4 Marks - Maths STD 12 Science Questions - Vidyadip

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Question 14 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.

Answer

Suppose, $\text{A}=\begin{bmatrix}1&-1&2\\0&2&-3\\3&-2&4\end{bmatrix}\text{B}=\begin{bmatrix}-2&0&1\\9&2&-3\\6&1&-2\end{bmatrix}$
$\text{A}\times\text{B}=\begin{bmatrix}1&-1&2\\0&2&-3\\3&-2&4\end{bmatrix}\begin{bmatrix}-2&0&1\\9&2&-3\\6&1&-2\end{bmatrix}$
$=\begin{bmatrix}-2-9+12&0-2+2&1+3-4\\0+18-18&0+4-3&0-6+6\\-6-18+24&0-4+4&3+6-8\end{bmatrix}$
$=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
Since, A × 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.

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Question 24 Marks

Solve the following system of equations by matrix method:

$\frac{2}{\text{x}}+\frac{3}{\text{y}}+\frac{10}{\text{z}}=4,\frac{4}{\text{x}}-\frac{6}{\text{y}}+\frac{5}{\text{z}}=1,\frac{6}{\text{x}}+\frac{9}{\text{y}}-\frac{20}{\text{z}}=2:\text{x},\text{y},\text{z}\neq0$

Answer

Let
$\frac{1}{\text{x}}=\text{u},\frac{1}{\text{y}}=\text{v},\frac{1}{\text{z}}=\text{w}$
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_ijbe the co-factors of a_ij in A
$\text{C}_{11}=75\\ \text{C}_{21}=150\\ \text{C}_{31}=75$
$\text{C}_{12}=110\\ \text{C}_{22}=-100\\ \text{C}_{32}=30$
$\text{C}_{13}=72\\ \text{C}_{23}=0\\ \text{C}_{33}=-24$
$\text{adj A}=\begin{bmatrix}75&110&72\\ 150&-100&0\\ 75&30&-24\end{bmatrix}^\text{T}=\begin{bmatrix}75&150&75\\ 110&-100&30\\ 72&0&-24\end{bmatrix}$
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

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Question 34 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.

Answer

Let 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^-1B
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

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Question 44 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.

Answer

Let 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 = [a_ij]. Then,
$\text{C}_{11}=(-1)^{1+1}\begin{vmatrix}2&1\\1&1\end{vmatrix}=1,$ $\text{C}_{12}=(-1)^{1+2}\begin{vmatrix}1&1\\5&1\end{vmatrix}=4,$ $\text{C}_{13}=(-1)^{1+3}\begin{vmatrix}1&2\\5&1\end{vmatrix}=-9$
$\text{C}_{21}=(-1)^{2+1}\begin{vmatrix}1&1\\1&1\end{vmatrix}=0,$ $\text{C}_{22}=(-1)^{2+2}\begin{vmatrix}1&1\\5&1\end{vmatrix}=-4,$ $\text{C}_{23}=(-1)^{2+3}\begin{vmatrix}1&1\\5&1\end{vmatrix}=4$
$\text{C}_{31}=(-1)^{3+1}\begin{vmatrix}1&1\\2&1\end{vmatrix}=-1,$ $\text{C}_{32}=(-1)^{3+2}\begin{vmatrix}1&1\\1&1\end{vmatrix}=0,$ $\text{C}_{33}=(-1)^{3+3}\begin{vmatrix}1&1\\1&2\end{vmatrix}=1$
$\text{adj }\text{A}=\begin{bmatrix}1&4&-9\\0&-4&4\\-1&0&1\end{bmatrix}^\text{T}$
$=\begin{bmatrix}1&0&-1\\4&-4&0\\-9&4&1\end{bmatrix}$
$\Rightarrow\text{A}^{-1}=\frac{1}{|\text{A}|}\text{adj }\text{A}$
$=\frac{1}{-4}\begin{bmatrix}1&0&-1\\4&-4&0\\-9&4&1\end{bmatrix}$

