Question 2015 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, $\ce{AX = B}$ 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, $\text{x}=\frac{7-16\text{k}}{11},\text{y}=\frac{\text{k}+3}{11},\text{z}=\text{k}$
View full question & answer→Question 2025 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 2035 Marks
If $\text{A}=\begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix},$ find $A^{-1}$ and prove that $A^2 - 4A - 5I = 0.$
Answer$\text{A}=\begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix}$
$\Rightarrow\ |\text{A}|=\text{A}=\begin{vmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{vmatrix}$
$=1(1-4)-2(2-4)+2(4-2)$
$=-3+4+4=5$
Since, $|\text{A}|\neq0$
Hence, A is invertible.
Now, $\text{A}^2=\begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix}\begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix}$
$=\begin{bmatrix} 1+4+4 & 2+2+4 & 2+4+2 \\ 2+2+4 & 4+1+4 & 4+2+2 \\ 2+4+2 & 4+2+2 & 1+4+4 \end{bmatrix}$
$=\begin{bmatrix} 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 \end{bmatrix}$
Now, $\text{A}^2-4\text{A}-5\text{I}=\begin{bmatrix} 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 \end{bmatrix}$
$=-4\begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix}-5\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$
$=\begin{bmatrix} 9-4-5 & 8-8-0 & 8-8-0 \\ 8-8-0 & 9-4-5 & 8-8-0 \\ 8-8-0 & 8-8-0 & 9-4-5 \end{bmatrix}$
$=\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}=0$
$\Rightarrow\ \text{A}^2=4\text{A}-5\text{I}=0\ \text{[Proved]}$
Again, $A^2 - 4A - 5I = 0$
$\Rightarrow A^{-1} (A^2 - 4A - 5I) = A^{-10} [$Pre$-$multiplying with $A^{-1}]$
$\Rightarrow A^{-1}A^2 - 4A^{-1}A - 5A^{-1} = 0$
$\Rightarrow A - 4I = 5A^{-1}$
$\Rightarrow\ 5\text{A}^{-1}=\begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix}-4\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$
$=\begin{bmatrix} 1-4 & 2-0 & 2-0 \\ 2-0 & 1-4 & 2-0 \\ 2-0 & 2-0 & 1-4 \end{bmatrix}=\begin{bmatrix} -3 & 2 & 2 \\ 2 & -3 & 2 \\ 2 & 2 & -3 \end{bmatrix}$
$\Rightarrow\ \text{A}^{-1}=\frac{1}{5}\begin{bmatrix} -3 & 2 & 2 \\ 2 & -3 & 2 \\ 2 & 2 & -3 \end{bmatrix}$
View full question & answer→Question 2045 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 2055 Marks
$\text{If A} = \begin{bmatrix}2&-1&1\\-1&2&-1\\1&-1&2\end{bmatrix} $ Verify that $A^3 - 6A^{2 }+ 9A + 4I = 0$ and hence find $A^{-1}.$
Answer$\text{Given:}\ \text{A}=\begin{bmatrix}2&-1&1\\-1&2&-1\\1&-1&2\end{bmatrix}$
$\therefore\ \text{A}^2=\begin{bmatrix}2&-1&1\\-1&2&-1\\1&-1&2\end{bmatrix}\begin{bmatrix}2&-1&1\\-1&2&-1\\1&-1&2\end{bmatrix}$
$\Rightarrow\ \text{A}^{2}=\begin{bmatrix}4+1+1&-2-2-1&2+1+2\\-2-2-1&1+4+1&-1-2-2\\2+1+2&-1-2-2&1+1+4\end{bmatrix}=\begin{bmatrix}6&-5&5\\-5&6&-5\\5&-5&6\end{bmatrix}$
$\text{Now}\ \text{A}^3=\text{A}^2\text{A}= \begin{bmatrix}6&-5&5\\-5&6&-5\\5&-5&6\end{bmatrix}\begin{bmatrix}2&-1&1\\-1&2&-1\\1&-1&2\end{bmatrix}$
$=\begin{bmatrix}12+5+5&-6-10-5&6+5+10\\-10-6-5&5+12+5&-5-6-10\\10+5+6&-5-10-6&5+5+12\end{bmatrix}=\begin{bmatrix}22&-21&21\\-21&22&-21\\21&-21&22\end{bmatrix}$
$\text{L.H.S.= A}^3-6\text{A}^2+9\text{A}-4\text{I}$
$=\begin{bmatrix}22&-21&21\\-21&22&-21\\21&-21&22\end{bmatrix}-6\begin{bmatrix}6&-5&5\\-5&6&-5\\5&-5&6\end{bmatrix}+9\begin{bmatrix}2&-1&1\\-1&2&-1\\1&-1&2\end{bmatrix}-4\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
$=\begin{bmatrix}22&-21&21\\-21&22&-21\\21&-21&22\end{bmatrix}-\begin{bmatrix}36&-30&30\\-30&36&-30\\30&-30&36\end{bmatrix}+\begin{bmatrix}18&-9&9\\-9&18&-9\\9&-9&18\end{bmatrix}-\begin{bmatrix}4&0&0\\0&4&0\\0&0&4\end{bmatrix}$
$=\begin{bmatrix}22-36&-21+30&21-30\\-21+30&22-36&-21+30\\21-30&-21+30&22-36\end{bmatrix}+\begin{bmatrix}18-4&-9-0&9-0\\-9-0&18-4&-9-0\\9-0&-9-0&18-4\end{bmatrix}$
$=\begin{bmatrix}-14&9&-9\\9&-14&9\\-9&9&-14\end{bmatrix}+\begin{bmatrix}14&-9&9\\-9&14&-9\\9&-9&14\end{bmatrix}=\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}=0$
$=\text{R.H.S.}$
Now, to find $A^{-1},$ multiplying $A^3 - 6A^{2 }+ 9A - 4I$ by $A^{-1}$
$\Rightarrow A^3A^{-1 }- 6A^2A^{-1} + 9AA^{-1} - 4I.A^{-1} = 0.A^{-1}$
$\Rightarrow A^2 - 6A + 9I - 4A^{-1} = 0 $
$\Rightarrow 4A^{-1} = A^2 - 6A + 9I$
$\Rightarrow\ \ 4\text{A}^{-1}=\begin{bmatrix}6&-5&5\\-5&6&-5\\5&-5&6\end{bmatrix}-6\begin{bmatrix}2&-1&1\\-1&2&-1\\1&-1&2\end{bmatrix}+9\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
$\Rightarrow\ \ 4\text{A}^{-1}=\begin{bmatrix}6&-5&5\\-5&6&-5\\5&-5&6\end{bmatrix}-\begin{bmatrix}12&-6&6\\-6&12&-6\\6&-6&12\end{bmatrix}+\begin{bmatrix}9&0&0\\0&9&0\\0&0&9\end{bmatrix}$
$\Rightarrow\ \ 4\text{A}^{-1}=\begin{bmatrix}6-12+9&-5+6+0&5-6+0\\-5+6+0&6-12+9&-5+6+0\\5-6+0&-5+6+0&6-12+9\end{bmatrix}=\begin{bmatrix}3&1&-1\\1&3&1\\-1&1&3\end{bmatrix}$
$\Rightarrow \ \text{A}^{-1}=\frac{1}{4}\begin{bmatrix}3&1&-1\\1&3&1\\-1&1&3\end{bmatrix}$
View full question & answer→Question 2065 Marks
Solve the following systems of linear equations by cramer's rule:
5x - 7y + z = 11,
6x - 8y - z = 15,
3x + 2y - 6z = 7
Answer$\text{D}=\begin{vmatrix}5&-7&1\\6&-8&-1\\3&2&-6 \end{vmatrix}$
$=5(48+2)+7(-36+3)+1(12+24)$
$=5(50)+7(-33)+1(36)=55$
$\text{D}_1=\begin{vmatrix}11&-7&1\\15&-8&-1\\7&2&-6 \end{vmatrix}$
$=11(48+2)+7(-90+7)+1(30+36)$
$=11(50)+7(-83)+1(86)=55$
$\text{D}_2=\begin{vmatrix}5&11&1\\6&15&-1\\3&7&-6 \end{vmatrix}$
$=5(-90+7)-11(-36+3)+1(42-45)$
$=5(-83)-11(-33)+1(-3)=-55$
$\text{D}_3=\begin{vmatrix}5&-7&11\\6&-8&15\\3&2&7 \end{vmatrix}$
