&-1\\ 3&-5&0\end{vmatrix}$
$=3{(0-5)}+1{(0+3)}-2{(0-6)}$
$=-15+3+12$
$=0$
Let $C _{ ij }$ be the co-factors of the elements $a _{ ij }$ in $A \left[ a _{ ij }\right]$. Then,
$\text{C}_{11}={(-1)}^{1+1}\begin{vmatrix}2&-1\\ -5&0\end{vmatrix}=-5,\\ \text{C}_{12}={(-1)}^{1+2}\begin{vmatrix}0&-1\\ 3&0\end{vmatrix}=-3,\\ \text{C}_{13}={(-1)}^{1+3}\begin{vmatrix}0&2\\ 3&-5\end{vmatrix}=-6$
$\text{C}_{21}={(-1)}^{2+1}\begin{vmatrix}-1&-2\\ -5&0\end{vmatrix}=10,\\ \text{C}_{22}={(-1)}^{2+2}\begin{vmatrix}3&-2\\ 3&0\end{vmatrix}=6,\\ \text{C}_{23}={(-1)}^{2+3}\begin{vmatrix}3&-1\\ 3&-5\end{vmatrix}=12$
$\text{C}_{31}={(-1)}^{3+1}\begin{vmatrix}-1&-2\\ 2&-1\end{vmatrix}=5,\\ \text{C}_{32}={(-1)}^{3+2}\begin{vmatrix}3&-2\\ 0&-1\end{vmatrix}=3,\\ \text{C}_{33}={(-1)}^{3+3}\begin{vmatrix}3&-1\\ 0&2\end{vmatrix}=6$
$\text{adj A}=\begin{bmatrix}-5&-3&-6\\ 10&6&12\\ 5&6&6\end{bmatrix}^\text{T}$
$=\begin{bmatrix}-5&10&5\\ -3&6&3\\ -6&12&6\end{bmatrix}$
$\text{(adj A) B}=\begin{bmatrix}-5&10&5\\ -3&6&3\\ -6&12&6\end{bmatrix}\begin{bmatrix}2\\ -1\\ 3\end{bmatrix}$
$=\begin{bmatrix}-10-10+15\\ -6-6+9\\ -12-12+18\end{bmatrix}$
$=\begin{bmatrix}-5\\ -3\\ -6\end{bmatrix}\neq0$
Hence, the given system of equations is consisent.
View full question & answer→Question 295 Marks
Solve the follwing system of equations by matrix method:
$x + y + z = 3$
$2x - y + z = -1$
$2x + y - 3z = -9$
AnswerThe above system can be written in matirx form as:
$\begin{bmatrix}1&1&1\\2&-1&1\\2&1&-3\end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}3\\-1\\-9\end{bmatrix}$
Or AX = B
Where,
$\text{A}=\begin{bmatrix}1&1&1\\2&-1&1\\2&1&-3\end{bmatrix},\text{X}=\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix},\text{B}=\begin{bmatrix}3\\-1\\-9\end{bmatrix}$
Since, $|\text{A}|=14\neq0$, the above system has a unique solution, given by
$X = A^{-1}B$
Let $C_{ij}$ be the co-factors of $a_{ij}$ in A
$\begin{matrix}\text{C}_{11}=2&\text{C}_{21}=4&\text{C}_{31}=2\\\text{C}_{12}=8&\text{C}_{22}=-5&\text{C}_{32}=1\\\text{C}_{13}=4&\text{C}_{23}=1&\text{C}_{33}=-3\end{matrix}$
$\text{Adj A}=\begin{bmatrix}2&8&4\\4&-5&1\\2&1&-3\end{bmatrix}^\text{T}=\begin{bmatrix}2&4&2\\8&-5&1\\4&1&-3\end{bmatrix}$
Now,
$\text{X}=\text{A}^{-1}\text{B}=\frac{1}{|\text{A}|}\times\text{Adj A}\times\text{B}$
$=\frac{1}{14}\begin{bmatrix}2&4&2\\8&-5&1\\4&1&-3\end{bmatrix}\begin{bmatrix}3\\-1\\-9\end{bmatrix}$
$=\frac{1}{14}\begin{bmatrix}-16\\20\\38\end{bmatrix}$
$\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}\frac{-8}{7}\\\frac{10}{7}\\\frac{19}{7}\end{bmatrix}$
Hence, $\text{x}=\frac{-8}{7},\text{y}=\frac{10}{7},\text{z}=\frac{19}{7}$
View full question & answer→Question 305 Marks
Solve the following system of equations by matrix method:
$3x + 4y + 2z = 8$
$2y - 3z = 3$
$x - 2y + 6z = -2$
Answer$\begin{bmatrix}3&4&2\\0&2&-3\\1&-2&6\end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}8\\3\\-2\end{bmatrix}$
or AX = B
$|\text{A}|=3(6)-4(3)+2(-2)$
$=18-12-4$
$=2\neq0$
Hence, the system has a unique solution, given by
$X = A^{-1}B$
$\begin{matrix}\text{C}_{11}=6&\text{C}_{21}=-28&\text{C}_{31}=-16\\\text{C}_{12}=-3&\text{C}_{22}=16&\text{C}_{32}=9\\\text{C}_{13}=-2&\text{C}_{23}=10&\text{C}_{33}=6\end{matrix}$
Next, $\text{X}=\text{A}^{-1}\text{B}=\frac{1}{|\text{A}|}(\text{adj A})\times\text{B}$
$=\frac{1}{2}\begin{bmatrix}6&-28&-16\\-3&16&9\\2&10&6\end{bmatrix}\begin{bmatrix}8\\3\\-2\end{bmatrix}$
$=\frac{1}{2}\begin{bmatrix}48-84+32\\-24+48-18\\-16+30-12\end{bmatrix}$
$=\frac{1}{2}\begin{bmatrix}-4\\6\\2\end{bmatrix}$
$\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}-2\\3\\1\end{bmatrix}$
Hence, $x = -2, y = 3, z = 1$
View full question & answer→Question 315 Marks
Two schools P and Q want to award their selected students on the values of Discipline, Politeness and Punctuality. The school P wants to award ₹ x each, ₹ y each and ₹ z each the three respectively values to its 3, 2 and 1 students with a total award money of ₹ 1,000. School Q wants to spend ₹ 1,500 to award its 4, 1 and 3 students on the respective values (by giving the same award money for three values as before). If the total amount of awards for one prize on each value is ₹ 600, using matrices, find the award money for each value. Apart from the above three values, suggest one more value for awards.
Answerx, y and z be prize amount per student for Discipline, Politeness and Punctuality respectively.
As per the data in the question, we get
3x + 2y + z = 1000
4x + y + 3z = 1500
x + y + z = 600
The above three simultaneous equations can be written in matrix form as:
$\begin{bmatrix}3&2&1\\4&1&3\\1&1&1\end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}1000\\1500\\600\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}3&2&1\\4&1&3\\1&1&1\end{bmatrix}^{-1}\begin{bmatrix}1000\\1500\\600\end{bmatrix}\ \dots(1)$
$\text{A}=\begin{bmatrix}3&2&1\\4&1&3\\1&1&1\end{bmatrix}$
$|\text{A}|=3(-2)-2(1)+1(3)=-5$
$\text{cof }\text{A}=\begin{bmatrix}-2&-1&3\\-1&2&-1\\5&-5&-5\end{bmatrix}$
$\text{adj }\text{A}=(\text{cof }\text{A})^\text{T}=\begin{bmatrix}-2&-1&5\\-1&2&-5\\3&-1&-5\end{bmatrix}$
$\text{A}^{-1}=\frac{\text{adj }\text{A}}{|\text{A}|}=\frac{\begin{bmatrix}-2&-1&5\\-1&2&-5\\3&-1&-5\end{bmatrix}}{-5}$
From (1)
$\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\frac{\begin{bmatrix}-2&-1&5\\-1&2&-5\\3&-1&-5\end{bmatrix}}{-5}\begin{bmatrix}1000\\1500\\600\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}-2&-1&5\\-1&2&-5\\3&-1&-5\end{bmatrix}\begin{bmatrix}-200\\-300\\-120\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}100\\200\\300\end{bmatrix}$
Excellence in extra-curricular activities should be another value considered for an award.
View full question & answer→Question 325 Marks
A company produces three product every day.Their production on a certain day is $45$ tons. It is found that the production of third product exceeds the production of first product by $8$ tons while the total production of first and third product is twice the production of second product. Determine the production level of each product using matrix method.
AnswerLet x, y and z be the production level of the first, second and third product, respectively.
According to the question,
$x + y + z = 45 .....(1)$
$-x + z = 8 .....(2)$
$x + y = 2y$ (Since the production of first and third products twice the production of second product)
$x - 2y + z = 0 .....(3)$
The given system of equation can be written in matrix form as follows:
$\begin{bmatrix}1&1&1\\-1&0&1\\1&-2&1\end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}45\\8\\0\end{bmatrix}$
$\text{AX}=\text{B}$
$\text{A}=\begin{bmatrix}1&1&1\\-1&0&1\\1&-2&1\end{bmatrix}\text{X}=\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}\text{B}=\begin{bmatrix}45\\8\\0\end{bmatrix}$
Now,
$|\text{A}|=1(-0+2)-1(-1-1)+1(2-0)$
$=2+2+2$
$=6$
Let $C _{ ij }$ be the co-factors of the elements $a _{ ij }$ in $A \left[ a _{ ij }\right]$. Then,
$\text{C}_{11}=(-1)^{1+1}\begin{vmatrix}0&1\\-2&1\end{vmatrix}=2,$ $\text{C}_{12}=(-1)^{1+2}\begin{vmatrix}-1&1\\1&1\end{vmatrix}=2,$ $\text{C}_{13}=(-1)^{1+3}\begin{vmatrix}-1&0\\1&-2\end{vmatrix}=2$
$\text{C}_{21}=(-1)^{2+1}\begin{vmatrix}1&1\\-2&1\end{vmatrix}=-3,$ $\text{C}_{22}=(-1)^{2+2}\begin{vmatrix}1&1\\1&1\end{vmatrix}=0,$ $\text{C}_{23}=(-1)^{2+3}\begin{vmatrix}1&1\\1&-2\end{vmatrix}=3$
$\text{C}_{31}=(-1)^{3+1}\begin{vmatrix}1&1\\0&1\end{vmatrix}=1,$ $\text{C}_{32}=(-1)^{3+2}\begin{vmatrix}1&1\\-1&1\end{vmatrix}=-2,$ $\text{C}_{33}=(-1)^{3+3}\begin{vmatrix}1&1\\-1&0\end{vmatrix}=1$
$\text{adj }\text{A}=\begin{bmatrix}2&2&2\\-3&0&3\\1&-2&1\end{bmatrix}^\text{T}$
$=\begin{bmatrix}2&-3&1\\2&0&-2\\2&3&1\end{bmatrix}$
$\Rightarrow\text{A}^{-1}=\frac{1}{|\text{A}|}\text{adj }\text{A}$
$=\frac{1}{6}\begin{bmatrix}2&-3&1\\2&0&-2\\2&3&1\end{bmatrix}$
$\text{X}=\text{A}^{-1}\text{B}$
$\Rightarrow\text{X}=\frac{1}{6}\begin{bmatrix}2&-3&1\\2&0&-2\\2&3&1\end{bmatrix}\begin{bmatrix}45\\8\\0\end{bmatrix}$
$\Rightarrow\text{X}=\frac{1}{6}\begin{bmatrix}90-24+0\\90+0+0\\90+24+0\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\frac{1}{6}\begin{bmatrix}66\\90\\114\end{bmatrix}$
$\therefore$ x = 11, y = 15 and z = 19
Thus, the production level of first, second and third product is 11, 15 and 19, respectively.