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Question 54 Marks

Solve the following system of equations by matrix method:

$\frac{2}{\text{x}}-\frac{3}{\text{y}}+\frac{3}{\text{z}}=10$

$\frac{1}{\text{x}}+\frac{1}{\text{y}}+\frac{1}{\text{z}}=10$

$\frac{3}{\text{x}}-\frac{1}{\text{y}}+\frac{2}{\text{z}}=13$

Answer

Let $\frac{1}{\text{x}}=\text{u},\frac{1}{\text{y}}=\text{v},\frac{1}{\text{z}}=\text{w}$
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, $\text{x}=\frac{1}{2},\text{y}=\frac{1}{3},\text{z}=\frac{1}{5}$

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Question 64 Marks

Solve the following system of equations by matrix method:

x - y + 2z = 7

3x + 4y - 5z = -5

2x - y + 3z = 12

Answer

The 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_ijbe the co-factors of a_ijin 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

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Question 74 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

Answer

Here,
$\text{A}=\begin{bmatrix}2&2&-4\\-4&2&-4\\2&-1&5\end{bmatrix}\text{and }\text{B}=\begin{bmatrix}1&-1&0\\2&3&4\\0&1&2\end{bmatrix}$
$\text{BA}=\begin{bmatrix}1&-1&0\\2&3&4\\0&1&2\end{bmatrix}\begin{bmatrix}2&2&-4\\-4&2&-4\\2&-1&5\end{bmatrix}$
$\Rightarrow\text{BA}=\begin{bmatrix}2+4+0&2-2+0&-4+4+0\\4-12+8&4+6-4&-8-12+20\\0-4+4&0+2-2&0-4+10\end{bmatrix}$
$=\begin{bmatrix}6&0&0\\0&6&0\\0&0&6\end{bmatrix}$
$\Rightarrow\text{BA}=6\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
$\Rightarrow\text{BA}=6\text{I}_3$
$\Rightarrow\text{B}\Big(\frac{1}{6}\text{A}\Big)=\text{I}_3$
$\Rightarrow\text{B}^{-1}=\frac{1}{6}\text{A}$
$\Rightarrow\text{B}^{-1}=\frac{1}{6}\begin{bmatrix}2&2&-4\\-4&2&-4\\2&-1&5\end{bmatrix}$
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^-1C
$\Rightarrow\text{X}=\frac{1}{6}\begin{bmatrix}2&2&-4\\-4&2&-4\\2&-1&5\end{bmatrix}\begin{bmatrix}3\\17\\7\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\frac{1}{6}\begin{bmatrix}6+34-28\\-12+34-28\\6-17+35\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\frac{1}{6}\begin{bmatrix}12\\-6\\24\end{bmatrix}$
$\therefore$ x = 2, y = -1 and z = 4.

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Question 84 Marks

Solve the following for x and y.

$\begin{bmatrix}3&-4\\9&2\end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\end{bmatrix}=\begin{bmatrix}10\\2\end{bmatrix}$

Answer

Here,
$\begin{bmatrix}3&-4\\9&2\end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\end{bmatrix}=\begin{bmatrix}10\\2\end{bmatrix}$
$\Rightarrow\begin{bmatrix}3\text{x}-4\text{y}\\9\text{x}-2\text{y}\end{bmatrix}=\begin{bmatrix}10\\2\end{bmatrix}$
$\Rightarrow3 \text{x}-4\text{y}=10\ \dots(1)$
$9\text{x}+2 \text{y}=2\ \dots(2)$
Solving both the equation, we get
$\text{x}=\frac{14}{21}$
$=\frac{2}{3}$
Substituting the value of x in eq. (1), we get
$3\times\frac{2}{3}-4\text{y}=10$
$\Rightarrow2-4\text{y}=10$
$\Rightarrow4 \text{y}=-8$
$\Rightarrow\text{y}=-2$
$\therefore\ \text{x}=\frac{2}{3}\text{ and }\text{y}=-2$