$=5(-56-30)+7(42-45)+11(12+24)$
$=5(-86)+7(-3)+11(36)=-55$
Now,
$\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{55}{55}=1$
$\text{y}=\frac{\text{D}_2}{\text{D}}=\frac{-55}{55}=-1$
$\text{z}=\frac{\text{D}_3}{\text{D}}=\frac{-55}{55}=-1$
$\therefore\text{x}=1,\text{ y}=-1$ and $\text{z}=-1$
View full question & answer→Question 2075 Marks
Using matrix method, solve the system of equation $3x + 2y - 2z = 3, x + 2y + 3z = 6$ and $2x - y + z = 2.$
AnswerGiven system of equation is:
$3x + 2y - 2z = 3$
$x + 2y + 3z = 6$
and $2x - y + z = 2$
or $AX = B$
i.e., $\begin{bmatrix}3&2&-2\\1&2&3\\2&-1&1\end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}3\\6\\2\end{bmatrix}$
$\therefore X = A^{-1}B$
For $A^{-1},$
Cofactors are
$\text{A}_{11}=5,\text{A}_{12}=5,\text{A}_{13}=-5,$
$\text{A}_{21}=0,\text{A}_{22}=7,\text{A}_{23}=7,$
$\text{A}_{31}=10,\text{A}_{32}=11\text{ and A}_{33}=4$
$\therefore\ \ \text{adj A}=\begin{bmatrix}5&5&-5\\0&7&7\\10&-11&4\end{bmatrix}^\text{T}=\begin{bmatrix}5&0&10\\5&7&-11\\-5&7&4 \end{bmatrix}$
$|\text{A}|=3(5)+2(5)+(-2)(-5)=35$
$\therefore\ \ \text{A}^{-1}=\frac{\text{adj A}}{|\text{A}|}=\frac{1}{35}\begin{bmatrix}5&0&10\\5&7&-11\\-5&7&4\end{bmatrix}$
Now, $X = A^{-1}B$
$\Rightarrow\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\frac{1}{35}\begin{bmatrix}5&0&10\\5&7&-11\\-5&7&4\end{bmatrix}\begin{bmatrix}3\\6\\2\end{bmatrix}=\frac{1}{35}$
$\begin{bmatrix}15+20\\15+45-22\\-15+42+8\end{bmatrix}=\frac{1}{35}\begin{bmatrix}35\\35\\35\end{bmatrix}=\begin{bmatrix}1\\1\\1\end{bmatrix}$
$\therefore x = 1, y = 1$ and $z = 1$
View full question & answer→Question 2085 Marks
If $a, b$ and $c$ are real numbers, and $\triangle=\begin{vmatrix}b+c&c+a&a+b\\c+a&a+b&b+c\\a+b&b+c&c+a\end{vmatrix}=0,$ Show that either $a + b + c = 0$ or $a = b = c.$
Answer$\triangle=\begin{vmatrix}b+c&c+a&a+b\\c+a&a+b&b+c\\a+b&b+c&c+a\end{vmatrix}$
Applying $R_{1 }\rightarrow R_1 + R_2 + R_3,$ we have:
$\triangle=\begin{vmatrix}2(a+b+c)&2(a+b+c)&2(a+b+c)\\c+a&a+b&b+c\\a+b&b+c&c+a\end{vmatrix}$
$=2(a+b+c)\begin{vmatrix}1&1&1\\c+a&a+b&b+c\\a+b&b+c&c+a\end{vmatrix}$
$=2(a+b+c)\begin{vmatrix}1&0&0\\c+a&a-b&b-a\\a+b&c-a&c-b\end{vmatrix}\ \text{C}_1\rightarrow\text{C}_2-\text{C}_1\ \text{and}\ \text{C}_3\rightarrow\text{C}_3-\text{C}_1$
Expanding along $R_1$, we have:
$\triangle = 2(a + b + c) (1) [(b - c) (c - b) - (b - a) (c - a)]$
$= 2(a + b + c) [-b^{2 }- c^2 + 2bc - bc + ba + ac - a^2]$
$= 2(a + b + c) [ab + bc + ca - a^2 - b^2 - c^2]$
It is given that $\triangle = 0.$
$(a + b + c) [ab + bc + ca - a^2 - b^2 - c^2] = 0$
$\Rightarrow$ Either $a + b + c = 0,$ or $ab + bc + ca - a^2 - b^2 - c^2 = 0.$
Now,
$ab + bc + ca - a^2 - b^2 - c^2 = 0$
$\Rightarrow -2ab - 2bc - 2ca + 2a^2 + 2b^2 + 2c^2 = 0$
$\Rightarrow (a - b)^2 + (b - c)^2 + (c - a)^2 = 0$
$\Rightarrow (a -b)^{2 }= (b - c)^2 = (c - a)^2 = 0 [(a - b)^2, (b - c)^2, (c - a)^2$ are non$-$negative$]$
$\Rightarrow (a - b) = (b - c) = (c - a) = 0$
$\Rightarrow a = b = c$
Hence, if $\triangle = 0,$ then either $a + b + c = 0$ or $a = b = c.$
View full question & answer→Question 2095 Marks
Let A $=\begin{bmatrix}1&-2&1\\-2&3&1\\1&1&5\end{bmatrix}.$ Verify that
- $[adj. A]^{-1} = adj.(A^{-1})$
- $(A^{-1})^{-1} = A$
AnswerA = $\begin{bmatrix}1&-2&1\\-2&3&1\\1&1&5\end{bmatrix}$
$\therefore|\text{A}| = 1(15 - 1) + 2(-10-1) + 1(-2-3)=14-22-5=13$
Now, $A_{11 }= 14, A_{12} = 11, A_{13} = -5 A_{21} = 11, A_{22} = 4, A_{23} = -3 A_{31} = -5, A_{32} = -3, A_{33} = -1$
$\therefore\text{adj. A} = \begin{bmatrix}14&11&-5\\11&4&-3\\-5&-3&-1\end{bmatrix}$ $\therefore\text{A}^{-1}=\frac{1}{|\text{A}|}(\text{adj. A})$ = $-\frac{1}{13}\begin{bmatrix}14&11&-5\\11&4&-3\\-5&-3&-1\end{bmatrix}=\frac{1}{13}\begin{bmatrix}-14&-11&5\\-11&-4&3\\5&3&1\end{bmatrix}$
- |adj. $A| = 14(- 4 - 9) - 11(- 11 - 15) - 5(- 33 + 20)$
$= 14(-13) -11(-26) -5(-13)$
$= -182 + 286 + 65 = 169$
We have,
adj. $($adj. $A) = \begin{bmatrix}-13&26&-13\\26&-39&-13\\-13&-13&-65\end{bmatrix}$
$\therefore\ [\text{adj. A}]^{-1}=\frac{1}{|\text{adj. A}|}(\text{adj. (adj. A))}$
= $\frac{1}{169}\begin{bmatrix}-13&26&-13\\26&-39&-13\\-13&-13&-65\end{bmatrix}$
= $\frac{1}{13}\begin{bmatrix}-1&2&-1\\2&-3&-1\\-1&-1&-5\end{bmatrix}$
Now, $A^{-1} = \frac{1}{13}\begin{bmatrix}-14&-11&5\\-11&-4&3\\5&3&1\end{bmatrix}=\begin{bmatrix}-\frac{14}{13}&-\frac{11}{13}&\frac{5}{13}\\-\frac{11}{13}&-\frac{4}{13}&\frac{3}{13}\\\frac{5}{13}&\frac{3}{13}&\frac{1}{13}\end{bmatrix}$
$\therefore\ \text{adj.}(\text{A}^{-1})=\begin{bmatrix}-\frac{4}{169}-\frac{9}{169}&-\bigg(-\frac{11}{169}-\frac{15}{169}\bigg)&-\frac{33}{169}+\frac{20}{169}\\-\bigg(-\frac{11}{169}-\frac{15}{169}\bigg)&-\frac{14}{169}-\frac{25}{169}&-\bigg(-\frac{42}{169}+\frac{55}{169}\bigg)\\-\frac{33}{169}+\frac{20}{169}&-\bigg(-\frac{42}{169}+\frac{55}{169}\bigg)&\frac{56}{169}-\frac{121}{169}\end{bmatrix}$
$=\frac{1}{169}\begin{bmatrix}-13&26&-13\\26&-39&-13\\-13&-13&-65\end{bmatrix}=\frac{1}{13}\begin{bmatrix}-1&2&-1\\2&-3&-1\\-1&-1&-5\end{bmatrix}$
Hence, $[adj. A]^{-1} = adj. (A^{-1}).$
- We have shown that:
$A^{-1 }= \frac{1}{13}\begin{bmatrix}-14&-11&5\\-11&-4&3\\5&3&1\end{bmatrix}$
And, $adj. A^{-1} = \frac{1}{13}\begin{bmatrix}-1&2&-1\\2&-3&-1\\-1&-1&-5\end{bmatrix}$
Now, $\bigg|\text{A}^{-1}\bigg|=\bigg(\frac{1}{13}\bigg)^3[-14\times(-13)+11\times(-26)+5\times(-13)]=\bigg(\frac{1}{13}\bigg)^3\times(-169)=-\frac{1}{13}$