View full question & answer→Question 335 Marks
Solve the following systems of homogeneous linear equations by matrix method:
$3x - y + 2z = 0$
$4x + 3y + 3z = 0$
$5x + 7y + 4z =0$
AnswerHere,$3x - y + 2z = 0 .....(1)$
$4x + 3y + 3z = 0 .....(2)$
$5x + 7y + 4z =0 .....(3)$
The given system of homogeneous equaions can be written in matrix form as follows:
$\begin{bmatrix}3&-1&2\\4&3&3\\5&7&4\end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix} $
$\text{AX}=\text{O}$
Here,
$\text{A}=\begin{bmatrix}3&-1&2\\4&3&3\\5&7&4\end{bmatrix},\text{X}=\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}\text{and }\text{O}=\begin{bmatrix}0\\0\\0\end{bmatrix} $
Now,
$|\text{A}|=\begin{vmatrix}3&-1&2\\4&3&3\\5&7&4\end{vmatrix}$
$=3(12-21)+1(16-15)+2(28-15)$
$=0$
$\therefore\ |\text{A}|\neq0$
So, the given system of homogeneous equations has non-trivial solutions.
Substituting z = k in eq. (1) & eq. (2), we get
$3x - y = -2k$ and $4x + 3y = -3k$
$\text{AX}=\text{B}$
Here,
$\text{A}=\begin{bmatrix}3&-1\\4&3\end{bmatrix},\text{X}=\begin{bmatrix}\text{x}\\\text{y}\end{bmatrix}\text{and }\text{B}=\begin{bmatrix}-2\text{k}\\-3\text{k}\end{bmatrix} $
$\Rightarrow\begin{bmatrix}3&-1\\4&3\end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\end{bmatrix}=\begin{bmatrix}-2\text{k}\\-3\text{k}\end{bmatrix} $
$|\text{A}|=\begin{vmatrix}3&-1\\4&3\end{vmatrix}$
$=(3\times3+4\times1)$
$=13$
So, $A^{-1}$ exists.
We have
$\text{adj }\text{A}=\begin{bmatrix}3&1\\-4&3\end{bmatrix}$
$\text{A}^{-1}=\frac{1}{|\text{A}|}\text{adj }\text{A}$
$\Rightarrow\text{A}^{-1}=\frac{1}{13}\begin{bmatrix}3&1\\-4&3\end{bmatrix}$
$\text{X}=\text{A}^{-1}\text{B}$
$\Rightarrow\begin{bmatrix}\text{x}\\\text{y}\end{bmatrix}=\frac{1}{13}\begin{bmatrix}3&1\\-4&3\end{bmatrix}\begin{bmatrix}-2\text{k}\\-3\text{k}\end{bmatrix}$
$=\frac{1}{13}\begin{bmatrix}-6\text{k}-3\text{k}\\8\text{k}-9\text{k}\end{bmatrix} $
Thus, $\text{x}=\frac{-9\text{k}}{13}, \text{y}=\frac{-\text{k}}{13}\text{ and }\text{z}=\text{k}$ (where k is any real number) satisfy the given system of equations.
View full question & answer→Question 345 Marks
Solve the following system of equations by matrix method:
$x - y + z = 2$
$2x - y = 0$
$2y - z = 1$
AnswerThe above system of equations can be written as:
$\begin{bmatrix}1&-1&1\\ 2&-1&0\\ 0&2&-1\end{bmatrix}\begin{bmatrix}\text{x}\\ \text{y}\\ \text{z}\end{bmatrix}=\begin{bmatrix}2\\ 0\\ 1\end{bmatrix}$
Or AX = B
$\text{|A|}=1{(1)}+1{(-2)}+1{(4)}=1-2+4=3\neq0$
So, the above system has a unique solution, given by
$\text{X}=\text{A}^{-1}\text{B}$
Let $C _{ ij }$ be the co-factors of $a _{ ij }$ in A
$\text{C}_{11}=1\\ \text{C}_{21}=1\\ \text{C}_{31}=+1$
$\text{C}_{12}=2\\ \text{C}_{22}=-1\\ \text{C}_{32}=2$
$\text{C}_{13}=4\\ \text{C}_{23}=-2\\ \text{C}_{33}=1$
$\text{adj A}=\begin{bmatrix}1&2&4\\ 1&-1&-2\\ +1&2&1\end{bmatrix}^\text{T}=\begin{bmatrix}1&1&+1\\ 2&-1&2\\ 4&-2&1\end{bmatrix}$
$\text{X}=\text{A}^{-1}\text{B}=\frac{1}{\text{|A|}}\text{(Adj A)}\times\text{B}$
$=\frac{1}{3}\begin{bmatrix}1&1&+1\\ 2&-1&2\\ 4&-2&1\end{bmatrix}\begin{bmatrix}2\\ 0\\ 1\end{bmatrix}$
$\begin{bmatrix}\text{x}\\ \text{y}\\ \text{z}\end{bmatrix}=\frac{1}{3}\begin{bmatrix}3\\ 6\\ 9\end{bmatrix}=\begin{bmatrix}1\\ 2\\ 3\end{bmatrix}$
Hence, x = 1, y = 2, z = 3
View full question & answer→Question 355 Marks
Solve the following system of equations by matrix method:$\frac{2}{\text{x}}-\frac{3}{\text{y}}+\frac{3}{\text{z}}=10$
$\frac{1}{\text{x}}+\frac{1}{\text{y}}+\frac{1}{\text{z}}=10$
$\frac{3}{\text{x}}-\frac{1}{\text{y}}+\frac{2}{\text{z}}=13$
AnswerLet $\frac{1}{\text{x}}=\text{u},\frac{1}{\text{y}}=\text{v},\frac{1}{\text{z}}=\text{w}$
2u - 3v + 3w = 10
u + v + w = 10
3u - v + 2w = 13
Which can be written as:
$\begin{bmatrix}2&-3&3\\ 1&1&1\\ 3&-2&2\end{bmatrix}\begin{bmatrix}\text{x}\\ \text{y}\\ \text{z}\end{bmatrix}=\begin{bmatrix}10\\ 10\\ 13\end{bmatrix}$
$\text{|A|}=2{(3)}+3{(-1)}+3{(-4)}$
$=6-3-12=-9\neq0$
Hence, the system has a unique solution, given by
$\text{X}=\text{A}^{-1}\times\text{B}$
$\text{C}_{11}=3\ \text{C}_{21}=3\ \text{C}_{31}=-6$
$\text{C}_{12}=1\ \text{C}_{22}=-5\ \text{C}_{32}=1$
$\text{C}_{13}=-4\ \text{C}_{23}=-7\ \text{C}_{33}=5$
$\text{X}=\frac{1}{\text{|A|}}\text{(Adj A)}\times\text{(B)}$
$=\frac{1}{-9}\begin{bmatrix}3&3&-6\\ 1&-5&1\\ -4&-7&5\end{bmatrix}\begin{bmatrix}10\\ 10\\ 13 \end{bmatrix}$
$\frac{1}{-9}\begin{bmatrix}30+30-78\\ 10-50+13\\ -40-70+65\end{bmatrix}$
$\begin{bmatrix}\text{u}\\ \text{v}\\ \text{w}\end{bmatrix}=\frac{-1}{9}\begin{bmatrix}-18\\ -27\\ -45\end{bmatrix}=\begin{bmatrix}2\\ 3\\ 5\end{bmatrix}$
Hence, $\text{x}=\frac{1}{2},\text{y}=\frac{1}{3},\text{z}=\frac{1}{5}$
View full question & answer→Question 365 Marks
A school wants to award its students for the values of Honesty, Regularity and Hard work with a total cash award of Rs. 6,000. Three times the award money for Hard work added to that given for honesty amounts to Rs. 11,000. The award money given for Honesty and Hard work together is double the one given for Regularity. Represent the above situation algebraically and find the award money for each value, using matrix method. Apart from these values, namely, Honesty, Regularity and Hard work, suggest one more value which the school must include for awards.