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Question 94 Marks

Solve the following system of equations by matrix method:
3x + 4y + 7z = 14
2x - y + 3z = 4
x + 2y - 3z = 0

Answer

The 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^-1B.
Let c_ijbe the co-factors of the elements a_ijin $\text{A}=\text{[a}_\text{ij}]$. Then,
$\text{C}_{11}{(-1)}^{1+1}{(-1)}=-1,\text{C}_{12}={(-1)}^{1+2}{(3)}=-3$
$\text{C}_{21}={(-1)}^{2+1}{(1)}=-1,\text{C}_{22}={(-1)}^{2+2}{(3)}=3$
$\text{adj}\ \text{A}=\begin{bmatrix}-1&-3\\ -1&3\end{bmatrix}^\text{T}$
$=\begin{bmatrix}-1&-1\\ -3&3\end{bmatrix}$
$\text{A}^{-1}=\frac{1}{\text{|A|}}\text{adj A}$
$=\frac{1}{-6}\begin{bmatrix}-1&-1\\ -3&3\end{bmatrix}$
$\text{X}=\text{A}^{-1}\text{B}$
$=\frac{1}{-6}\begin{bmatrix}-1&-1\\ -3&3\end{bmatrix}\begin{bmatrix}19\\ 23\end{bmatrix}$
$=\frac{1}{-6}\begin{bmatrix}-19-23\\ -57+69\end{bmatrix}$
$=\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\begin{bmatrix}\frac{-42}{-6}\\ \frac{12}{-6}\end{bmatrix}$

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Question 104 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.

Answer

x, 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.

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Question 114 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

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Question 124 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

Answer

This 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

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Question 134 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

Answer

Here,
$\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

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Question 144 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?

Answer

Let 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.

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Question 154 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

Answer

Here,
$\text{A}=\begin{bmatrix}2&-3&5\\ 3&2&-4\\ 1&1&-2\end{bmatrix}$
$\text{|A|}=\begin{bmatrix}2&-3&5\\ 3&2&-4\\ 1&1&-2\end{bmatrix}$
$=2{(-4+4)}+3{(-6+4)}+5{(3-2)}$
$=0-6+5$
$=-1$
Let C_ijbe the co-factors of the elements a_ijin 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

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Question 164 Marks

Solve the following system of equations by matrix method:
5x +3y + z = 16
2x + y +3z = 19
x + 2y + 4z = 25

Answer

Here,
$\text{A}=\begin{bmatrix}5&3&1\\ 2&1&3\\ 1&2&4\end{bmatrix}$
$\text{|A|}=\begin{vmatrix}5&3&1\\ 2&1&3\\ 1&2&4\end{vmatrix}$
$=5{(4-6)}-3{(8-3)}+1{(4-1)}$
$=-10-15+3$
$=-22$
Let C_ijbe 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

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Question 174 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?

Answer

Let 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.

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Question 184 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

Answer

This 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_ijbe the co-factors of a_ijin 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 = k

Then, 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, $\text{x}=\frac{7-16\text{k}}{11},\text{y}=\frac{\text{k}+3}{11},\text{z}=\text{k}$

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Question 194 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.

Answer

From 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.

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Question 204 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.

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Question 214 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, $\text{x}=\frac{-5\text{k}}{11},\ \text{y}=\frac{12\text{k}}{11},\ \text{z}=\text{k}$

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Question 224 Marks

Solve the following system of equations by matrix method:

8x + 4y + 3z = 18

2x + y + z = 5

x + 2y + z = 5

Answer

The 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

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Question 234 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.