$\therefore\ (\text{A}^{-1})^{-1}=\frac{\text{adj. A}^{-1}}{|\text{A}^{-1}|}=\frac{1}{\bigg(-\frac{1}{13}\bigg)}\times\frac{1}{13}\begin{bmatrix}-1&2&-1\\2&-3&-1\\-1&-1&-5\end{bmatrix}=\begin{bmatrix}1&-2&1\\-2&3&1\\1&1&5\end{bmatrix}=\text{A}$
$\therefore\ (\text{A}^{-1})^{-1}=\text{A}$ View full question & answer→Question 2105 Marks
Prove that: $\begin{vmatrix}\text{a}&\text{a}+\text{b}&\text{a}+2\text{b} \\\text{a}+2\text{b}&\text{a}&\text{a}+\text{b}\\\text{a}+\text{b}&\text{a}+2\text{b}&\text{a} \end{vmatrix}=9(\text{a}+\text{b})\text{b}^2$
AnswerLet $\text{L.H.S}=\begin{vmatrix}\text{a}&\text{a}+\text{b}&\text{a}+2\text{b} \\\text{a}+2\text{b}&\text{a}&\text{a}+\text{b}\\\text{a}+\text{b}&\text{a}+2\text{b}&\text{a} \end{vmatrix}$
$=\begin{vmatrix}3\text{a}+3\text{b}&3\text{a}+3\text{b}&3\text{a}+3\text{b} \\\text{a}+2\text{b}&\text{a}&\text{a}+\text{b}\\\text{a}+\text{b}&\text{a}+2\text{b}&\text{a} \end{vmatrix} [$Applying $R_1 \rightarrow R_2 + R_2 + R_3]$
$=3(\text{a}+\text{b})\begin{vmatrix}1&1&1\\\text{a}+2\text{b}&\text{a}&\text{a}+\text{b}\\\text{a}+\text{b}&\text{a}+2\text{b}&\text{a} \end{vmatrix} [$Taking out $3(a + b)$ common from $R_1]$
$=3(\text{a}+\text{b})\begin{vmatrix}0&0&1\\2\text{b}&-\text{b}&\text{a}+\text{b}\\-\text{b}&2\text{b}&\text{a} \end{vmatrix} [$Applying $C_1 \rightarrow C_1 - C_2$ and $C_2 \rightarrow C_2 - C_3]$
$=3(\text{a}+\text{b})\text{b}^2\begin{vmatrix}0&0&1\\2&-1&\text{a}+\text{b}\\-1&2&\text{a} \end{vmatrix} [$Taking out $b$ common from $C_1$ and $C_2]$
$=3(\text{a}+\text{b})\text{b}^2\times3$
$=9(\text{a}+\text{b})\text{b}^2$
$=\text{R.H.S}$
View full question & answer→Question 2115 Marks
Show that $\begin{vmatrix}\text{x}+1&\text{x}+2&\text{x}+\text{a}\\\text{x}+2&\text{x}+3&\text{x}+\text{b}\\\text{x}+3&\text{x}+4&\text{x}+\text{c}\end{vmatrix}=0$ where $a, b, c$ are in $A.P.$
Answer$2\text{b}=\text{a}+\text{c}$
$\text{L.H.S}=\begin{vmatrix}\text{x}+1&\text{x}+2&\text{x}+\text{a}\\\text{x}+2&\text{x}+3&\text{x}+\text{b}\\\text{x}+3&\text{x}+4&\text{x}+\text{c}\end{vmatrix} [$Applying $R_2 = 2R_2]$
$=\frac{1}{2}\begin{vmatrix}\text{x}+1&\text{x}+2&\text{x}+\text{a}\\2\text{x}+4&2\text{x}+6&2\text{x}+2\text{b}\\\text{x}+3&\text{x}+4&\text{x}+\text{c}\end{vmatrix}$
$=\frac{1}{2}\begin{vmatrix}\text{x}+1&\text{x}+2&\text{x}+\text{a}\\0&0&0\\\text{x}+3&\text{x}+4&\text{x}+\text{c}\end{vmatrix}$ $[\because2\text{b}=\text{a}+\text{c}]$
$[$Applying $R_2 \rightarrow R_2 - (R_1 + R_3)]$
$=0$
$=\text{R.H.S}$
View full question & answer→Question 2125 Marks
Prove that: $\begin{vmatrix} 1&\text{a}&\text{bc}\\1&\text{b}&\text{ca}\\1&\text{c}&\text{ab}\end{vmatrix}=\begin{vmatrix} 1&\text{a}&\text{a}^2\\1&\text{b}&\text{b}^2\\1&\text{c}&\text{c}^2\end{vmatrix}$
Answer$\begin{vmatrix} 1&\text{a}&\text{bc}\\1&\text{b}&\text{ca}\\1&\text{c}&\text{ab}\end{vmatrix}$
Apply $R_1 \rightarrow R_1a, R_2 \rightarrow R_2b, R_3 \rightarrow R_3c$
$=\frac{1}{\text{abc}}\begin{vmatrix} \text{a}&\text{a}^2&\text{abc}\\\text{b}&\text{b}^2&\text{cab}\\\text{c}&\text{c}^2&\text{abc}\end{vmatrix}$
$=\frac{\text{abc}}{\text{abc}}\begin{vmatrix} \text{a}&\text{a}^2&1\\\text{b}&\text{b}^2&1\\\text{c}&\text{c}^2&1\end{vmatrix}$
$=-\begin{vmatrix} \text{a}&1&\text{a}^2\\\text{b}&1&\text{b}^2\\\text{c}&1&\text{c}^2\end{vmatrix}$
$=\begin{vmatrix} 1&\text{a}&\text{a}^2\\1&\text{b}&\text{b}^2\\1&\text{c}&\text{c}^2\end{vmatrix}$
View full question & answer→Question 2135 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 2145 Marks
Solve the following systems of linear equations by cramer's rule:
$x + y = 5,$
$y + z = 3,$
$x + z = 4$
AnswerLet $\text{D}=\begin{vmatrix}1&1&0\\0&1&1\\1&0&1\end{vmatrix}$
Expanding along $R_1$
$=1(1)-1(-1)+0(-1)$
$=1+1+0=2$
Also $\text{D}_1=\begin{vmatrix}5&1&0\\3&1&1\\4&0&1\end{vmatrix}$
Expanding along $R_1$
$=5(1)-1(-1)+0(-4)$
$=5+1+0=6$
Again $\text{D}_2=\begin{vmatrix}1&5&0\\0&3&1\\1&4&1\end{vmatrix}$
Expanding along $R_1$
$=1(-1)-5(-1)+0(-3)$
$=-1+5+0=4$
Also $\text{D}_3=\begin{vmatrix}1&1&5\\0&1&3\\1&0&4\end{vmatrix}$
$=1(4)-1(-3)+5(-1)$
$=4+3-5=2$
Now $\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{6}{2}=3$
$\text{y}=\frac{\text{D}_2}{\text{D}}=\frac{4}{2}=2$
$\text{z}=\frac{\text{D}_3}{\text{D}}=\frac{2}{2}=1$
Hence, $x = 3, y = 2, z = 1$
View full question & answer→Question 2155 Marks
Evaluate the following integrals:
$\int^\limits{\frac{\pi}{2}}_{0}\frac{\cos\text{x}}{\big(\cos\frac{\text{x}}{2}+\sin\frac{\text{x}}{2}\big)^\text{n}}\text{ dx}$
AnswerLet $\text{I}=\int^\limits{\frac{\pi}{2}}_{0}\frac{\cos\text{x}}{\big(\cos\frac{\text{x}}{2}+\sin\frac{\text{x}}{2}\big)^\text{n}}\text{ dx}$
$=\int^\limits{\frac{\pi}{2}}_{0}\frac{\cos^2\frac{\text{x}}{2}-\sin^2\frac{\text{x}}{2}}{\big(\cos\frac{\text{x}}{2}+\sin\frac{\text{x}}{2}\big)^\text{n}}\text{ dx}$
$=\int^\limits{\frac{\pi}{2}}_{0}\frac{\big(\cos\frac{\text{x}}{2}+\sin\frac{\text{x}}{2}\big)\big(\cos\frac{\text{x}}{2}+\sin\frac{\text{x}}{2}\big)}{\big(\cos\frac{\text{x}}{2}+\sin\frac{\text{x}}{2}\big)^{\text{n}}}\text{ dx}$
$=\int^\limits{\frac{\pi}{2}}_{0}\frac{\big(\cos\frac{\text{x}}{2}-\sin\frac{\text{x}}{2}\big)}{\big(\cos\frac{\text{x}}{2}+\sin\frac{\text{x}}{2}\big)^{\text{n}-1}}\text{ dx}$
Put $\cos\frac{\text{x}}{2}+\sin\frac{\text{x}}{2}=\text{z}$
$\therefore\ \Big(-\sin\frac{\text{x}}{2}\times\frac{1}{2}+\cos\frac{\text{x}}{2}\times\frac{1}{2}\Big)\text{dx}=\text{dz}$