AnswerLet the award money given for Honesty, Regularity and Hard work be x, y and z respectively. Since total cash award is Rs. 6,000. $\therefore$ x + y + z = Rs. 6000 .....(1) Three times the award money for Hard work and Honesty is Rs 11,000. $\therefore$ x + 3z = Rs. 11000 $\Rightarrow$ x + 0y + 3z = 11000 .....(2) Award money for Honesty and Hard work is double the one given for regularity. $\therefore$ x + z = 2y $\Rightarrow$ x - 2y + z = 0 .....(3)The above system can be written in matrix form as:
$\begin{bmatrix}1&1&1\\1&0&3\\1&-2&1\end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}6000\\11000\\0\end{bmatrix}$ or AX = B, where $\text{A}=\begin{bmatrix}1&1&1\\1&0&3\\1&-2&1\end{bmatrix}\text{X}=\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}\text{and }\text{B}=\begin{bmatrix}6000\\11000\\0\end{bmatrix}$ $|\text{A}|=6\neq0$ Thus, A is non-singular. Hence, it is invertible. $\text{Adj }\text{A}=\begin{bmatrix}6&-3&3\\2&0&-2\\-2&3&-1\end{bmatrix}$ $\therefore\ \text{A}^{-1}=\frac{1}{|\text{A}|}(\text{adj }\text{A})=\frac{1}{6}\begin{bmatrix}6&-3&3\\2&0&-2\\-2&3&-1\end{bmatrix}$ $\text{X}=\text{A}^{-1}\text{B}=\frac{1}{6}\begin{bmatrix}6&-3&3\\2&0&-2\\-2&3&-1\end{bmatrix}\begin{bmatrix}6000\\11000\\0\end{bmatrix}$ $\Rightarrow\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}500\\2000\\3500\end{bmatrix}$ Hence, x = 500, y = 2000 and z = 3500. Thus, award money given for Honesty, Regularity and Hardwork are Rs. 500, Rs. 2000 and Rs. 3500 respectively. School can include sincerity for awards.
View full question & answer→Question 375 Marks
Solve the following systems of homogeneous linear equations by matrix method:
$x + y - 6z = 0$
$x - y + 2z = 0$
$-3x + y + 2z = 0$
AnswerThe given system of homogeneous equations can be written in matrix form as follows:
$\begin{bmatrix}1&1&-6\\1&-1&2\\-3&1&2\end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix} $
$\text{AX}=\text{O}$
Here,
$\text{A}=\begin{bmatrix}1&1&-6\\1&-1&2\\-3&1&2\end{bmatrix},\text{X}=\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}\text{and }\text{O}=\begin{bmatrix}0\\0\\0\end{bmatrix} $
Now,
$|\text{A}|=\begin{vmatrix}1&1&-6\\1&-1&2\\-3&1&2\end{vmatrix}$
$=1(-2-2)-1(2+6)-6(1-3)$
$=-4-8+12$
$=0$
$\therefore\ |\text{A}|=0$
So, the given system of homogeneous equations has non-trivial solution.
Substituting z = k in eq. (1) and eq. (2), we get
x + y = 6k and x - y = -2k
$\text{AX}=\text{B}$
Here,
$\text{A}=\begin{bmatrix}1&1\\1&-1\end{bmatrix},\text{X}=\begin{bmatrix}\text{x}\\\text{y}\end{bmatrix}\text{and }\text{B}=\begin{bmatrix}6\text{k}\\-2\text{k}\end{bmatrix} $
$\Rightarrow\begin{bmatrix}1&1\\1&-1\end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\end{bmatrix}=\begin{bmatrix}6\text{k}\\-2\text{k}\end{bmatrix} $
Now,
$|\text{A}|=\begin{vmatrix}1&1\\1&-1\end{vmatrix}$
$=(1\times-1-1\times1)$
$=-2$
So,$ A^{-1}$ exists.
We have
$\text{adj }\text{A}=\begin{bmatrix}-1&-1\\-1&1\end{bmatrix}$
$\text{A}^{-1}=\frac{1}{|\text{A}|}\text{adj }\text{A}$
$\Rightarrow\text{A}^{-1}=\frac{1}{-2}\begin{bmatrix}-1&-1\\-1&1\end{bmatrix}$
$\text{X}=\text{A}^{-1}\text{B}$
$\Rightarrow\begin{bmatrix}\text{x}\\\text{y}\end{bmatrix}=\frac{1}{-2}\begin{bmatrix}-1&-1\\-1&1\end{bmatrix}\begin{bmatrix}6\text{k}\\-2\text{k}\end{bmatrix} $
$=\frac{1}{-2}\begin{bmatrix}-6\text{k}+2\text{k}\\-6\text{k}-2\text{k}\end{bmatrix}$
Thus, x = 2k, y = 4k and z = k (where k is any real number) satisfy the given system of equations.
View full question & answer→Question 385 Marks
Solve the following system of equations by matrix method:
x + y - z = 3
2x + 3y + z = 10
3x - y -7z = 1
AnswerHere,
$\text{A}=\begin{bmatrix}1&1&-1\\ 2&1&1\\ 3&-1&-7\end{bmatrix}$
$\text{|A|}=\begin{vmatrix}1&1&-1\\ 2&3&1\\ 3&-1&-7\end{vmatrix}$
$= 1 (-21 + 1) -1 (-14 - 3 ) - 1 (-2 - 9)$
$= -20 + 17 + 11$
$= 8$
$\text{C}_{11}=(-1)^{1+1}\begin{vmatrix}3&1\\ -1&-7\end{vmatrix}=-20,\\ \text{C}_{12}=(-1)^{1+2}\begin{vmatrix}2&1\\ 3&-7\end{vmatrix}=17,\\ \text{C}_{13}=(-1)^{1+3}\begin{vmatrix}2&3\\ 3&-1\end{vmatrix}=-11$
$\text{C}_{21}=(-1)^{2+1}\begin{vmatrix}1&-1\\ -1&-7\end{vmatrix}=8,\\ \text{C}_{22}=(-1)^{2+2}\begin{vmatrix}1&-1\\ 3&-7\end{vmatrix}=-4,\\ \text{C}_{23}(-1)^{2+3}\begin{vmatrix}1&1\\ 3&-1\end{vmatrix}=4$
$\text{C}_{31}=(-1)^{3+1}\begin{vmatrix}1&-1\\ 3&1\end{vmatrix}=4,\\ \text{C}_{32}=(-1)^{3+2}\begin{vmatrix}1&-1\\ 2&1\end{vmatrix}=-3,\\ \text{C}_{33}(-1)^{3+3}\begin{vmatrix}1&1\\ 2&3\end{vmatrix}=1$
$\text{C}_{31}=(-1)^{3+1}\begin{vmatrix}1&-1\\ 3&1\end{vmatrix}=4,\\ \text{C}_{32}=(-1)^{3+2}\begin{vmatrix}1&-1\\ 2&1\end{vmatrix}=-3,\\ \text{C}_{33}(-1)^{3+3}\begin{vmatrix}1&1\\ 2&3\end{vmatrix}=1$
$\text{adj A}=\begin{bmatrix}-20&17&-11\\ 8&-4&4\\ 4&-3&1\end{bmatrix}^\text{T}$
$=\begin{bmatrix}-20&8&4\\ 17&-4&-3\\ -11&4&1\end{bmatrix}$
$\Rightarrow\text{A}^{-1}=\frac{1}{\text{|A|}}\text{adj A}$
$=\frac{1}{8}\begin{bmatrix}-20&8&4\\ 17&-4&-3\\ -11&4&1\end{bmatrix}$
$\text{X}=\text{A}^{-1}\text{B}$
$\Rightarrow\begin{bmatrix}\text{x}\\ \text{y}\\ \text{z}\end{bmatrix}=\frac{1}{8}\begin{bmatrix}-20&8&4\\ 17&-4&-3\\ -11&4&1\end{bmatrix}\begin{bmatrix}3\\ 10\\ 1\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\ \text{y}\\ \text{z}\end{bmatrix}=\frac{1}{8}\begin{bmatrix}-60+80+4\\ 51-40-3\\ -33+40+1\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\ \text{y}\\ \text{z}\end{bmatrix}=\frac{1}{8}\begin{bmatrix}24\\ 8\\ 8\end{bmatrix}$
$\Rightarrow\text{x}=\frac{24}{8},\text{y}=\frac{8}{8}\ \text{and z}=\frac{8}{8}$
$\therefore\text{x}=3,\text{y}=1\text{ and }\text{z}=1$
View full question & answer→Question 395 Marks
If $\text{A}=\begin{bmatrix}1&-2&0\\ 2&1&3\\ 0&-2&1\end{bmatrix}$, find $A^{-1}.$ Using $A^{-1},$ solve the system of linear equations:
$x - 2y = 10, 2x + y + 3z = 8, -2y + z = 7$
AnswerHere, $\text{A}=\begin{bmatrix}1&-2&0\\ 2&1&3\\ 0&-2&1\end{bmatrix}$ $\text{|A|}=1{(1+6)}+2{(2-0)}+0{(-4-0)}$ $=7+4+0$ $=11$
Let $C_{ij}$ be the co-factors of the elements $a_{ij}$ in $A = [a_{ij}]$,
Then, $\text{C}_{11}={(-1)}^{1+1}\begin{vmatrix}1&3\\ -2&1\end{vmatrix}=7,\\ \text{C}_{12}={(-1)}^{1+2}\begin{vmatrix}2&3\\ 0&1\end{vmatrix}=-2,\\ \text{C}_{13}={(-1)}^{1+3}\begin{vmatrix}2&1\\ 0&-2\end{vmatrix}=-4$
$\text{C}_{21}={(-1)}^{2+1}\begin{vmatrix}-2&0\\ -2&1\end{vmatrix}=2,\\ \text{C}_{22}={(-1)}^{2+2}\begin{vmatrix}1&0\\ 0&1\end{vmatrix}=2,\\ \text{C}_{23}={(-1)}^{2+3}\begin{vmatrix}1&-2\\ 0&-2\end{vmatrix}=2$ $\text{C}_{31}={(-1)}^{3+1}\begin{vmatrix}-2&0\\ 1&3\end{vmatrix}=-6,\\ \text{C}_{32}={(-1)}^{3+2}\begin{vmatrix}1&0\\ 2&3\end{vmatrix}=-3,\\ \text{C}_{23}={(-1)}^{3+3}\begin{vmatrix}1&-2\\ 2&1\end{vmatrix}=5$
$\therefore\text{adj A}=\begin{bmatrix}7&-2&-4\\ 2&1&2\\ -6&-3&5\end{bmatrix}^\text{T}$
$=\begin{bmatrix}7&2&-6\\ -2&1&-3\\ -4&2&5\end{bmatrix}$ $\Rightarrow\text{A}^{-1}=\frac{1}{\text{|A|}}\text{adj A}$