Answer

Let 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 elements a_ij in A = [a_ij]. Then,
$\text{C}_{11}=(-1)^{1+1}17=17,\text{C}_{12}=(-1)^{1+2}26=-26$
$\text{C}_{21}=(-1)^{2+1}1=-1,\text{C}_{22}=(-1)^{2+2}2=2$
$\text{adj }\text{A}=\begin{bmatrix}17&-26\\-1&2\end{bmatrix}^\text{T}$
$=\begin{bmatrix}17&-1\\-26&2\end{bmatrix}$
$\Rightarrow\text{A}^{-1}=\frac{1}{|\text{A}|}\text{adj }\text{A}$
$=\frac{1}{8}\begin{bmatrix}17&-1\\-26&2\end{bmatrix}$
$\text{X}=\text{A}^{-1}\text{B}$
$\Rightarrow\begin{bmatrix}\text{x}\\\text{y}\end{bmatrix}=\frac{1}{8}\begin{bmatrix}17&-1\\-26&2\end{bmatrix}\begin{bmatrix}7000\\110000\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\\text{y}\end{bmatrix}=\frac{1}{8}\begin{bmatrix}119000-110000\\-182000+220000\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\\text{y}\end{bmatrix}=\frac{1}{8}\begin{bmatrix}9000\\38000\end{bmatrix}$
$\Rightarrow\text{x}=\frac{9000}{8}\text{and }\text{y}=\frac{38000}{8}$
$\therefore\ \text{x}=1125\text{ and }\text{y}=4750.$

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Question 244 Marks

Solve the following system of equations by matrix method:
x + y + z = 6
x + 2z = 7
3x + y + z = 12

Answer

Here,
$\text{A}=\begin{bmatrix}1&1&1\\ 1&0&2\\ 3&1&1\end{bmatrix}$
$\text{|A|}=\begin{vmatrix}1&1&1\\ 1&0&2\\ 3&1&1\end{vmatrix}$
$=1{(0-2)}-1{(1-6)}+1{(1-0)}$
$=-2+5+1$
$=4$
Let C_ijbe the co-factors of the elements a_ij in A [a_ij]. 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

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Question 254 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

Answer

Here,

$\text{A}=\begin{bmatrix}1&2&0\\-2&-1&-2\\0&-1&1\end{bmatrix}$

$|\text{A}| = 1(-1-2)+2(2)$

$= -3 + 4$

$= 1$

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^-1B

$\Rightarrow\text{X}=\begin{bmatrix}-3&2&2\\-2&1&1\\-4&2&3\end{bmatrix}\begin{bmatrix}10\\8\\7\end{bmatrix}$

$\Rightarrow\text{X}=\begin{bmatrix}-30+16+14\\-20+8+7\\-40+16+21\end{bmatrix}$

$\Rightarrow\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}0\\-5\\-3\end{bmatrix}$

$\therefore$ x = 0, y = -5, and z = -3.

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Question 264 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

Answer

Here,
$\text{A}=\begin{bmatrix}3&-4&2\\ 2&3&5\\ 1&0&1\end{bmatrix}$
$\text{|A|}=3{(3-0)}+4{(2-5)}+2{(0-3)}$
$=9-12-6$
$=-9$
Let C_ijbe the co-factors of the elements a_ij in A = [a_ij]. 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

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Question 274 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.

Answer

We 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[a_ij]. 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.

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Question 284 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

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Question 294 Marks

Show that the following system of linear equation is inconsistent:
3x − y − 2z = 2
2y − z = −1
3x − 5y = 3

Answer

The 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 [a_ij]. 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.

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Question 304 Marks

Solve the follwing system of equations by matrix method:
x + y + z = 3
2x - y + z = -1
2x + y - 3z = -9

Answer

The 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^-1B
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, $\text{x}=\frac{-8}{7},\text{y}=\frac{10}{7},\text{z}=\frac{19}{7}$

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Question 314 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^-1B
$\begin{matrix}\text{C}_{11}=6&\text{C}_{21}=-28&\text{C}_{31}=-16\\\text{C}_{12}=-3&\text{C}_{22}=16&\text{C}_{32}=9\\\text{C}_{13}=-2&\text{C}_{23}=10&\text{C}_{33}=6\end{matrix}$
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

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Question 324 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.