$\Big(\cos\frac{\text{x}}{2}-\sin\frac{\text{x}}{2}\Big)\text{dx}=2\text{dz}$
When $\text{x}\rightarrow0,\text{ z}\rightarrow1$
When $\text{x}\rightarrow\frac{\pi}{2},\text{ z}\rightarrow\sqrt{2}$ $\Big(\text{z}=\cos\frac{\pi}{4}+\sin\frac{\pi}{4}=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}\Big)$
$\therefore\ \text{I}=2\int^\limits{\sqrt{2}}_{1}\frac{\text{dz}}{\text{z}^{\text{n}-1}}$
$=2\times\Big[\frac{\text{z}^{\text{n}-1}}{2-\text{n}}\Big]^{\sqrt{2}}_1$
$=\frac{2}{(2-\text{n})}\Big[\big(\sqrt{2}\big)^{2-\text{n}}-1\Big]$
$=\frac{2}{(2-\text{n})}\bigg(2^{\frac{2}{(2-\text{n})}}-1\bigg)$
$=\frac{2}{(2-\text{n})}\Big(2^{1-\frac{\text{n}}{2}}-1\Big)$
View full question & answer→Question 2165 Marks
Find the inverse of the following matrices: $\begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos\alpha & \sin\alpha \\ 0 & \sin\alpha & -\cos\alpha \end{bmatrix}$
AnswerHere, $\text{A}=\begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos\alpha & \sin\alpha \\ 0 & \sin\alpha & -\cos\alpha \end{bmatrix}$
Expanding using first column, we get
$|\text{A}|=1\begin{vmatrix}\cos\alpha & \sin\alpha \\ \sin\alpha & -\cos\alpha \end{vmatrix}-0+0$
$=-\cos^2\alpha+\sin^2\alpha$
$=-(\cos^2\alpha+\sin^2\alpha)$
$|\text{A}|=-1\neq0$
Therefore, $A^{-1}$ exists
Cofactors of a are:
$\text{C}_{11}=-1,\ \text{C}_{21}=0,\ \text{C}_{31}=0$
$\text{C}_{12}=0,\ \text{C}_{22}=-\cos\alpha,\ \text{C}_{32}=-\sin\alpha$
$\text{C}_{13}=0,\ \text{C}_{23}=-\sin\alpha,\ \text{C}_{33}=\cos\alpha$
$\therefore\ \text{adj A}=\begin{bmatrix}\text{C}_{11} & \text{C}_{12} & \text{C}_{13} \\ \text{C}_{21} & \text{C}_{22} & \text{C}_{23} \\ \text{C}_{31} & \text{C}_{32} &\text{C}_{33} \end{bmatrix}$
$=\begin{bmatrix}-1 & 0 & 0 \\ 0 & -\cos\alpha & -\sin\alpha \\ 0 & -\sin\alpha & \cos\alpha \end{bmatrix}^\text{T}$
$=\begin{bmatrix}-1 & 0 & 0 \\ 0 & -\cos\alpha & -\sin\alpha \\ 0 & -\sin\alpha & \cos\alpha \end{bmatrix}$
Now, $\text{A}^{-1}=\frac{1}{|\text{A}|}\text{ adj A}$
$\therefore\ \text{A}^{-1}=\frac{1}{(-1)}\begin{bmatrix}-1 & 0 & 0 \\ 0 & -\cos\alpha & -\sin\alpha \\ 0 & -\sin\alpha & \cos\alpha \end{bmatrix}$
Hence, $\text{A}^{-1}=\begin{bmatrix}-1 & 0 & 0 \\ 0 & -\cos\alpha & -\sin\alpha \\ 0 & -\sin\alpha & \cos\alpha \end{bmatrix}$
View full question & answer→Question 2175 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 elements $a_{ij}$ in $A = [a_{ij}].$ Then,
$\text{C}_{11}=(-1)^{1+1}17=17,\text{C}_{12}=(-1)^{1+2}26=-26$
$\text{C}_{21}=(-1)^{2+1}1=-1,\text{C}_{22}=(-1)^{2+2}2=2$
$\text{adj }\text{A}=\begin{bmatrix}17&-26\\-1&2\end{bmatrix}^\text{T}$
$=\begin{bmatrix}17&-1\\-26&2\end{bmatrix}$
$\Rightarrow\text{A}^{-1}=\frac{1}{|\text{A}|}\text{adj }\text{A}$
$=\frac{1}{8}\begin{bmatrix}17&-1\\-26&2\end{bmatrix}$
$\text{X}=\text{A}^{-1}\text{B}$
$\Rightarrow\begin{bmatrix}\text{x}\\\text{y}\end{bmatrix}=\frac{1}{8}\begin{bmatrix}17&-1\\-26&2\end{bmatrix}\begin{bmatrix}7000\\110000\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\\text{y}\end{bmatrix}=\frac{1}{8}\begin{bmatrix}119000-110000\\-182000+220000\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\\text{y}\end{bmatrix}=\frac{1}{8}\begin{bmatrix}9000\\38000\end{bmatrix}$
$\Rightarrow\text{x}=\frac{9000}{8}\text{and }\text{y}=\frac{38000}{8}$
$\therefore\ \text{x}=1125\text{ and }\text{y}=4750.$
View full question & answer→Question 2185 Marks
Solve the following systems of linear equations by cramer's rule:
x + y = 1,
x + z = -6,
x - y - 2z = 3
AnswerThese equations can be written as
x + y + 0z = 1
x + 0y + z = -6
x - y - 2z = 3
$\text{D}=\begin{vmatrix}1&1&0\\1&0&1\\1&-1&-2 \end{vmatrix}$
$=1(0+1)-1(-2-1)+0(-1-0)=4$
$\text{D}_1=\begin{vmatrix}1&1&0\\-6&0&1\\3&-1&-2 \end{vmatrix}$
$=1(0+1)-1(12-3)+0(6-0)=-8$
$\text{D}_2=\begin{vmatrix}1&1&0\\1&-6&1\\1&3&-2 \end{vmatrix}$
$=1(12-3)-1(-2-1)+0(3+6)=12$
$\text{D}_3=\begin{vmatrix}1&1&0\\1&0&-6\\1&-1&3 \end{vmatrix}$
$=1(0-6)-1(3+6)+1(-1-0)=-16$
Now,
$\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{-8}{4}=-2$
$\text{y}=\frac{\text{D}_2}{\text{D}}=\frac{12}{4}=3$
$\text{z}=\frac{\text{D}_3}{\text{D}}=\frac{-16}{4}=-4$
$\therefore\text{x}=-2,\text{y}=3$ and $\text{z}=-4$
View full question & answer→Question 2195 Marks
By using properties of determinants, show that: $\begin{vmatrix}x&x^2&yz\\y&y^2&zx\\z&z^2&xy\end{vmatrix}=(x-y)(y-z)(z-x)(xy+yz+zx)$
Answer$\text{Let}\ \triangle=\begin{vmatrix}x&x^2&yz\\y&y^2&zx\\z&z^2&xy\end{vmatrix}.$
Applying $R_2\rightarrow\text{R}_2-\text{R}_1 \ \text{and R}_3\rightarrow\text{R}_3-\text{R}_1,$ we have:
$\triangle=\begin{vmatrix}x&x^2&yz\\y-x&y^2-x^2&zx-yz\\z-x&z^2-x^2&xy-yz\end{vmatrix}$
$=\begin{vmatrix}x&x^2&yz\\-(x-y)&-(x-y)(x+y)&z(x-y)\ z-x)&(z-x)(z+x)&-y(z-x)\end{vmatrix}$
$=(x-y)(z-x)\begin{vmatrix}x&x^2&yz\\-1&-x-y&z\\1&z+x&-y\end{vmatrix}$
Applying $R_3\rightarrow\text{R}_3+\text{R}_2,$ we have:
$\triangle=(x-y)(z-x)\begin{vmatrix}x&x^2&yz\\-1&-x-y&z\\0&z-y&z-y\end{vmatrix}$
$\triangle=(x-y)(z-x)(z-y)\begin{vmatrix}x&x^2&yz\\-1&-x-y&z\\0&1&1\end{vmatrix}$
Expanding along $R_3,$ we have:
$\triangle=\left[(x-y)(z-x)(z y)\right]\Bigg[(-1)\begin{vmatrix}x&yz\\-1&z\end{vmatrix}+1\begin{vmatrix}x&x^2\\-1&-x-y\end{vmatrix}\Bigg]$
$=(x-y)(z-x)(z-y)\left[(-xz-yz)+(-x^2-xy+x^2)\right]$
$=-(x-y)(z-x)(z-y)(xy+yz+zx)$
$=(x-y)(y-z)(z-x)(xy+yz+zx)$
Hence, the given result is proved.