$=\frac{1}{11}\begin{bmatrix}7&2&-6\\ -2&1&-3\\ -4&2&5\end{bmatrix}$ or,
$\text{AX = B}$ where, $\text{A}=\begin{bmatrix}1&-2&0\\ 2&1&3\\ 0&-2&1\end{bmatrix},\text{X}=\begin{bmatrix}\text{x}\\ \text{y}\\ \text{z}\end{bmatrix}\text{and }\text{B}=\begin{bmatrix}10\\ 8\\ 7\end{bmatrix}$
Now, $\therefore\text{X}=\text{A}^{-1}\text{B}$
$\Rightarrow\text{X}=\frac{1}{11}\begin{bmatrix}7&2&-6\\ -2&1&-3\\ -4&2&5\end{bmatrix}\begin{bmatrix}10\\ 8\\ 7\end{bmatrix}$
$\Rightarrow\text{X}=\frac{1}{11}\begin{bmatrix}70+16-42\ \\ -20+8-21\\ -40+16+35\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\ \text{y}\\ \text{z}\end{bmatrix}=\frac{1}{11}\begin{bmatrix}44\\ -33\\ 11\end{bmatrix}$ $\therefore$ $x = 4, y = -3$ and $z = 1$
View full question & answer→Question 405 Marks
Solve the following system of equations by matrix method:
$2x + 6y = 2$
$3x - z = -8$
$2x - y + z = -3$
AnswerHere,
$\text{A}=\begin{bmatrix}2&6&0\\ 3&0&-1\\ 2&-1&1\end{bmatrix}$
$\text{|A|}=\begin{vmatrix}2&6&0\\ 3&0&-1\\ 2&-1&1\end{vmatrix}$
$=2{(0-1)}-1{(3+2)}+0{(-3+0)}$
$=-2-30$
$=-32$
Let $C _{ ij }$ be the co-factors of the elemennts $a _{ ij }$ in $A \left[ a _{ ij }\right]$. Then,
$\text{C}_{11}=-{(-1)}^{1+1}\begin{vmatrix}0&-1\\ -1&1\end{vmatrix}=-1\\ \text{C}_{12}={(-1)}^{1+2}\begin{vmatrix}3&-1\\ 2&1\end{vmatrix}=-5,\\ \text{C}_{13}={(-1)}^{1+3}\begin{vmatrix}3&0\\ 2&-1\end{vmatrix}=-3 $
$\text{C}_{21}=-{(-1)}^{2+1}\begin{vmatrix}6&0\\ -1&1\end{vmatrix}=-6,\\ \text{C}_{22}={(-1)}^{2+2}\begin{vmatrix}2&0\\ 2&1\end{vmatrix}=2,\\ \text{C}_{23}={(-1)}^{2+3}\begin{vmatrix}2&6\\ 2&-1\end{vmatrix}=14 $
$\text{C}_{31}=-{(-1)}^{3+1}\begin{vmatrix}6&0\\ 0&-1\end{vmatrix}=-6,\\ \text{C}_{32}={(-1)}^{3+2}\begin{vmatrix}2&0\\ 3&-1\end{vmatrix}=2,\\ \text{C}_{33}={(-1)}^{3+3}\begin{vmatrix}2&6\\ 3&0\end{vmatrix}=-18 $
$\text{adj A}=\begin{bmatrix}-1&-5&-3\\ -6&2&14\\ -6&2&-18\end{bmatrix}^\text{T}$
$=\begin{bmatrix}-1&-6&-6\\ -5&2&2\\ -3&14&-18\end{bmatrix}$
$\Rightarrow\text{A}^{-1}=\frac{1}{\text{|A|}}\text{adj A}$
$=\frac{1}{-32}\begin{bmatrix}-1&-6&-6\\ -5&2&2\\ -3&14&-18\end{bmatrix}$
$\text{X}=\text{A}^{-1}\text{B}$
$\Rightarrow\begin{bmatrix}\text{x}\\ \text{y}\\ \text{z}\end{bmatrix}=\frac{1}{-32}\begin{bmatrix}-1&-6&-6\\ -5&2&2\\ -3&14&-18\end{bmatrix}\begin{bmatrix}2\\ -8\\ -3\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\ \text{y}\\ \text{z}\end{bmatrix}=\frac{1}{-32}\begin{bmatrix}-2+48+18\\ -10-16-6\\ -6-112+54\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\ \text{y}\\ \text{z}\end{bmatrix}=\frac{1}{-32}\begin{bmatrix}64\\ -32\\ -64\end{bmatrix}$
$\Rightarrow\text{x}=\frac{64}{-32},\text{y}=\frac{-32}{-32}\text{ and }\text{z}=\frac{-64}{-32}$
$\therefore$ x = -2, y = 1 and z = 2
View full question & answer→Question 415 Marks
Show that the following system of linear equation is inconsistent:
4x − 5y − 2z = 2
5x − 4y + 2z = −2
2x + 2y + 8z = −1
AnswerThe above system can be written as:
$\begin{bmatrix}4&-5&-2\\ 5&-4&2\\ 2&2&8\end{bmatrix}\begin{bmatrix}\text{x}\\ \text{y}\\ \text{z}\end{bmatrix}=\begin{bmatrix}2\\ -2\\ -1\end{bmatrix}$
or $\text{AX = B}$
$\text{|A|}=4{(-36)}+5{(36)}-2{(18)}$
$=-144+180-36$
$=0$
So, A is singular and the above system will be inconsisent, if
$\text{(adj A)}\times\text{B}\neq0$
$\text{C}_{11}=-36\\ \text{C}_{21}=36\\ \text{C}_{31}=-18$
$\text{C}_{12}=-36\\ \text{C}_{22}=36\\ \text{C}_{32}=-18$
$\text{C}_{13}=18\\ \text{C}_{23}=-18\\ \text{C}_{33}=9$
$(\text{adj A})=\begin{bmatrix}-36&-36&18\\ 36&36&-18\\ -18&-18&9\end{bmatrix}^\text{T}=\begin{bmatrix}-36&36&-18\\ -36&36&-18\\\ 18&-18&9\end{bmatrix}$
$(\text{adj A})\times\text{B}=\begin{bmatrix}-36&36&-18\\ -36&36&-18\\ 18&-18&9\end{bmatrix}\begin{bmatrix}2\\ -2\\ -1\end{bmatrix}$$=\begin{bmatrix}-72-72+18\\ -72-72+18\\ 36+36-9\end{bmatrix}\neq0$
Hence, the above system is inconsistent.
View full question & answer→Question 425 Marks
Given $\text{A}=\begin{bmatrix}2&2&-4\\-4&2&-4\\2&-1&5\end{bmatrix},\text{B}=\begin{bmatrix}1&-1&0\\2&3&4\\0&1&2\end{bmatrix}$ , find BA and use this to solve the system of equations $y + 2z = 7, x - y = 3, 2x + 3y + 4z = 17$
AnswerHere,
$\text{A}=\begin{bmatrix}2&2&-4\\-4&2&-4\\2&-1&5\end{bmatrix}\text{and }\text{B}=\begin{bmatrix}1&-1&0\\2&3&4\\0&1&2\end{bmatrix}$
$\text{BA}=\begin{bmatrix}1&-1&0\\2&3&4\\0&1&2\end{bmatrix}\begin{bmatrix}2&2&-4\\-4&2&-4\\2&-1&5\end{bmatrix}$
$\Rightarrow\text{BA}=\begin{bmatrix}2+4+0&2-2+0&-4+4+0\\4-12+8&4+6-4&-8-12+20\\0-4+4&0+2-2&0-4+10\end{bmatrix}$
$=\begin{bmatrix}6&0&0\\0&6&0\\0&0&6\end{bmatrix}$
$\Rightarrow\text{BA}=6\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
$\Rightarrow\text{BA}=6\text{I}_3$
$\Rightarrow\text{B}\Big(\frac{1}{6}\text{A}\Big)=\text{I}_3$
$\Rightarrow\text{B}^{-1}=\frac{1}{6}\text{A}$
$\Rightarrow\text{B}^{-1}=\frac{1}{6}\begin{bmatrix}2&2&-4\\-4&2&-4\\2&-1&5\end{bmatrix}$
Now, BX = C
where, $\text{B}=\begin{bmatrix}1&-1&0\\2&3&4\\0&1&2\end{bmatrix},\text{X}=\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}\text{and }\text{C}=\begin{bmatrix}3\\17\\7\end{bmatrix}$
$\therefore$ $X = B^{-1}C$
$\Rightarrow\text{X}=\frac{1}{6}\begin{bmatrix}2&2&-4\\-4&2&-4\\2&-1&5\end{bmatrix}\begin{bmatrix}3\\17\\7\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\frac{1}{6}\begin{bmatrix}6+34-28\\-12+34-28\\6-17+35\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\frac{1}{6}\begin{bmatrix}12\\-6\\24\end{bmatrix}$
$\therefore$ $x = 2, y = -1$ and $z = 4.$
View full question & answer→Question 435 Marks
Show that the following system of linear equations is consistent and also find solutions:x - y + z = 3
2x + y - z = 2
-x - 2y + 2z = 1
AnswerHere,
$\text{x}-\text{y}+\text{z}=3\ \dots(1)$
$2\text{x}+\text{y}-\text{z}=2\ \dots(2)$
$-\text{x}-2\text{y}+2\text{z}=1 \ \dots(3)$
Or, $\text{AX}=\text{B}$
Where,
$\text{A}=\begin{bmatrix}1&-1&1\\ 2&1&-1\\ -1&-2&2\end{bmatrix},\text{X}=\begin{bmatrix}\text{x}\\ \text{y}\\ \text{z}\end{bmatrix}\text{and }\text{B}=\begin{bmatrix}3\\ 2\\ 1\end{bmatrix}$
$\begin{bmatrix}1&-1&1\\ 2&1&-1\\ -1&-2&2\end{bmatrix}\begin{bmatrix}\text{x}\\ \text{y}\\ \text{z}\end{bmatrix}=\begin{bmatrix}3\\ 2\\ 1\end{bmatrix}$
$\text{|A|}=\begin{vmatrix}1&-1&1\\ 2&1&-1\\ -1&-2&2\end{vmatrix}$
View full question & answer→Question 445 Marks
Solve the following system of equations by matrix method:
$6x - 12y + 25z = 4$
$4x + 15y - 20z = 3$
$2x + 18y + 15z = 10$
AnswerHere,
$\text{A}=\begin{bmatrix}6&-12&25\\ 4&15&-20\\ 2&18&15\end{bmatrix}$
$\text{|A|}=\begin{vmatrix}6&-12&25\\ 4&15&-20\\ 2&18&15\end{vmatrix}$
$=6(225+360)+12(60+40)+25(72-30)$
$= 3510 + 1200 + 1050$
$= 5760$
Let $C _{ ij }$ be the co-factors of the elements $a _{ ij }$ in $A \left[ a _{ ij }\right]$. Then,