Answer

x, 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.

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Question 334 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.

Answer

Let 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 = [a_ij]. 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.

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Question 344 Marks

Solve the following system of equations by matrix method:

x - y + z = 2

2x - y = 0

2y - z = 1

Answer

The 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_ijbe 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

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Question 354 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.

Answer

Let 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.

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Question 364 Marks

Solve the following systems of homogeneous linear equations by matrix method:

x + y - 6z = 0

x - y + 2z = 0

-3x + y + 2z = 0

Answer

The 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.

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Question 374 Marks

Solve the following system of equations by matrix method:

x + y - z = 3

2x + 3y + z = 10

3x - y -7z = 1

Answer

Here,
$\text{A}=\begin{bmatrix}1&1&-1\\ 2&1&1\\ 3&-1&-7\end{bmatrix}$
$\text{|A|}=\begin{vmatrix}1&1&-1\\ 2&3&1\\ 3&-1&-7\end{vmatrix}$
$= 1 (-21 + 1) -1 (-14 - 3 ) - 1 (-2 - 9)$
$= -20 + 17 + 11$
$= 8$
$\text{C}_{11}=(-1)^{1+1}\begin{vmatrix}3&1\\ -1&-7\end{vmatrix}=-20,\\ \text{C}_{12}=(-1)^{1+2}\begin{vmatrix}2&1\\ 3&-7\end{vmatrix}=17,\\ \text{C}_{13}=(-1)^{1+3}\begin{vmatrix}2&3\\ 3&-1\end{vmatrix}=-11$
$\text{C}_{21}=(-1)^{2+1}\begin{vmatrix}1&-1\\ -1&-7\end{vmatrix}=8,\\ \text{C}_{22}=(-1)^{2+2}\begin{vmatrix}1&-1\\ 3&-7\end{vmatrix}=-4,\\ \text{C}_{23}(-1)^{2+3}\begin{vmatrix}1&1\\ 3&-1\end{vmatrix}=4$
$\text{C}_{31}=(-1)^{3+1}\begin{vmatrix}1&-1\\ 3&1\end{vmatrix}=4,\\ \text{C}_{32}=(-1)^{3+2}\begin{vmatrix}1&-1\\ 2&1\end{vmatrix}=-3,\\ \text{C}_{33}(-1)^{3+3}\begin{vmatrix}1&1\\ 2&3\end{vmatrix}=1$
$\text{C}_{31}=(-1)^{3+1}\begin{vmatrix}1&-1\\ 3&1\end{vmatrix}=4,\\ \text{C}_{32}=(-1)^{3+2}\begin{vmatrix}1&-1\\ 2&1\end{vmatrix}=-3,\\ \text{C}_{33}(-1)^{3+3}\begin{vmatrix}1&1\\ 2&3\end{vmatrix}=1$
$\text{adj A}=\begin{bmatrix}-20&17&-11\\ 8&-4&4\\ 4&-3&1\end{bmatrix}^\text{T}$
$=\begin{bmatrix}-20&8&4\\ 17&-4&-3\\ -11&4&1\end{bmatrix}$
$\Rightarrow\text{A}^{-1}=\frac{1}{\text{|A|}}\text{adj A}$
$=\frac{1}{8}\begin{bmatrix}-20&8&4\\ 17&-4&-3\\ -11&4&1\end{bmatrix}$
$\text{X}=\text{A}^{-1}\text{B}$
$\Rightarrow\begin{bmatrix}\text{x}\\ \text{y}\\ \text{z}\end{bmatrix}=\frac{1}{8}\begin{bmatrix}-20&8&4\\ 17&-4&-3\\ -11&4&1\end{bmatrix}\begin{bmatrix}3\\ 10\\ 1\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\ \text{y}\\ \text{z}\end{bmatrix}=\frac{1}{8}\begin{bmatrix}-60+80+4\\ 51-40-3\\ -33+40+1\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\ \text{y}\\ \text{z}\end{bmatrix}=\frac{1}{8}\begin{bmatrix}24\\ 8\\ 8\end{bmatrix}$
$\Rightarrow\text{x}=\frac{24}{8},\text{y}=\frac{8}{8}\ \text{and z}=\frac{8}{8}$
$\therefore\text{x}=3,\text{y}=1\text{ and }\text{z}=1$