View full question & answer→Question 2205 Marks
Solve the following systems of linear equations by cramer's rule$:$
$x + y + z + w = 2,$
$x - 2y + 2z + 2w = -6,$
$2x + y - 2z + 2w = -5,$
$3x - y + 3z - 3w = -3$
AnswerHere, $\text{D}=\begin{vmatrix}1&1&1&1\\1&-2&2&2\\2&1&-2&2\\3&-1&3&-3\end{vmatrix}$
$\therefore\text{D}=\begin{vmatrix}1&0&0&0\\1&-3&1&1\\2&-1&-4&0\\3&-4&0&-6\end{vmatrix}=1\begin{vmatrix}-3&1&1\\-1&-4&0\\-4&0&-6\end{vmatrix}$
$[C_2 \rightarrow C_2 - C_1, C_3 \rightarrow C_3 - C_1, C_4 \rightarrow C_4 - C_1]$
$\therefore\text{D}=\begin{vmatrix}0&0&1\\-1&-4&0\\-22&6&-6\end{vmatrix} [C_1 \rightarrow C_1 + 3C_3, C_2 \rightarrow C_2 - C_3]$
$=1(-6-88)=-94$
$\text{D}_1=\begin{vmatrix}2&1&1&1\\-6&-2&2&2\\-5&1&-2&2\\-3&-1&3&-3\end{vmatrix}=188$
$\text{D}_2=\begin{vmatrix}1&2&1&1\\1&-6&2&2\\2&-5&-2&2\\3&-3&3&-3\end{vmatrix}=-282$
$\text{D}_3=\begin{vmatrix}1&1&2&1\\1&-2&-6&2\\2&-1&-5&-2\\3&-1&-3&-3\end{vmatrix}=-141$
$\text{D}_4=\begin{vmatrix}1&1&1&2\\1&-2&2&-6\\2&1&-2&-5\\3&-1&3&-3\end{vmatrix}=47$
Now,
$\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{188}{-94}=-2$
$\text{y}=\frac{\text{D}_2}{\text{D}}=\frac{-282}{-94}=3$
$\text{z}=\frac{\text{D}_3}{\text{D}}=\frac{-141}{94}=\frac{3}{2}$
$\text{w}=\frac{\text{D}_4}{\text{D}}=\frac{47}{-94}=-\frac{1}{2}$
Hence, $\text{x}=-2,\text{y}=3,\text{z}=\frac{3}{2},\text{w}=-\frac{1}{2}$
View full question & answer→Question 2215 Marks
$\begin{vmatrix}\text{b}+\text{c}&\text{a}&\text{a}\\\text{b}&\text{c}+\text{a}&\text{b}\\\text{c}&\text{c}&\text{a}+\text{b}\end{vmatrix}=4\text{abc}$
Answer$\text{L.H.S}=\begin{vmatrix}\text{b}+\text{c}&\text{a}&\text{a}\\\text{b}&\text{c}+\text{a}&\text{b}\\\text{c}&\text{c}&\text{a}+\text{b}\end{vmatrix}$
$=\begin{vmatrix}0&-2\text{c}&-2\text{b}\\\text{b}&\text{c}+\text{a}&\text{b}\\\text{c}&\text{c}&\text{a}+\text{b}\end{vmatrix} [$Applying $R_1 \rightarrow R_1 - (R_2 + R_3)]$
$=\begin{vmatrix}0&-2\text{c}&-2\text{b}\\\text{b}&\text{c}+\text{a}-\text{b}&0\\\text{c}&0&\text{a}+\text{b}-\text{c}\end{vmatrix} [$Applying $C_2 \rightarrow C_2 - C_1$ and $C_3 \rightarrow C_3 - C_1]$
$=0\begin{vmatrix}\text{c}+\text{a}-\text{b}&0\\0&\text{a}+\text{b}-\text{c}\end{vmatrix}-(-2\text{c})\begin{vmatrix}\text{b}&0\\\text{c}&\text{a}+\text{b}-\text{c}\end{vmatrix}-2\text{b}\begin{vmatrix}\text{b}&\text{c}+\text{a}-\text{b}\\\text{c}&0\end{vmatrix}$
$=2\text{c}[\text{b}(\text{a}+\text{b}-\text{c})-0]-2\text{b}[0-\text{c}(\text{c}+\text{a}-\text{b})]$
$=2\text{bc}[\text{a}+\text{b}-\text{c}]-2\text{bc}[\text{b}-\text{c}-\text{a}]$
$=2\text{bc}[(\text{a}+\text{b}-\text{c})-(\text{b}-\text{c}-\text{a})]$
$=4\text{abc}$
$=\text{R.H.S}$
View full question & answer→Question 2225 Marks
If $x + y + z = 0,$ then prove that $\begin{vmatrix}\text{xa}&\text{yb}&\text{zc}\\\text{yc}&\text{za}&\text{xb}\\\text{zb}&\text{xc}&\text{ya}\end{vmatrix}=\text{xyz}\begin{vmatrix}\text{a}&\text{b}&\text{c}\\\text{c}&\text{a}&\text{b}\\\text{b}&\text{c}&\text{a}\end{vmatrix},$
AnswerSince, $x + y + z = 0,$ also we have to prove
$\begin{vmatrix}\text{xa}&\text{yb}&\text{zc}\\\text{yc}&\text{za}&\text{xb}\\\text{zb}&\text{xc}&\text{ya}\end{vmatrix}=\text{xyz}\begin{vmatrix}\text{a}&\text{b}&\text{c}\\\text{c}&\text{a}&\text{b}\\\text{b}&\text{c}&\text{a}\end{vmatrix}$
$\text{L.H.S}=\begin{vmatrix}\text{xa}&\text{yb}&\text{zc}\\\text{yc}&\text{za}&\text{xb}\\\text{zb}&\text{xc}&\text{ya}\end{vmatrix}$
$=\text{xa}(\text{za}\cdot\text{ya}-\text{xb}\cdot\text{xc})-\text{yb}(\text{yc}\cdot\text{ya}-\text{xb}\cdot\text{zb})+\text{zc}(\text{yc}\cdot\text{xc}-\text{za}\cdot\text{zb})$
$=\text{xa}(\text{a}^2\text{yz}-\text{x}^2\text{bc})\cdot-\text{yb}(\text{y}^2\text{ac}-\text{b}^2\text{xz})+\text{zc}(\text{c}^2\text{xy}-\text{z}^2\text{ab})$
$=\text{xyza}^3-\text{x}^3\text{abc}-\text{y}^3\text{abc}+\text{b}^3\text{xyz}+\text{c}^3\text{xyz}-\text{z}^3\text{abc}$
$=\text{xyz}\big(\text{a}^3+\text{b}^3+\text{c}^3\big)-\text{abc}\big(\text{x}^3+\text{y}^3+\text{z}^3\big)$
$=\text{xyz}\big(\text{a}^3+\text{b}^3+\text{c}^3\big)-\text{abc}(3\text{xyz})$ $\big[\because\ \text{x}+\text{y}+\text{z}=0\Rightarrow\ \text{x}^3+\text{y}^3+\text{z}^3-3\text{xyz}\big]$
$=\text{xyz}\big(\text{a}^3+\text{b}^3+\text{c}^3-3\text{abc}\big)\ \ \dots(1)$
Now, $\text{R.H.S}=\text{xyz}\begin{vmatrix}\text{a}&\text{b}&\text{c}\\\text{c}&\text{a}&\text{b}\\\text{b}&\text{c}&\text{a}\end{vmatrix}=\text{xyz}\begin{vmatrix}\text{a}+\text{b}+\text{c}&\text{b}&\text{c}\\\text{a}+\text{b}+\text{c}&\text{a}&\text{b}\\\text{a}+\text{b}+\text{c}&\text{c}&\text{a}\end{vmatrix}$ $[\text{C}_1\rightarrow\text{C}_1+\text{C}_2+\text{C}_3]$
$=\text{xyz}(\text{a}+\text{b}+\text{c})\begin{vmatrix}1&\text{b}&\text{c}\\1&\text{a}&\text{b}\\1&\text{c}&\text{a}\end{vmatrix}$ $\big[\text{Taking }(\text{a}+\text{b}+\text{c})\text{ common from C}_1\big]$
$=\text{xyz }(\text{a}+\text{b}+\text{c})\begin{vmatrix}0&\text{b}-\text{c}&\text{c}-\text{a}\\0&\text{a}-\text{c}&\text{b}-\text{a}\\1&\text{c}&\text{a}\end{vmatrix}$ $\big[\text{R}_1\rightarrow\text{R}_1-\text{R}_3\text{ and R}_2\rightarrow\text{R}_2-\text{R}_3\big]$
Expanding along $C_1,$
$=\text{xyz}(\text{a}+\text{b}+\text{c})\big[1(\text{b}-\text{c})(\text{b}-\text{a})-(\text{a}-\text{c})(\text{c}-\text{a})\big]$
$=\text{xyz}(\text{a}+\text{b}+\text{c})(\text{b}^2-\text{ab}-\text{bc}+\text{ac}+\text{a}^2+\text{c}^2-2\text{ac})$
$=\text{xyz}(\text{a}+\text{b}+\text{c})(\text{a}^2+\text{b}^2+\text{c}^2-\text{ab}-\text{bc}-\text{ca})$
$=\text{xyz}(\text{a}^3+\text{b}^3+\text{c}^3-3\text{abc})\ \ \dots(2)$
From $(1)$ and $(2),$ we get
$\text{L.H.S = R.H.S}$
View full question & answer→Question 2235 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 2245 Marks
Show that the following systems of linear equations has infinite number of solutions and solve:
x - y + z = 3,
2x + y - z = 2,
-x - 2y + 2z = 1
AnswerWe have,
$\text{D}=\begin{vmatrix}1&-1&1\\2&1&-1\\-1&-2&2\end{vmatrix}=\begin{vmatrix}0&-1&0\\3&1&0\\-3&-2&0\end{vmatrix}=0$
$\text{D}_1=\begin{vmatrix}3&-1&1\\2&1&-1\\1&-2&2\end{vmatrix}=\begin{vmatrix}3&-1&0\\2&1&0\\1&-2&0\end{vmatrix}=0$
$\text{D}_2=\begin{vmatrix}1&3&1\\2&2&-1\\-1&1&2\end{vmatrix}=\begin{vmatrix}1&0&0\\2&-4&-3\\-1&4&3\end{vmatrix}=1(-12+12)=0$
$\text{D}_3=\begin{vmatrix}1&-1&3\\2&1&2\\-1&-2&1\end{vmatrix}=\begin{vmatrix}1&0&0\\2&3&-4\\-1&-3&4\end{vmatrix}=1(12-12)=0$
$\therefore\text{D}=\text{D}_1=\text{D}_2=\text{D}_3=0$
So, either the system is consistent with infinitely many solutions or it is inconsistent.