$\text{C}_{11}={(-1)}^{1+1}\begin{vmatrix}15&-20\\ 18&15\end{vmatrix}=585,\\ \text{C}_{12}={(-1)}^{1+2}\begin{vmatrix}4&-20\\ 2&15\end{vmatrix}=-100,\\ \text{C}_{13}={(-1)}^{1+3} \begin{vmatrix}4&15\\ 2&18\end{vmatrix}=42$
$\text{C}_{21}={(-1)}^{2+1}\begin{vmatrix}-12&-25\\ 18&15\end{vmatrix}=630,\\ \text{C}_{22}={(-1)}^{2+2}\begin{vmatrix}6&-25\\ 2&15\end{vmatrix}=-40,\\ \text{C}_{23}={(-1)}^{2+3} \begin{vmatrix}6&-12\\ 2&18\end{vmatrix}=132$
$\text{C}_{31}={(-1)}^{3+1}\begin{vmatrix}-12&25\\ 15&-20\end{vmatrix}=-135,\\ \text{C}_{32}={(-1)}^{3+2}\begin{vmatrix}6&-25\\ 4&-20\end{vmatrix}=220,\\ \text{C}_{33}={(-1)}^{3+3} \begin{vmatrix}6&-12\\ 4&15\end{vmatrix}=138$
$\text{adj A}\begin{bmatrix}585&-100&42\\ 630&40&-132\\ -135&220&138\end{bmatrix}^\text{T}$
$=\begin{bmatrix}585&630&-135\\ -100&40&220\\ 42&-132&138\end{bmatrix}$
$\Rightarrow\text{A}^{-1}\frac{1}{\text{|A|}}\text{adj}\ \text{A}$
$=\frac{1}{5760}\begin{bmatrix}585&630&-135\\ -100&40&220\\ 42&-132&138\end{bmatrix}$
$\text{X}=\text{A}^{-1}\text{B}$
$\Rightarrow\begin{bmatrix}\text{x}\\ \text{y}\\ \text{z}\end{bmatrix}=\frac{1}{5760}\begin{bmatrix}585&630&-135\\ -100&40&220\\ 42&-132&138\end{bmatrix}\begin{bmatrix}4\\ 3\\ 10\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\ \text{y}\\ \text{z}\end{bmatrix}=\frac{1}{5760}\begin{bmatrix}2340+1890-1350\\ -400+120+2200\\ 168-396+1380\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\ \text{y}\\ \text{z}\end{bmatrix}=\frac{1}{5760}\begin{bmatrix}2880\\ 1920\\ 1152\end{bmatrix}$
$\Rightarrow\text{x}=\frac{2880}{5760},\text{y}=\frac{1920}{5760}\text{ and }\text{z}=\frac{1152}{5760}$
$\therefore\text{x}=\frac{1}{2},\text{y}=\frac{1}{3}\text{ and }\text{z}=\frac{1}{5}$
View full question & answer→Question 455 Marks
The prices of three commodities $P, Q$ and $R$ are $R s . x, y$ and $z$ per unit respectively. A purchases 4 units of $R$ and sells 3 units of $P$ and 5 units of $Q$. $B$ purchases 3 units of $Q$ and sells 2 units of $P$ and $1$ unit of R. $C$ purchases 1 unit of $P$ and sells 4 units of $Q$ and 6 units of $R$. In the process $A, B$ and $C$ earn $Rs. 6000, Rs. 5000$ and $Rs. 13000$ respectively. If selling the units is positive earning and buying the units is negative earnings, find the price per unit of three commodities by using matrix method.
AnswerThe prices of three commodities P, Q and R are Rs. x, Rs. y and Rs. z per unit, respectively.
According to the question,
$3x + 5y - 4z = 6000$
$2x - 3y + z = 5000$
$-x + 4y + 6z = 13000$
The given system of equations can be written in matrix form as follows:
$\begin{bmatrix}3&5&-4\\2&-3&1\\-1&4&6\end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}6000\\5000\\13000\end{bmatrix}$
$\text{AX}=\text{B}$
Here,
$\text{A}=\begin{bmatrix}3&5&-4\\2&-3&1\\-1&4&6\end{bmatrix}\text{X}=\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}\text{B}=\begin{bmatrix}6000\\5000\\13000\end{bmatrix}$
Now,
$|\text{A}|=3(-18-4)-5(12+1)-4(8-3)$
$=-66-65-20$
$=-151\neq0$
So, $A^{-1}$ exists.
Let $C _{ ij }$ be the co-factors of the elements $a _{ ij }$ in $A \left[ a _{ ij }\right]$. Then,
$\text{C}_{11}=(-1)^{1+1}\begin{vmatrix}-3&1\\4&6\end{vmatrix}=-22,$ $\text{C}_{12}=(-1)^{1+2}\begin{vmatrix}2&1\\-1&6\end{vmatrix}=-13,$ $\text{C}_{13}=(-1)^{1+3}\begin{vmatrix}2&-3\\-1&4\end{vmatrix}=5$
$\text{C}_{21}=(-1)^{2+1}\begin{vmatrix}5&-4\\4&6\end{vmatrix}=-46,$ $\text{C}_{22}=(-1)^{2+2}\begin{vmatrix}3&-4\\-1&6\end{vmatrix}=14,$ $\text{C}_{23}=(-1)^{2+3}\begin{vmatrix}3&5\\-1&4\end{vmatrix}=-17$
$\text{C}_{31}=(-1)^{3+1}\begin{vmatrix}5&-4\\-3&1\end{vmatrix}=-7,$ $\text{C}_{32}=(-1)^{3+2}\begin{vmatrix}3&-4\\2&1\end{vmatrix}=-11,$ $\text{C}_{33}=(-1)^{3+3}\begin{vmatrix}3&5\\2&-3\end{vmatrix}=-19$
$\text{adj }\text{A}=\begin{bmatrix}-22&-13&5\\-46&14&-17\\-7&-11&-19\end{bmatrix}^\text{T}$
$=\begin{bmatrix}-22&-46&-7\\-13&14&-11\\5&-17&-19\end{bmatrix}$
$\Rightarrow\text{A}^{-1}=\frac{1}{|\text{A}|}\text{adj }\text{A}$
$=\frac{1}{-151}\begin{bmatrix}-22&-46&-7\\-13&14&-11\\5&-17&-19\end{bmatrix}$
$\text{X}=\text{A}^{-1}\text{B}$
$\Rightarrow\text{X}=\frac{1}{-151}\begin{bmatrix}-22&-46&-7\\-13&14&-11\\5&-17&-19\end{bmatrix}\begin{bmatrix}6000\\5000\\13000\end{bmatrix}$
$\Rightarrow\text{X}=\frac{1}{-151}\begin{bmatrix}-453000\\-151000\\-302000\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}3000\\1000\\2000\end{bmatrix}$
Thus, the prices of the three commodities P, Q and R are Rs. 3000, Rs. 1000 and Rs. 2000 per unit, respectively.
View full question & answer→Question 465 Marks
$\text{A}=\begin{bmatrix}1&-2&0\\ 2&1&3\\ 0&-2&1\end{bmatrix}\text{and }\text{B}=\begin{bmatrix}7&2&-6\\ -2&1&-3\\ -4&2&5\end{bmatrix}$, find AB. Hence, solve the system of equations:
x - 2y = 10, 2x + y + 3z = 8 and -2y + z = 7
Answer$\text{A}=\begin{bmatrix}1&-2&0\\2&1&3\\0&-2&1\end{bmatrix}\text{and }\text{B}=\begin{bmatrix}7&2&-6\\-2&1&-3\\-4&2&5\end{bmatrix}$
$\text{A}\times\text{B}=\begin{bmatrix}11&0&0\\ 0&11&0\\\ 0&0&11\end{bmatrix}$
AB = 11I, where I is a 3 × 3 unit matrix
$\text{A}^{-1}=\frac{1}{11}\text{B}$ [By def. of inverse]Or
Or $\frac{1}{11}=\begin{bmatrix}7&2&-6\\ -2&1&-3\\ -4&2&5\end{bmatrix}$
Now, the given system of equations can be written as:
$\begin{bmatrix}1&-2&0\\ 2&1&3\\ 0&-2&1\end{bmatrix}\begin{bmatrix}\text{x}\\ \text{y}\\ \text{z}\end{bmatrix}=\begin{bmatrix}10\\ 8\\ 7\end{bmatrix}$
Or $\text{AX = B}$
$\text{X = A}^{-1}\text{B}$
Or $=\frac{1}{11}\begin{bmatrix}7&2&-6\\-2&1&-3\\-4&2&5\end{bmatrix}\begin{bmatrix}10\\8\\7\end{bmatrix}$
$\begin{bmatrix}\text{x}\\ \text{y}\\ \text{z}\end{bmatrix}=\frac{1}{11}\begin{bmatrix}44\\ -33\\ 11\end{bmatrix}=\begin{bmatrix}4\\ -3\\ 1\end{bmatrix}$
Hence, x = 4, y = -3, z = 1
View full question & answer→Question 475 Marks
Show that the following system of linear equation is inconsistent:
x + y − 2z = 5
x − 2y + z = −2
−2x + y + z = 4
AnswerThe above system can be written as:
$\begin{bmatrix}1&1&-2\\ 1&-2&1\\ -2&1&1\end{bmatrix}\begin{bmatrix}\text{x}\\ \text{y}\\ \text{z}\end{bmatrix}=\begin{bmatrix}5\\ -2\\ 4\end{bmatrix}$
or $\text{AX = B}$
$\text{|A|}=1{(-3)}-1{(-3)}=-3-3+6=0$
So, A is singular. Now the system can be inconsistent, if
$(\text{adj A})\times\text{B}\neq0$
$\text{C}_{11}=-3\\ \text{C}_{21}=-3\\ \text{C}_{31}=-3$
$\text{C}_{12}=-3\\ \text{C}_{22}=-3\\ \text{C}_{32}=-3$
$\text{C}_{13}=-3\\ \text{C}_{23}=-3\\ \text{C}_{33}=-3$
$(\text{adj A})=\begin{bmatrix}-3&-3&-3\\ -3&-3&-3\\ -3&-3&-3\end{bmatrix}^\text{T}=\begin{bmatrix}-3&-3&-3\\ -3&-3&-3\\ -3&-3&-3\end{bmatrix}$
$(\text{adj A})\times\text{(B)}=\begin{bmatrix}-3&-3&-3\\ -3&-3&-3\\ -3&-3&-3\end{bmatrix}\begin{bmatrix}5\\ -2\\ 4\end{bmatrix}=\begin{bmatrix}-15+6-12\\ -15+6-12\\ -15+6-12\end{bmatrix}$
$=\begin{bmatrix}-21\\ -21\\ -12\end{bmatrix}$
$\neq0$
Hence, the given system is inconsistent.