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Question 384 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

Answer

Here,

$\text{A}=\begin{bmatrix}1&-2&0\\ 2&1&3\\ 0&-2&1\end{bmatrix}$

$\text{|A|}=1{(1+6)}+2{(2-0)}+0{(-4-0)}$

$=7+4+0$

$=11$

Let C_ijbe 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

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Question 394 Marks

Solve the following systems of homogeneous linear equations by matrix method:

x + y + z = 0

x - y - 5z = 0

x + 2y + 4z = 0

Answer

x + 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

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Question 404 Marks

Solve the following system of equations by matrix method:
2x + 6y = 2
3x - z = -8
2x - y + z = -3

Answer

Here,
$\text{A}=\begin{bmatrix}2&6&0\\ 3&0&-1\\ 2&-1&1\end{bmatrix}$
$\text{|A|}=\begin{vmatrix}2&6&0\\ 3&0&-1\\ 2&-1&1\end{vmatrix}$
$=2{(0-1)}-1{(3+2)}+0{(-3+0)}$
$=-2-30$
$=-32$
Let C_ij be the co-factors of the elemennts a_ij in A [a_ij]. 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

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Question 414 Marks

Show that the following system of linear equation is inconsistent:
4x − 5y − 2z = 2
5x − 4y + 2z = −2
2x + 2y + 8z = −1

Answer

The 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.

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Question 424 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

Answer

Here,
$\text{x}-\text{y}+\text{z}=3\ \dots(1)$
$2\text{x}+\text{y}-\text{z}=2\ \dots(2)$
$-\text{x}-2\text{y}+2\text{z}=1 \ \dots(3)$
Or, $\text{AX}=\text{B}$
Where,
$\text{A}=\begin{bmatrix}1&-1&1\\ 2&1&-1\\ -1&-2&2\end{bmatrix},\text{X}=\begin{bmatrix}\text{x}\\ \text{y}\\ \text{z}\end{bmatrix}\text{and }\text{B}=\begin{bmatrix}3\\ 2\\ 1\end{bmatrix}$
$\begin{bmatrix}1&-1&1\\ 2&1&-1\\ -1&-2&2\end{bmatrix}\begin{bmatrix}\text{x}\\ \text{y}\\ \text{z}\end{bmatrix}=\begin{bmatrix}3\\ 2\\ 1\end{bmatrix}$
$\text{|A|}=\begin{vmatrix}1&-1&1\\ 2&1&-1\\ -1&-2&2\end{vmatrix}$

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Question 434 Marks

Solve the following system of equations by matrix method:
6x - 12y + 25z = 4
4x + 15y - 20z = 3
2x + 18y + 15z = 10