Consider the first two equations, written as
x - y = 3 - z
2x + y = 2 + z
to solve these equation, written as
Here,
$\text{D}=\begin{vmatrix}1&-1\\2&1\end{vmatrix}=1+2=3$
$\text{D}_1=\begin{vmatrix}3-\text{z}&-1\\2+\text{z}&1\end{vmatrix}=(3-\text{z})+(2+\text{z})=5$
$\text{D}_2=\begin{vmatrix}1&3-\text{z}\\1&2+\text{z}\end{vmatrix}=(2+\text{z})-(6-2\text{z})=-4+3\text{z}$
$\therefore\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{5}{3}$
$\text{y}=\frac{\text{D}_2}{\text{D}}=\frac{-4+3\text{z}}{3}$
Let z = k, then the equations have the solution.
$\text{x}=\frac{5}{3},\text{ y}=\frac{-4+3\text{k}}{3},\text{ z}=\text{ k}$
View full question & answer→Question 2255 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 2265 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 $A = [a_{ij}].$ 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 $z = -1$
View full question & answer→Question 2275 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 $A[a_{ij}].$ 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.
View full question & answer→Question 2285 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 $A [a_{ij}].$ 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 $z = -5$
View full question & answer→Question 2295 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 2305 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, $\text{x}=\text{A}^{-1}\text{B}=\frac{1}{|\text{A}|}(\text{adj }\text{A})\text{B}$
$=\frac{1}{-2}\begin{bmatrix}-1&-1\\-1&1\end{bmatrix}\begin{bmatrix}-\text{k}\\5\text{k}\end{bmatrix}$
$\begin{bmatrix}\text{x}\\\text{y}\end{bmatrix}=\Big(\frac{1}{-2}\Big)\begin{bmatrix}\text{k}-5\text{k}\\\text{k}+5\text{k}\end{bmatrix}=\begin{bmatrix}2\text{k}\\-3\text{k}\end{bmatrix}$
$x = 2k, y = -3k, z = k$
View full question & answer→Question 2315 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 $A [a_{ij}]$. 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.
View full question & answer→Question 2325 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 2335 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 2345 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 2355 Marks
Compute the adjoint of the following matrices:
$\begin{bmatrix} 1 & 2 & 5 \\ 2 & 3 & 1 \\ -1 & 1 & 1 \end{bmatrix}$
Verify that (adjoint A)A = |A|I = A (adjoint A) for the above matrices.
Answer$\text{B}=\begin{bmatrix} 1 & 2 & 5 \\ 2 & 3 & 1 \\ -1 & 1 & 1 \end{bmatrix}$
Now,
$\text{C}_{11}=\begin{vmatrix}3 & 1 \\1 & 1 \end{vmatrix}=2,\text{C}_{12}=-\begin{vmatrix}2 & 1 \\1 & 1 \end{vmatrix}=-3$
$\text{and C}_{13}=\begin{vmatrix}2 & 3 \\ -1 & 1 \end{vmatrix}=5$
$\text{C}_{21}=\begin{vmatrix}2 & 5 \\1 & 1 \end{vmatrix}=3,\text{C}_{22}=-\begin{vmatrix}1 & 5 \\-1 & 1 \end{vmatrix}=-6$
$\text{and C}_{23}=\begin{vmatrix}1 & 2 \\ -1 & 1 \end{vmatrix}=-3$
$\text{C}_{31}=\begin{vmatrix}2 & 5 \\3 & 1 \end{vmatrix}=13,\text{C}_{32}=-\begin{vmatrix}1 & 5 \\2 & 1 \end{vmatrix}=9$
$\text{and C}_{33}=\begin{vmatrix}1 & 2 \\ 2 & 3 \end{vmatrix}=-1$
$\therefore\ \text{adjoint B}=\begin{bmatrix}2 & -3 & 5 \\3 & 6 & -3 \\ -13 & 9 & -1 \end{bmatrix}^\text{T}$
$=\begin{bmatrix}2 & 3 & -13 \\-3 & 6 & 9 \\ 5 & -3 & -1\end{bmatrix}$
$\text{(adj B)B}=\begin{bmatrix}21 & 0 & 0 \\0 & 21 & 0 \\ 0 & 0 & 21 \end{bmatrix}$
and |B| = 21
$\therefore\ |\text{B}|\text{I}=\begin{bmatrix}21 & 0 & 0 \\0 & 21 & 0 \\ 0 & 0 & 21 \end{bmatrix}$
and $\text{B(adjoint B)}=\begin{bmatrix}21 & 0 & 0 \\0 & 21 & 0 \\ 0 & 0 & 21 \end{bmatrix}$
Thus, $\text{(adjoint A)A}=|\text{A}|\text{ I}=\text{A(adjoint A)}$
View full question & answer→Question 2365 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 $A = [a_{ij}].$ 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.
View full question & answer→Question 2375 Marks
Solve the following systems of linear equations by cramer's rule:
2x - 3y - 4z = 29,
-2x + 5y - z = -15,
3x - y + 5z = -11
AnswerGiven, 2x - 3y - 4z = 29
-2x + 5y - z = -15
3x - y + 5z = -11
$\text{D}=\begin{vmatrix}2&-3&-4\\-2&5&-1\\3&-1&5 \end{vmatrix}$
$=2(25-1)+3(-10+3)-4(2-15)$
$=2(24)+3(-7)-4(-13)=79$
$\text{D}_1=\begin{vmatrix}29&-3&-4\\-15&5&-1\\-11&-1&5 \end{vmatrix}$
$=29(25-1)+3(-72-11)-4(15+55)$
$=29(24)+3(-86)-4(70)=158$
$\text{D}_2=\begin{vmatrix}2&29&-4\\-2&-15&-1\\3&-11&5 \end{vmatrix}$
$=2(-75-11)-29(-10+3)-4(22+45)$
$=2(-86)-29(-7)-4(67)=-237$
$\text{D}_3=\begin{vmatrix}2&-3&29\\-2&5&-15\\3&-1&-11 \end{vmatrix}$
$=2(-55-15)+3(22+45)+29(2-15)$
$=2(-70)+3(67)+29(-13)=-316$
Now,
$\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{158}{79}=2$
$\text{y}=\frac{\text{D}_2}{\text{D}}=\frac{-237}{79}=-3$
$\text{z}=\frac{\text{D}_3}{\text{D}}=\frac{-316}{79}=-4$
$\therefore\text{x}=2,\text{y}=-3$ and $\text{z}=-4$
View full question & answer→Question 2385 Marks
Verify A (adj. A) = (adj. A) A = |A|I.