View full question & answer→Question 485 Marks
Show that the following system of linear equation is inconsistent:
$2x + 5y = 7$
$6x + 15y = 13$
AnswerThe given system of equations can be expresed as follows:
$\text{AX = B}$
Here,
$\text{A}=\begin{bmatrix}2&5\\ 6&15\end{bmatrix},\text{X}=\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}\text{and }\text{B}=\begin{bmatrix}7\\ 13\end{bmatrix}$
Now,
$\text{|A|}=\begin{vmatrix}2&5\\ 6&15\end{vmatrix}$
$={(30-30)}$
$=0$
Let $C _{ ij }$ be the co-factors of the elements $a _{ ij }$ in $A \left[ a _{ ij }\right]$. Then,
$\text{C}_{11}=-1^{1+1}{(15)}=15,\\ \text{C}_{12}=-1^{1+2}{(6)}=-6\\ $
$\text{C}_{21}=-1^{2+1}{(5)}=-5,\\ \text{C}_{22}=-1^{2+2}{(6)}=2\\ $
$\text{adj A}=\begin{bmatrix}15&-6\\ -5&2\end{bmatrix}^\text{T}$
$=\begin{bmatrix}15&-5\\ -6&2\end{bmatrix}$
$(\text{adj A) B}=\begin{bmatrix}15&-5\\ -6&2\end{bmatrix}\begin{bmatrix}7\\ 13\end{bmatrix}$
$=\begin{bmatrix}105-65\\ -42+26\end{bmatrix}$
$=\begin{bmatrix}40\\ -16\end{bmatrix}\neq0$
Hence, the given system of equations is inconsitent.
View full question & answer→Question 495 Marks
A shopkeeper has $3$ varieties of pens ' $A$ ',$ 'B'$ and $'C'$. Meenu purchased $1$ pen of each variety for a total of$ Rs 21.$ Jeevan purchased $4$ pens of $'A'$ variety $3$ pens of ' $B$ ' variety and $2 $ pens of ' $C$ ' variety for $Rs 60 $. While Shikha purchased $6$ pens of 'A' variety, $2$ pens of $'B'$ variety and $3$ pens of $'C'$ variety for $Rs 70 $. Using matrix method, find cost of each variety of pen.
AnswerAs there are $3$ varieties of pen $A, B$ and $C$
Meenu purchased $1$ pen of each variety which costs her Rs$. 21$
Therefore,
$A + B + C = 2$
Similarly,
For Jeevan
$4A + 3B + 2C = 60$
For Shikha
$6A + 2B + 3C = 70$
$\begin{bmatrix}1&1&1\\4&3&2\\6&2&3\end{bmatrix}\begin{bmatrix}\text{A}\\\text{B}\\\text{C}\end{bmatrix}=\begin{bmatrix}21\\60\\70\end{bmatrix}$
where, $\text{P}=\begin{bmatrix}1&1&1\\4&3&2\\6&2&3\end{bmatrix},\text{Q}=\begin{bmatrix}21\\60\\70\end{bmatrix}$
$|\text{P}|=1(9-4)-1(12-12)+1(8-18)$
$=-5\neq0$
$\therefore $ $P^{-1}$ exists
$\text{X}=\text{P}^{-1}\text{Q}$
$\begin{matrix}\text{C}_{11}=5&\text{C}_{12}=0&\text{C}_{13}=-10\\\text{C}_{21}=-1&\text{C}_{22}=-3&\text{C}_{23}=4\\\text{C}_{31}=-1&\text{C}_{32}=2&\text{C}_{33}=-1\end{matrix}$
$\text{adj }\text{P}=\begin{bmatrix}5&0&-10\\-1&-3&4\\-1&2&-1\end{bmatrix}^\text{T}=\begin{bmatrix}5&-1&-1\\0&-3&2\\-10&4&-1\end{bmatrix}$
$\text{P}^{-1}=\frac{1}{-5}\begin{bmatrix}5&-1&-1\\0&-3&2\\-10&4&-1\end{bmatrix}$
$\text{X}=\text{P}^{-1}\text{Q}$
$=\frac{1}{-5}\begin{bmatrix}5&-1&-1\\0&-3&2\\-10&4&-1\end{bmatrix}\begin{bmatrix}21\\60\\70\end{bmatrix}$
$=\frac{1}{-5}\begin{bmatrix}105-60-70\\0-180+140\\-210+240-70\end{bmatrix}$
$=\frac{1}{-5}\begin{bmatrix}-25\\-40\\-40\end{bmatrix}$
$\therefore\ \text{X} = \begin{bmatrix}5\\8\\8\end{bmatrix}$
Therefore, cost of A variety of pens = Rs. 5
Cost of B variety of pens $= Rs. 8$
Cost of C variety of pens$ = Rs. 8$
View full question & answer→Question 505 Marks
Solve the following system of equations by matrix method:
$3x + y = 7$
$5x + 3y = 12$
AnswerThe given system of equations can be written in matrix form as follows:
$\begin{bmatrix}3&1\\ 5&3\end{bmatrix}\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\begin{bmatrix}7\\ 12\end{bmatrix}$
AX = B
Here,
$\text{A}=\begin{bmatrix}3&1\\ 5&3\end{bmatrix},\text{x}=\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}\text{ and }\text{B}=\begin{bmatrix}7\\ 12\end{bmatrix}$
Now,
$\text{|A|}=\begin{vmatrix}3&1\\ 5&3\end{vmatrix}$
$=9-5$
$=4\neq0$
So, the given system has a unique solution given by $X = A^{-1} B.$
Let $C _{ ij }$ be the co-factors of the elements $a _{ ij }$ in $A \left[ a _{ ij }\right]$. Then,
$\text{C}_{11}=-{(-1)}^{1+1}{(3)}=3,\text{C}_{12}={(-1)}^{1+2}{(5)}=-5$
$\text{C}_{21}={(-1)}^{2+1}{(1)}=-1,\text{C}_{22}=-{(-1)}^{2+2}{(3)}=3$
$\text{adj A}=\begin{bmatrix}3&-5\\ -1&3\end{bmatrix}^\text{T}$
$=\begin{bmatrix}3&-1\\-5&3\end{bmatrix}$
$\text{A}^{-1}=\frac{1}{|\text{A}|}\text{adj A}$
$=\frac{1}{4}\begin{bmatrix}3&-1\\ -5&3\end{bmatrix}$
$X = A^{-1}B$
$=\frac{1}{4}\begin{bmatrix}3&-1\\ -5&3\end{bmatrix}\begin{bmatrix}7\\ 12\end{bmatrix}$
$=\frac{1}{4}\begin{bmatrix}21-12\\ -35+36\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\begin{bmatrix}\frac{9}{4}\\ \frac{1}{4}\end{bmatrix}$
$\therefore\ \text{x}=\frac{9}{4}\text{ and }\text{y}=\frac{1}{4}$
View full question & answer→Question 515 Marks
Solve the follwing system of equations by matrix method:
$3x + 4y - 5 = 0$
$x - y + 3 = 0$
AnswerThe given system of equations can be written in matrix form as follow:
$\begin{bmatrix}3&4\\ 1&-1\end{bmatrix}\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\begin{bmatrix}5\\ -3\end{bmatrix}$
AX = B
Here,
$\text{A}=\begin{bmatrix}3&4\\ 1&-1\end{bmatrix},\text{x}=\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}\text{ and }\text{ B}=\begin{bmatrix}5\\ -3\end{bmatrix}$
Now,
$\text{|A|}=\begin{vmatrix}3&4\\ 1&-1\end{vmatrix}$
$=-3-4$
$=-7\neq0$
So, the given system has a unique solution given by $X = A^{-1} B.$
Let $C _{ ij }$ be the co-factors of the elements $a _{ ij }$ in $A \left[ a _{ ij }\right]$. Then,
$\text{C}_{11}=(-1)^{1+1}(-1)=-1,\text{C}_{12}=(-1)^{1+2}(1)=-1$
$\text{C}_{21}=(-1)^{2+1(4)}=-4,\text{C}_{22}=-{(-1)}^{2+2}{(3)}={(3)}$
$\text{adj A}=\begin{bmatrix}-1&-1\\ -4&3\end{bmatrix}^{T}=\begin{bmatrix}-1&-4\\ -1&3\end{bmatrix}$
$\text{A}^{-1}=\frac{1}{\text{|A|}}\text{adj A}$
$=\frac{1}{-7}\begin{bmatrix}-1&-4\\ -1&3\end{bmatrix}$
$X = A^{-1} B$
$=\frac{1}{-7}\begin{bmatrix}-1&-4\\ -1&3\end{bmatrix}\begin{bmatrix}5\\ -3\end{bmatrix}$
$=\frac{1}{-7}\begin{bmatrix}-5+12\\ -5-9\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\begin{bmatrix}\frac{7}{-7}\\ \frac{-14}{-7}\end{bmatrix}$
$\therefore$ x = -1 and y = 2
View full question & answer→Question 525 Marks
Show that the following system of linear equation is inconsistent:
4x − 2y = 3
6x − 3y = 5
AnswerThis system can be written as:
$\begin{bmatrix}4&-2\\ 6&-3\end{bmatrix}\begin{bmatrix} \text{x}\\ \text{y}\end{bmatrix}=\begin{bmatrix}3\\ 5\end{bmatrix}$
or $\text{AX = B}$
$\text{|A|}=-12+12=0$
So, A is singular, Now system will be inconsisent, if
$(\text{adj A})\times\text{B}\neq0$
$\text{C}_{11}=-3$
$\text{C}_{12}=-6$
$\text{C}_{21}=2$
$\text{C}_{22}=4$
$\text{adj A}=\begin{bmatrix}-3&-6\\ 2&4\end{bmatrix}^\text{T}\begin{bmatrix}-3&2\\ -6&4\end{bmatrix}$
$(\text{adj A})\times(\text{B})=\begin{bmatrix}-3&2\\ -6&4\end{bmatrix}\begin{bmatrix}3\\ 5\end{bmatrix}$
$=\begin{bmatrix}-9+10\\ -18+20\end{bmatrix}$
$=\begin{bmatrix}1\\ 2\end{bmatrix}$
$\neq0$
Hence, the above system is inconsisent.