Answer

Here,
$\text{A}=\begin{bmatrix}6&-12&25\\ 4&15&-20\\ 2&18&15\end{bmatrix}$
$\text{|A|}=\begin{vmatrix}6&-12&25\\ 4&15&-20\\ 2&18&15\end{vmatrix}$
$=6(225+360)+12(60+40)+25(72-30)$
$= 3510 + 1200 + 1050$
$= 5760$
Let C_ij be the co-factors of the elements a_ij in A [a_ij]. Then,
$\text{C}_{11}={(-1)}^{1+1}\begin{vmatrix}15&-20\\ 18&15\end{vmatrix}=585,\\ \text{C}_{12}={(-1)}^{1+2}\begin{vmatrix}4&-20\\ 2&15\end{vmatrix}=-100,\\ \text{C}_{13}={(-1)}^{1+3} \begin{vmatrix}4&15\\ 2&18\end{vmatrix}=42$
$\text{C}_{21}={(-1)}^{2+1}\begin{vmatrix}-12&-25\\ 18&15\end{vmatrix}=630,\\ \text{C}_{22}={(-1)}^{2+2}\begin{vmatrix}6&-25\\ 2&15\end{vmatrix}=-40,\\ \text{C}_{23}={(-1)}^{2+3} \begin{vmatrix}6&-12\\ 2&18\end{vmatrix}=132$
$\text{C}_{31}={(-1)}^{3+1}\begin{vmatrix}-12&25\\ 15&-20\end{vmatrix}=-135,\\ \text{C}_{32}={(-1)}^{3+2}\begin{vmatrix}6&-25\\ 4&-20\end{vmatrix}=220,\\ \text{C}_{33}={(-1)}^{3+3} \begin{vmatrix}6&-12\\ 4&15\end{vmatrix}=138$
$\text{adj A}\begin{bmatrix}585&-100&42\\ 630&40&-132\\ -135&220&138\end{bmatrix}^\text{T}$
$=\begin{bmatrix}585&630&-135\\ -100&40&220\\ 42&-132&138\end{bmatrix}$
$\Rightarrow\text{A}^{-1}\frac{1}{\text{|A|}}\text{adj}\ \text{A}$
$=\frac{1}{5760}\begin{bmatrix}585&630&-135\\ -100&40&220\\ 42&-132&138\end{bmatrix}$
$\text{X}=\text{A}^{-1}\text{B}$
$\Rightarrow\begin{bmatrix}\text{x}\\ \text{y}\\ \text{z}\end{bmatrix}=\frac{1}{5760}\begin{bmatrix}585&630&-135\\ -100&40&220\\ 42&-132&138\end{bmatrix}\begin{bmatrix}4\\ 3\\ 10\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\ \text{y}\\ \text{z}\end{bmatrix}=\frac{1}{5760}\begin{bmatrix}2340+1890-1350\\ -400+120+2200\\ 168-396+1380\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\ \text{y}\\ \text{z}\end{bmatrix}=\frac{1}{5760}\begin{bmatrix}2880\\ 1920\\ 1152\end{bmatrix}$
$\Rightarrow\text{x}=\frac{2880}{5760},\text{y}=\frac{1920}{5760}\text{ and }\text{z}=\frac{1152}{5760}$
$\therefore\text{x}=\frac{1}{2},\text{y}=\frac{1}{3}\text{ and }\text{z}=\frac{1}{5}$

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Question 444 Marks

The prices of three commodities P, Q and R are Rs. 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.

Answer

The 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 = [a_ij]. 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.

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Question 454 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

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Question 464 Marks

Show that the following system of linear equation is inconsistent:
x + y − 2z = 5
x − 2y + z = −2
−2x + y + z = 4

Answer

The 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.

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Question 474 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.

Answer

As 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

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Question 484 Marks

Solve the following system of equations by matrix method:

3x + y = 7

5x + 3y = 12

Answer

The 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 = [a_ij]. 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^-1B
$=\frac{1}{4}\begin{bmatrix}3&-1\\ -5&3\end{bmatrix}\begin{bmatrix}7\\ 12\end{bmatrix}$
$=\frac{1}{4}\begin{bmatrix}21-12\\ -35+36\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\begin{bmatrix}\frac{9}{4}\\ \frac{1}{4}\end{bmatrix}$
$\therefore\ \text{x}=\frac{9}{4}\text{ and }\text{y}=\frac{1}{4}$

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Question 494 Marks

Solve the follwing system of equations by matrix method:
3x + 4y - 5 = 0
x - y + 3 = 0

Answer

The 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_ijbe the cofactors of the elements a_ij in A = [a_ij]. 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

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Question 504 Marks

Solve the following systems of homogeneous linear equations by matrix method:

2x + 3y - z = 0

x - y - 2z = 0

3x + y + 3z = 0

Answer

2x + 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

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