$\begin{bmatrix}1&-1&2\\3&0&-2\\1&0&3\end{bmatrix}$
Answer$\text{Let A}=\begin{bmatrix}1&-1&2\\3&0&-2\\1&0&3\end{bmatrix}$ $\Rightarrow\ |\text{A}|=\begin{vmatrix}1&-1&2\\3&0&-2\\1&0&3\end{vmatrix}$
$\therefore\ \text{A}_{11}=+\begin{vmatrix}0&-2\\0&3\end{vmatrix}=+0+0=0$
$\text{A}_{12}=-\begin{vmatrix}3&-2\\1&3\end{vmatrix}=-(9+2)=-11$
$\text{A}_{13}=+\begin{vmatrix}3&0\\1&0\end{vmatrix}=+(0-0)=0$
$\text{A}_{21}=-\begin{vmatrix}-1&2\\0&3\end{vmatrix}=-(-3-0)=3$
$\text{A}_{22}=+\begin{vmatrix}1&2\\1&3\end{vmatrix}=3-2=1$
$\text{A}_{23}=-\begin{vmatrix}1&-2\\1&0\end{vmatrix}=-(0+1)=-1$
$\text{A}_{31}=+\begin{vmatrix}-1&2\\0&-2\end{vmatrix}=2-0=2$
$\text{A}_{32}=-\begin{vmatrix}1&2\\3&-2\end{vmatrix}=-(-2-6)=8$
$\text{A}_{33}=+\begin{vmatrix}1&-1\\3&0\end{vmatrix}=3+0=3$
$\therefore\ \text{adj.A}=\begin{bmatrix}\text{A}_{11}&\text{A}_{12}&\text{A}_{13}\\\text{A}_{21}&\text{A}_{22}&\text{A}_{23}\\\text{A}_{31}&\text{A}_{32}&\text{A}_{33}\end{bmatrix}$
$=\begin{bmatrix}0&-11&0\\3&1&-1\\2&8&3\end{bmatrix}=\begin{bmatrix}0&3&2\\-11&1&8\\0&-1&3\end{bmatrix}$
$\therefore\ \text{A.(adj. A)}=\begin{bmatrix}1&-1&2\\3&0&-2\\1&0&3\end{bmatrix}\begin{bmatrix}0&3&2\\-11&1&8\\0&-1&3\end{bmatrix}$
$=\begin{bmatrix}0+11+0&3-1-2&2-8+6\\0-0-0&9+0+2&6+0-6\\0+0+0&3+0-3&2+0+9\end{bmatrix}$
$=\begin{bmatrix}11&0&0\\0&11&0\\0&0&11\end{bmatrix} \dots\dots(1)$
$\text{Again}\ \ \text{(adj. A). A}=\begin{bmatrix}0&3&2\\-11&1&8\\0&-1&3\end{bmatrix}\begin{bmatrix}1&-1&2\\3&0&-2\\1&0&3\end{bmatrix}$
$=\begin{bmatrix}0+9+2&0+0+0&0-6+6\\-11+3+8&11+0+0&-22-2+24\\0-3+3&0-0+0&0+2+9\end{bmatrix}$
$=\begin{bmatrix}11&0&0\\0&11&0\\0&0&11\end{bmatrix} \dots\dots(2)$
$\text{And}\ \ |\text{A}|=\begin{vmatrix}1&-1&2\\3&0&-2\\1&0&3\end{vmatrix}$
$=1(0-0)-(-1)(9+2)+2(0-0)=0+11+0=11$
$\text{Also}\ \ | \text{A|I}=\text{|A|I}_3=11\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}=\begin{bmatrix}11&0&0\\0&11&0\\0&0&11\end{bmatrix} \dots\dots(3)$
$\therefore$ From eq. (1), (2) and (3) A.(adj. A) = (adj. A).A = |A|I
View full question & answer→Question 2395 Marks
Find the matrix $X$ satisfying the matrix equation. $\text{X}\begin{bmatrix}5 & 3 \\-1 & -2 \end{bmatrix}\begin{bmatrix}14 & 7 \\7 & 7 \end{bmatrix}$
Answer$\text{X}\begin{bmatrix}5 & 3 \\-1 & -2 \end{bmatrix}\begin{bmatrix}14 & 7 \\7 & 7 \end{bmatrix}=-10+3=-7\neq0$
Hence, $A$ is invertible.
If $C_{ij}$ is a cofactor of $a_{ij}$ in $A,$ then $C_{11} = -2, C_{12} = 1, C_{21} = -3$ and $C_{22} = 5.$
Now, $\text{adj A}\begin{bmatrix}-2 & 1 \\-3 & 5 \end{bmatrix}^\text{T}=\begin{bmatrix}-2 & -3 \\1 & 5 \end{bmatrix}$
$\Rightarrow\ \text{A}^{-1}=\frac{1}{|\text{A}|}\text{ adj A}=\frac{-1}{7}\begin{bmatrix}-2 & -3 \\1 & 5 \end{bmatrix}$
Let: $\text{B}=\begin{bmatrix}14 & 7 \\7 & 7 \end{bmatrix}$
$\Rightarrow\ \text{B}=\begin{bmatrix}14 & 7 \\7 & 7 \end{bmatrix}=98-49=49\neq0$
Hence, $B$ is invertible.
The given matrix equation becomes $XA = B.$
$\Rightarrow \ce{(XA)A^{-1} = BA^{-1}}$
$\Rightarrow\ \text{X(AA}^{-1})=\text{B}=\begin{bmatrix}14 & 7 \\7 & 7 \end{bmatrix}\times\frac{-1}{7}\times\begin{bmatrix}-2 & -3 \\1 & 5 \end{bmatrix}$
$\Rightarrow\ \text{X}=\frac{-1}{7}\begin{bmatrix}-28+7 & -42+35 \\ -14+7 & -21+35 \end{bmatrix}$
$\Rightarrow\ \text{X}=\frac{-1}{7}\begin{bmatrix}-21 & -7 \\ -7 & 14 \end{bmatrix}$
$\Rightarrow\ \text{X}=\begin{bmatrix} 3 & 1 \\1 & -2 \end{bmatrix}$
View full question & answer→Question 2405 Marks
If $\text{A}=\begin{vmatrix}1&2&0\\-2&-1&-2\\0&-1&1\end{vmatrix},$ then find the value of $A^{-1}.$
Using $^{A-1}.$ solve the system of equations $x - 2y = 10, 2x - y - z = 8$ and $-2y + z = 7.$
AnswerWe have, $\text{A}=\begin{vmatrix}1&2&0\\-2&-1&-2\\0&-1&1\end{vmatrix}\ \ \dots(\text{i})$
$\therefore\ |\text{A}|=1(-3)-2(-2)+0=1\neq0$
Now, $\text{A}_{11}=-3,\text{A}_{12}=2,\text{A}_{13}=2,$ $\text{A}_{21}=-2,\text{A}_{22}=1,\text{A}_{23}=1,$ $\text{A}_{31}=-4,\text{A}_{32}=2$ and $A_{33}=3$
$\therefore\ \ \text{adj (A)}=\begin{vmatrix}-3&2&2\\-2&1&1\\-4&2&3\end{vmatrix}^\text{T}=\begin{vmatrix}-3&-2&-4\\2&1&2\\2&1&3\end{vmatrix}$
$\therefore\ \ \text{A}^{-1}=\frac{\text{adj A}}{|\text{A}|}$
$=\frac{1}{1}\begin{vmatrix}-3&-2&-4\\2&1&2\\2&1&3\end{vmatrix}$
$\Rightarrow\ \text{A}^{-1}=\begin{vmatrix}-3&-2&-4\\2&1&2\\2&1&3\end{vmatrix}\ \ \dots(\text{i})$
Also, we have the system of linear equations as
$x - 2y = 10$
$2x - y - z = 8$
and $-2y + z = 7$
In the from of $CX = D$.