View full question & answer→Question 535 Marks
Solve the following systems of homogeneous linear equations by matrix method:
2x + 3y - z = 0
x - y - 2z = 0
3x + y + 3z = 0
Answer2x + 3y - z = 0 x - y - 2z = 0 3x + y + 3z = 0 Hence, $\text{A}=\begin{bmatrix}2&3&-1\\1&-1&-2\\3&1&3\end{bmatrix}$ $|\text{A}|=\begin{bmatrix}2&3&-1\\1&-1&-2\\3&1&3\end{bmatrix}$ $=2(-3+2)-3(3+6)-1(4)$$=-2-27-4$
$\neq0$
Hence, the system has only trivial solutions given by x = y = z = 0
View full question & answer→Question 545 Marks
Solve the follwing system of equations by matrix method:
$5x + 2y = 3$
$3x + 2y = 5$
AnswerThe above system can be written in matrix form as: $\begin{bmatrix}5&2\\ 3&2\end{bmatrix}\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\begin{bmatrix}3\\ 2\end{bmatrix}$ Or $\text{AX = B}$
Where, $\text{A}=\begin{bmatrix}5&2\\ 3&2\end{bmatrix},\text{X}=\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix},\text{B}=\begin{bmatrix}3\\ 5\end{bmatrix}$Now, $\text{|A|}=10-6=4\neq0$
So, the above system has a unique solution, given by $\text{X}=\text{A}^{-1}\text{B}$
Let $C_{ij}$ be the co factor of $a_{ij}$ in A, then$\text{C}_{11} = 2,\text{C}_{12} = -3$
$\text{C}_{21} = -2,\text{C}_{22} = 5$
Also, $\text{Adj A}=\begin{bmatrix}2&-3\\ -2&5\end{bmatrix}^\text{T}=\begin{bmatrix}2&-2\\ -3&5\end{bmatrix}$ $\therefore\text{A}^{-1}=\frac{1}{|\text{A}|}\text{adj A}=\frac{1}{4}\begin{bmatrix}2&-2\\ -3&5\end{bmatrix}$
Now,$ X = A^{-1}B$ $=\frac{1}{4}\begin{bmatrix}2&-2\\ -3&5\end{bmatrix}\begin{bmatrix}3\\ 5\end{bmatrix}$ $=\frac{1}{4}\begin{bmatrix}-4\\ 16\end{bmatrix}$ $\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\begin{bmatrix}-1\\ 4\end{bmatrix}$ Hence, x = -1 y = 4
View full question & answer→Question 555 Marks
Show that the following system of linear equations is consistent and also find solution:
$6x + 4y = 2$
$9x + 6y =3$
AnswerHere,
$6\text{x}+4\text{y}=2\ \dots(1)$
$9\text{x}+6\text{y}\ \dots(2)$
AX = B
Here,
$\text{A}=\begin{bmatrix}6&4\\ 9&6\end{bmatrix},\text{X}\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}\text{and }\text{B}=\begin{bmatrix}2\\ 3\end{bmatrix}$
$\begin{bmatrix}6&4\\ 9&6\end{bmatrix}\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\begin{bmatrix}2\\ 3\end{bmatrix}$
$\text{|A|}=\begin{vmatrix}6&4\\ 9&6\end{vmatrix}$
$= 36 -36$
$= 0$
So, A is singular. Thus, the given system of equations is either inconsistent or it is consistent with infinitely many solutions because (adj A)B ≠ 0 or (adj A) = 0.
Let $C _{ ij }$ be the co-factors of the elements $a _{ ij }$ in $A \left[ a _{ ij }\right]$. Then,
$\text{C}_{11}=6,\\ \text{C}_{12}=-9,\\ \text{C}_{21}=-4,\\ \text{C}_{22}=6$
$\text{adj A}=\begin{bmatrix}6&-9\\ -4&6\end{bmatrix}^\text{T}$
$=\begin{bmatrix}6&-4\\ -9&6\end{bmatrix}$
$\text{(adj A) B}=\begin{bmatrix}6&-4\\ -9&6\end{bmatrix}\begin{bmatrix}2\\ 3\end{bmatrix}$
$=\begin{bmatrix}12-12\\ -18+18\end{bmatrix}$
$=\begin{bmatrix}0\\ 0\end{bmatrix}$
If |A| = 0 and (adj A) B = 0, then the system is consistent and has infinitely many solutions.
Thus, AX = B has infinitely many solutions.
Substituting y = k in the eq. (1), we get
$6\text{x}+4\text{k}=2$
$\Rightarrow6\text{x}=2-4\text{k}$
$\Rightarrow\text{x}=\frac{2-\text{4k}}{6}$
$\Rightarrow\text{x}=\frac{1-\text{2k}}{3}$
$\therefore \text{x}=\frac{1-2\text{k}}{3}\text{ and }\text{y}=\text{k}$
These values of x and y satisfy the third equation.
Thus, $\text{x}=\frac{1-2\text{k}}{3}$ and y = k ( where k is a real number ) satisfy the given system of equations.
View full question & answer→Question 565 Marks
Show that the following system of linear equations is consistent and also find solution:
$2x + 3y = 5$
$6x + 9y = 15$
Answer$2\text{x}+3\text{y}=5\dots(1)$
$6\text{x}+9\text{y}=15\dots(2)$
Or , AX = B
Where,
$\text{A}=\begin{bmatrix}2&3\\ 6&9\end{bmatrix},\text{x}=\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}\text{and }\text{B}=\begin{bmatrix}5\\ 15\end{bmatrix}$
$\begin{bmatrix}2&3\\ 6&9\end{bmatrix}\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\begin{bmatrix}5\\ 15\end{bmatrix}$
$\therefore\ |\text{A}|=\begin{vmatrix}2&3\\6&9\end{vmatrix}$
$=18-18$
$=0$
So, A is singular. Thus, the given system of equations is either inconsistent or it is consistent with infinitely many solutions because $(\text{adj A})\text{B}\neq0\text{ or }(\text{adj A})=0$.
$C_{11} = 9, C_{12} = -6, C_{21} = -3$ and $C_{22} = 2$
$\therefore\ \text{adj A}=\begin{bmatrix}9&-6\\-3&2\end{bmatrix}^\text{T}$
$=\begin{bmatrix}2&-3\\-6&9\end{bmatrix}$
$\Rightarrow(\text{adj A})\text{B}=\begin{bmatrix}9&-3\\-6&2\end{bmatrix}\begin{bmatrix}5\\15\end{bmatrix}$
$=\begin{bmatrix}45-45\\-30+30\end{bmatrix}$
$=\begin{bmatrix}0\\0\end{bmatrix}$
If |A| = 0 and (adj A) B = 0, then the system is consistent and has infinitely many solutions.
Thus, AX = B has infinitely many solutions.
Substituting y = k in eq. (1), we get
$2\text{x} + 3\text{k}=5$
$\Rightarrow2\text{x}=5-3\text{k}$
$\Rightarrow\text{x}=\frac{5-3\text{k}}{2}\text{ and }\text{y}=\text{k}$
These values of x and y satisfy the third equation.
Thus, $\Rightarrow\text{x}=\frac{5-3\text{k}}{2}\text{ and }\text{y}=\text{k}$ (where k is a real number) satisfy the given system of equations.
View full question & answer→Question 575 Marks
Solve the following systems of homogeneous linear equations by matrix method:
2x - y + z = 0
3x + 2y - z = 0
x + 4y + 3z = 0
AnswerThe given system of homogeneous equations can be written in matrix form as follows:
$\begin{bmatrix}2&-1&1\\3&2&-1\\1&4&3\end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}$
or, $\text{AX}=\text{O}$
where, $\text{A}=\begin{bmatrix}2&-1&1\\3&2&-1\\1&4&3\end{bmatrix},\text{X}=\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}\text{and }\text{O}=\begin{bmatrix}0\\0\\0\end{bmatrix}$
$|\text{A}|=\begin{vmatrix}2&-1&1\\3&2&-1\\1&4&3\end{vmatrix}$
$=2(6+4)+1(9+1)+1(12-2)$
$=40$
$\therefore\ |\text{A}|\neq0$
So, the given system has only trivial solution, which is given below:
x = y = z = 0
View full question & answer→Question 585 Marks
Show that the following system of linear equation is inconsistent:
$2x + 3y = 5$
$6x + 9y = 10$
AnswerThe given system of equations can be expresesed as follows:
$\text{AX = B}$
Here,
$\text{A}=\begin{bmatrix}2&3\\ 6&9\end{bmatrix},\text{X}=\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}\text{and }\text{B}=\begin{bmatrix}5\\ 10\end{bmatrix}$
Now,
$\text{|A|}=\begin{vmatrix}2&3\\ 6&9\end{vmatrix}$
$={(18-18)}$
$=0$
Let $C _{ ij }$ be the co-factors of the elements $a _{ ij }$ in $A \left[ a _{ ij }\right]$. Then,
$\text{C}_{11}={(-1)}^{1+1}{(9)}=9,\\ \text{C}_{12}={(-1)}^{1+2}{(6)}=-6$
$\text{C}_{21}={(-1)}^{2+1}{(3)}=-3,\\ \text{C}_{22}={(-1)}^{2+2}{(6)}=2$
$\text{adj A}=\begin{bmatrix}9&-6\\ -3&2\end{bmatrix}^\text{T}$
$=\begin{bmatrix}9&-3\\ -6&2\end{bmatrix}$
$\text{(adj A) = B}=\begin{bmatrix}9&-3\\ -6&2\end{bmatrix}\begin{bmatrix}5\\ 10\end{bmatrix}$
$=\begin{bmatrix}45-30\\ -30+30\end{bmatrix}$
$=\begin{bmatrix}15\\ -10\end{bmatrix}\neq0$
Hence, the given system of equations is inconsistent.