$\begin{bmatrix}1&2&0\\2&-1&-1\\0&-2&1\end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}10\\8\\7\end{bmatrix}$
where, $\text{C}=\begin{bmatrix}1&-2&0\\2&-1&-1\\0&-2&1\end{bmatrix}\text{X}=\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}$ and $D=\begin{bmatrix}10\\8\\7\end{bmatrix}$
We know that, $\big(\text{A}^\text{T}\big)^{-1}=\big(\text{A}^{-1}\big)^\text{T}$
$\therefore\ \text{C}^{\text{T}}=\begin{vmatrix}1&2&0\\-2&-1&-2\\0&-1&1\end{vmatrix}=\text{A} [$using Eq.$(i)]$
$\therefore X = C^{-1} D$
$\Rightarrow\ \begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}-3&2&2\\-2&1&1\\-4&2&3\end{bmatrix}\begin{bmatrix}10\\8\\7\end{bmatrix}$
$\begin{bmatrix}-30+16+14\\-20+8+7\\-40+16+21\end{bmatrix}=\begin{bmatrix}0\\-5\\-3\end{bmatrix}$
$\therefore x = 0, y = -5$ and $z = -3$
View full question & answer→Question 2415 Marks
Prove that $\begin{vmatrix}a^2&bc&ac+c^2\\a^2+ab&b^2&ac\\ab&b^2+bc&c^2\end{vmatrix}=4a^2b^2c^2$
Answer$\triangle=\begin{vmatrix}a^2&bc&ac+c^2\\a^2+ab&b^2&ac\\ab&b^2+bc&c^2\end{vmatrix}$
Taking out common factors $a, b,$ and $c$ from $C_1, C_2,$ and $C_3$_, we have:
$\triangle=abc\begin{vmatrix}a&c&a+c\\a+b&b&a\\b&b+c&c\end{vmatrix}$
Applying $R_2 \rightarrow R_2 + R_1$ and $R_3 \rightarrow R_3 - R_1$, we have:
$\triangle=abc\begin{vmatrix}a&c&a+c\\b&b-c&-c\\b-a&b&-a\end{vmatrix}$
Applying $R_2 \rightarrow R_2 + R_1$ we have:
$\triangle=abc\begin{vmatrix}a&c&a+c\\a+b&b&a\\b-a&b&-a\end{vmatrix}$
Applying $R_3 \rightarrow R_3 + R_2,$ we have:
$\triangle=abc\begin{vmatrix}a&c&a+c\\a+b&b&a\\2b&2b&0\end{vmatrix}$
$\triangle=2ab^2c\begin{vmatrix}a&c&a+c\\a+b&b&a\\1&1&0\end{vmatrix}$
$\triangle=2ab^2c\begin{vmatrix}a&c-a&a+c\\a+b&-a&a\\1&0&0\end{vmatrix}$
Expanding along $R_3$, we have:
$\triangle = 2ab^2c [a(c -a) + a(a + c)]$
$= 2ab^2c [ac - a^2 + a^{2 }+ ac]$
$= 2ab^2c (2ac)$
$= 4a^2b^2c^2$
Hence, the given result is proved.
View full question & answer→Question 2425 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 2435 Marks
Solve the following systems of linear equations by cramer's rule:
x - 4y - z = 11,
2x - 5y + 2z = 39,
-3x + 2y + z = 1
AnswerGiven, x - 4y - z = 11
2x - 5y + 2z = 39
-3x + 2y + z = 1
$\text{D}=\begin{vmatrix}1&-4&-1\\2&-5&2\\-3&2&1\end{vmatrix}$
$=1(-5-4)-(-4)(2+6)+(-1)(4-15)$
$=1(-9)-(-4)(8)+(-1)(-11)=34$
$\text{D}_1=\begin{vmatrix}11&-4&-1\\39&-5&2\\1&2&1\end{vmatrix}$
$=11(-5-4)-(-4)(39-2)+(-1)(78+5)$
$=11(-9)-(-4)(37)+(-1)(83)=-34$
$\text{D}_2=\begin{vmatrix}1&11&-1\\2&39&2\\-3&1&1\end{vmatrix}$
$=1(39-2)-11(2+6)+(-1)(2+117)$
$=1(37)-11(8)+(-1)(119)=-170$
$\text{D}_3=\begin{vmatrix}1&-4&11\\2&-5&39\\-3&2&1\end{vmatrix}$
$=1(-5-78)-(-4)(2+117)+11(4-15)$
$=1(-83)-(-4)(119)+11(-11)=272$
Now,
$\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{-34}{34}=-1$
$\text{y}=\frac{\text{D}_2}{\text{D}}=\frac{-170}{34}=-5$
$\text{z}=\frac{\text{D}_3}{\text{D}}=\frac{272}{34}=8$
$\therefore\text{x}=-1,\text{ y}=-5$ and $\text{z}=8$
View full question & answer→Question 2445 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 2455 Marks
If $\text{A}=\begin{bmatrix} 1 & -2 & 3 \\ 0 & -1 & 4 \\ -2 & 2 & 1 \end{bmatrix},$ find $(A^T)^{-1}.$
AnswerWe know that $(AT)^{-1} = (A^{-1})T.$
$\text{A}=\begin{bmatrix} 1 & -2 & 3 \\ 0 & -1 & 4 \\ -2 & 2 & 1 \end{bmatrix}$
$\text{A}^{-1}=\frac{1}{|\text{A}|}\text{ adj A}$
Now,
$\text{A}=\begin{vmatrix} 1 & -2 & 3 \\ 0 & -1 & 4 \\ -2 & 2 & 1 \end{vmatrix}$
$= 1(-1 - 8) - 2(-8 + 3)$
$= -9 + 10$
$= 1$
Now, to find Adj $A$
$A_{11} = (-1)^{1+1} (-9) = -9, A_{21} = (-1)^{2+1} (-8) = 8, A_{31} = (-1)^{3+1} (-5) = -5$
$A_{12} = (-1)^{1+2} (8) = -8, A_{22} = (-1)^{2+2} (7) = 7, A_{32} = (-1)^{3+2} (4) = -4$
$A_{13} = (-1)^{1+3} (-2) = -2, A_{23} = (-1)^{2+3} (-2) = 2, A_{33} = (-1)^{3+3} (-1) = -1$
Therefore,
$\text{A}=\begin{bmatrix} 1 & -2 & 3 \\ 0 & -1 & 4 \\ -2 & 2 & 1 \end{bmatrix}$
Thus,
$\text{A}^{-1}=\begin{bmatrix} -9 & 8 &-5 \\ -8 & 7 & -4 \\ -2 & 2 & -1 \end{bmatrix}$
$(\text{A}^\text{T})^{-1}=(\text{A}^{-1})^\text{T}$
$=\begin{bmatrix} -9 & 8 &-5 \\ -8 & 7 & -4 \\ -2 & 2 & -1 \end{bmatrix}^\text{T}$
$=\begin{bmatrix} -9 & 8 &-2 \\ -8 & 7 & 2 \\ -5 & 4 & -1 \end{bmatrix}$
Hence, $(\text{A}^\text{T})^{-1}=\begin{bmatrix} -9 & -8 &-2 \\ 8 & 7 & 2 \\ -5 & -4 & -1 \end{bmatrix}$
View full question & answer→Question 2465 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,
$\text{A}=\begin{bmatrix}1&1&-6\\1&-1&2\\-3&1&2\end{bmatrix},\text{X}=\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}$ 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,
$\text{A}=\begin{bmatrix}1&1\\1&-1\end{bmatrix},\text{X}=\begin{bmatrix}\text{x}\\\text{y}\end{bmatrix}$ 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 2475 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 2485 Marks
Evaluate the following:
$\begin{vmatrix}0&\text{xy}^2&\text{xz}^2\\\text{x}^2\text{y}&0&\text{yz}^2\\\text{x}^2\text{z}&\text{zy}^2&0\end{vmatrix}$
AnswerLet $\triangle=\begin{vmatrix}0&\text{xy}^2&\text{xz}^2\\\text{x}^2\text{y}&0&\text{yz}^2\\\text{x}^2\text{z}&\text{zy}^2&0\end{vmatrix}$
$\triangle=\begin{vmatrix}0&\text{xy}^2&\text{xz}^2\\\text{x}^2\text{y}&0&\text{yz}^2\\\text{x}^2\text{z}&\text{zy}^2&0\end{vmatrix}$
$=\text{x}^2\text{y}^2\text{z}^2\begin{vmatrix}0&\text{x}&\text{x}\\\text{y}&0&\text{y}\\\text{z}&\text{z}&0\end{vmatrix}$
$[$Taking $x^2$ common from from $C_1, y^2$ common from $C_2$ and $z^2$ common from $C_3]$
$=\text{x}^3\text{y}^3\text{z}^3\begin{vmatrix}0&0&1\\1&-1&1\\1&1&0\end{vmatrix}$
$[$Applying $C_2 \rightarrow C_2 - C_3]$
$=\text{x}^3\text{y}^3\text{z}^3(1+1)$ [Expanding along first row]
$=2\text{x}^3\text{y}^3\text{z}^3$
$\therefore\ \triangle=2\text{x}^3\text{y}^3\text{z}^3$
View full question & answer→Question 2495 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 2505 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 $A [a_{ij}].$ 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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