View full question & answer→Question 595 Marks
Solve the following system of equations by matrix method:
$5x + 7y + 2 = 0$
$4x + 6y + 3 = 0$
AnswerThe given system of equations can be written in matrix from as follws: $\begin{bmatrix}5&7\\ 4&6\end{bmatrix}\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\begin{bmatrix}-2\\ -3\end{bmatrix}$ $\text{AX = B}$Here,
$\text{A}=\begin{bmatrix}5&7\\ 4&6\end{bmatrix},\text{X=}\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}$ And$\text{ B}=\begin{bmatrix}-2\\ -3\end{bmatrix}$ Now, $|\text{A}|=\begin{bmatrix}5&7\\ 4&6\end{bmatrix}\\ $ $=30-28$ $=2\neq0$ The given system has a unique solution given by $\text{X}=\text{A}^{-1 }\text{B.}$ Let $C _{ ij }$ be the co-factors of the elements $a _{ ij }$ in $A \left[ a _{ ij }\right]$. Then,
$\text{C}_{11}=(-1)^{1+1}(6)=6,\text{C}_{12}=(-1)^{1+2}(4)=-4$ $\text{C}_{21}=(-1)^{2+1}(7)=-7,\text{C}_{22}=(-1)^{2+2}(5)$ $\text{adj}\ \text{A}=\begin{bmatrix}6&-4\\ -7&5\end{bmatrix}=\begin{bmatrix}6&-7\\ -4&5\end{bmatrix}$ $\text{A}^{-1}=\frac{1}{\text{|A|}}\text{ adj}\text{ A}$ $\Rightarrow \text{A}^{-1}=\frac{1}{2}\begin{bmatrix}6&-7\\ -4&5\end{bmatrix}$ $\text{X}=\text{A}^{-1}\text{B}$ $=\frac{1}{2}\begin{bmatrix}6&-7\\ -4&5\end{bmatrix}\begin{bmatrix}-2\\-3\end{bmatrix} $ $=\frac{1}{2}\begin{bmatrix}-12+21\\ 8-15\end{bmatrix} $ $\Rightarrow\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\begin{bmatrix}\frac{9}{2}\\ \frac{-7}{2}\end{bmatrix}$ $\therefore\text{X}=\frac{9}{2}$ And $\text{ y }=\frac{-7}{2}$
View full question & answer→Question 605 Marks
Solve the following for x and y.$\begin{bmatrix}3&-4\\9&2\end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\end{bmatrix}=\begin{bmatrix}10\\2\end{bmatrix}$
AnswerHere,
$\begin{bmatrix}3&-4\\9&2\end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\end{bmatrix}=\begin{bmatrix}10\\2\end{bmatrix}$
$\Rightarrow\begin{bmatrix}3\text{x}-4\text{y}\\9\text{x}-2\text{y}\end{bmatrix}=\begin{bmatrix}10\\2\end{bmatrix}$
$\Rightarrow3 \text{x}-4\text{y}=10\ \dots(1)$
$9\text{x}+2 \text{y}=2\ \dots(2)$
Solving both the equation, we get
$\text{x}=\frac{14}{21}$
$=\frac{2}{3}$
Substituting the value of x in eq. (1), we get
$3\times\frac{2}{3}-4\text{y}=10$
$\Rightarrow2-4\text{y}=10$
$\Rightarrow4 \text{y}=-8$
$\Rightarrow\text{y}=-2$
$\therefore\ \text{x}=\frac{2}{3}\text{ and }\text{y}=-2$
View full question & answer→Question 615 Marks
Solve the following systems of homogeneous linear equations by matrix method:
3x + y - 2z = 0
x + y + z = 0
x - 2y + z = 0
AnswerThe given system of homogeneous equations can be written in matrix form as follows:
$\begin{bmatrix}3&1&-2\\1&1&1\\1&-2&1\end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}$
or, $\text{AX}=\text{O}$
where, $\text{A}=\begin{bmatrix}3&1&-2\\1&1&1\\1&-2&1\end{bmatrix},\text{X}=\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}\text{and }\text{O}=\begin{bmatrix}0\\0\\0\end{bmatrix}$
Now,
$|\text{A}|=\begin{vmatrix}3&1&-2\\1&1&1\\1&-2&1\end{vmatrix}$
$=3(1+2)-1(1-1)-2(-2-1)$
$=9-0+6$
$=15\neq0$
So, the given system has only trivial solution, which is given below:
x = y = z = 0
View full question & answer→Question 625 Marks
Solve the following system of equations by matrix method:
$3x + y = 19$
$3x - y = 23$
AnswerThe given system of equations can be written in matrix form as follows:
$\begin{bmatrix}3&1\\ 3&-1\end{bmatrix}\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\begin{bmatrix}19\\ 23\end{bmatrix}$
AX = B
Here,
$\text{A}=\begin{bmatrix}3&1\\ 3&-1\end{bmatrix},\text{x}=\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}\text{ and }\text{B}=\begin{bmatrix}19\\ 23\end{bmatrix}$
Now,
$\text{|A|}=\begin{vmatrix}3&1\\ 3&-1\end{vmatrix}$
$= - 3 - 3$
$=-6\neq0$
So, the given system has a unique solution given by $X = A^{-1}B.$
Let $C _{ ij }$ be the co-factors of the elements $a _{ ij }$ in $A \left[ a _{ ij }\right]$. Then,
$\text{C}_{11}=(-1)^{1+1}(-1)=-1,\text{C}_{12}=(-1)^{1+2}(3)=-3$
$\text{C}_{21}=(-1)^{2+1}(1)=-1,\text{C}_{22}=(-1)^{2+2}(3)=3$
$\text{adj A}=\begin{bmatrix}-1&-3\\ -1&3\end{bmatrix}^\text{T}$
$=\begin{bmatrix}-1&-1\\ -3&3\end{bmatrix}$
View full question & answer→Question 635 Marks
Show that the following system of linear equations is consistent and also find solution:
$2x + 2y − 2z = 1$
$4x + 4y − z = 2$
$6x + 6y + 2z = 3$
AnswerThis system can be written as: $\begin{bmatrix}2&2&-2\\ 4&4&-1\\ 6&6&2\end{bmatrix}\begin{bmatrix}\text{x}\\ \text{y}\\ \text{z}\end{bmatrix}=\begin{bmatrix}1\\ 2\\ 3\end{bmatrix}$ or $\text{AX = B}$ $\text{|A|}=2{(14)}-2(14)-2{(0)}=0$ So, A is singular and the system has either no solution or infinite solutions according as $\text{(Adj A)}\times\text{(B)}\neq0$ or $\text{(Adj A)}\times\text{(B)}=0$
Let $C _{ ij }$ be the co-factors of the elements $a _{ ij }$ in $A \left[ a _{ ij }\right]$. Then,
A $\text{C}_{11}=14\\ \text{C}_{21}=-16\\ \text{C}_{31}=6$ $\text{C}_{12}=-14\\ \text{C}_{22}=16\\ \text{C}_{32}=-6$ $\text{C}_{1}=0\\ \text{C}_{23}=0\\ \text{C}_{33}=0$ $\text{adj A}=\begin{bmatrix}14&-14&0\\ -16&16&0\\ 0&0&0\end{bmatrix}^\text{T}=\begin{bmatrix}14&-16&6\\ -14&16&-6\\ 0&0&0\end{bmatrix}$ $(\text{adj A})\times\text{B}=\begin{bmatrix}14&-16&0\\ -14&16&-6\\ 0&0&0\end{bmatrix}\begin{bmatrix}1\\ 2\\ 3\end{bmatrix}=\begin{bmatrix}14-32+18\\ -14+32-18\\ 0+0+0\end{bmatrix}=\begin{bmatrix}0\\ 0\\ 0\end{bmatrix}$ So, $\text{AX}=\text{B}$ has infinite solutions. Now, let z = k So, 2x + 2y = 1 + 2k 4x + 4y = 2 + k which can be written as: $\begin{bmatrix}2&2\\ 4&4\end{bmatrix}\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\begin{bmatrix}1+2\text{k}\\ 2+\text{k}\end{bmatrix}$ or $\text{AX = B}$|A| = 0, z = 0
Again, $2\text{x}+2\text{y}=1$ $4\text{x}+4\text{y}=2$ Let $\text{y = k}$ $2\text{x}=1-2\text{k}$ $\text{x}=\frac{1}{2}-\text{k}$ Hence, $\text{x}=\frac{1}{2}-\text{k}$ $\text{y}=\text{k}$ $\text{z}=0$
View full question & answer→Question 645 Marks
Solve the following system of equations by matrix method:
$3x + 7y = 4$
$x + 2y = -1$
AnswerThe above system can be written in matrix form as:$\begin{bmatrix}3&7\\ 1&2\end{bmatrix}\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\begin{bmatrix}4\\ -1\end{bmatrix}$
Or AX = BWhere $\text{A}=\begin{bmatrix}3&7\\ 1&2\end{bmatrix},\text{x}=\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}\text{ and }\text{B}=\begin{bmatrix}4\\ -1\end{bmatrix}$
Now,$\text{|A|}=-1\neq0$
So, the above system has a unique solution, given by $X=A^{-1} B$ Now, let $C_{i j}$ be the co-factors of $a_{i j}$ in $A$ $\text{C}_{11} = 2,\text{C}_{12} = -1$$\text{C}_{21} = -7,\text{C}_{22} = 3$
$\text{Adj A}=\begin{bmatrix}2&-1\\ -7&3\end{bmatrix}^\text{T}=\begin{bmatrix}2&-7\\ -1&3\end{bmatrix}$
$\therefore\text{A}^{-1}=\frac{1}{\text{|A|}}.\text{adj A}=\frac{1}{(-1)}\begin{bmatrix}2&-7\\ -1&3\end{bmatrix}$
Now, $X = A^{-1}$ B$\Rightarrow\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\frac{1}{-1}\begin{bmatrix}2&-7\\ -1&3\end{bmatrix}\begin{bmatrix}4\\ -1\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\frac{1}{-1}\begin{bmatrix}15\\ -7\end{bmatrix}=\begin{bmatrix}-15\\ 7\end{bmatrix}$
Hence, x = -15 y = 7
View full question & answer→