Class 12: Determinants (Test Problems)

$\displaystyle \textbf{Question 1. }\text{Let }A\text{ be the non-singular square matrix of order }3\times3,\text{ then }|\text{adj}A|\text{ is equal to}$
$\displaystyle \text{(a) }|A|\qquad \text{(b) }|A|^2\qquad \text{(c) }|A|^3\qquad \text{(d) }3|A|$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For a square matrix of order }n,\ |\text{adj}A|=|A|^{n-1}$
$\displaystyle \text{Here, }n=3$
$\displaystyle \therefore |\text{adj}A|=|A|^{3-1}=|A|^2$
$\displaystyle \therefore \text{Correct option is (b).}$
$\\$

$\displaystyle \textbf{Question 2. }\text{If }A\text{ is an invertible matrix of order }2,\text{ then }\det(A^{-1})\text{ is equal to}$
$\displaystyle \text{(a) }\det(A)\qquad \text{(b) }\frac{1}{\det(A)}\qquad \text{(c) }1\qquad \text{(d) zero}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We know that }\det(A^{-1})=\frac{1}{\det(A)}$
$\displaystyle \therefore \text{Correct option is (b).}$
$\\$

$\displaystyle \textbf{Question 3. }\text{A square matrix }A\text{ is said to be non-singular, if}$
$\displaystyle \text{(a) }|A|=0\qquad \text{(b) }|A|\ne0\qquad \text{(c) }|A|=-1\qquad \text{(d) }|A|=1$
$\displaystyle \text{Answer:}$
$\displaystyle \text{A square matrix }A\text{ is non-singular if }|A|\ne0.$
$\displaystyle \therefore \text{Correct option is (b).}$
$\\$

$\displaystyle \textbf{Question 4. }\text{If }A\text{ and }B\text{ are square matrices of same order, then}$
$\displaystyle \text{(a) }|AB|=|A|\cdot|B|\qquad \text{(b) }|AB|\ne|A|\cdot|B|$
$\displaystyle \text{(c) }|AB|=\frac{|A|}{|B|},\ |B|\ne0\qquad \text{(d) }|AB|=\frac{|B|}{|A|},\ |A|\ne0$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For square matrices of same order, }|AB|=|A|\cdot|B|.$
$\displaystyle \therefore \text{Correct option is (a).}$
$\\$

$\displaystyle \textbf{Question 5. }\text{The adjoint of the matrix }A=\begin{bmatrix}1&2\\3&4\end{bmatrix}\text{ is}$
$\displaystyle \text{(a) }\begin{bmatrix}4&2\\3&1\end{bmatrix}\qquad \text{(b) }\begin{bmatrix}-4&2\\3&-1\end{bmatrix}$
$\displaystyle \text{(c) }\begin{bmatrix}4&-2\\-3&1\end{bmatrix}\qquad \text{(d) }\begin{bmatrix}1&-2\\-3&4\end{bmatrix}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For }A=\begin{bmatrix}a&b\\c&d\end{bmatrix},\ \text{adj}A=\begin{bmatrix}d&-b\\-c&a\end{bmatrix}$
$\displaystyle \therefore \text{adj}A=\begin{bmatrix}4&-2\\-3&1\end{bmatrix}$
$\displaystyle \therefore \text{Correct option is (c).}$
$\\$

$\displaystyle \textbf{Question 6. }\text{If }A=\begin{bmatrix}2&2\\4&0\end{bmatrix}\text{ and }B=\begin{bmatrix}1&1\\2&0\end{bmatrix},\text{ then}$
$\displaystyle \text{(a) }|A|=|B|\qquad \text{(b) }|A|=2|B|$
$\displaystyle \text{(c) }|A|=2^2|B|\qquad \text{(d) }|A|=-|B|$
$\displaystyle \text{Answer:}$
$\displaystyle |A|=\begin{vmatrix}2&2\\4&0\end{vmatrix}=2\cdot0-2\cdot4=-8$
$\displaystyle |B|=\begin{vmatrix}1&1\\2&0\end{vmatrix}=1\cdot0-1\cdot2=-2$
$\displaystyle \therefore |A|=4|B|=2^2|B|$
$\displaystyle \therefore \text{Correct option is (c).}$
$\\$

$\displaystyle \textbf{Question 7. }\text{If }\begin{vmatrix}2&4\\5&1\end{vmatrix}=\begin{vmatrix}2x&4\\6&x\end{vmatrix},\text{ then }x\text{ is}$
$\displaystyle \text{(a) }\pm\sqrt3\qquad \text{(b) }\pm\sqrt2\qquad \text{(c) }1\qquad \text{(d) }0$
$\displaystyle \text{Answer:}$
$\displaystyle \begin{vmatrix}2&4\\5&1\end{vmatrix}=2\cdot1-4\cdot5=-18$
$\displaystyle \begin{vmatrix}2x&4\\6&x\end{vmatrix}=2x^2-24$
$\displaystyle \therefore 2x^2-24=-18$
$\displaystyle \therefore 2x^2=6$
$\displaystyle \therefore x^2=3$
$\displaystyle \therefore x=\pm\sqrt3$
$\displaystyle \therefore \text{Correct option is (a).}$
$\\$

$\displaystyle \textbf{Question 8. }\text{The value of }\begin{vmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{vmatrix}\text{ is}$
$\displaystyle \text{(a) }0\qquad \text{(b) }1\qquad \text{(c) }2\qquad \text{(d) }5$
$\displaystyle \text{Answer:}$
$\displaystyle \begin{vmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{vmatrix}$
$\displaystyle =\cos^2\theta-(-\sin\theta)(\sin\theta)$
$\displaystyle =\cos^2\theta+\sin^2\theta=1$
$\displaystyle \therefore \text{Correct option is (b).}$
$\\$

$\displaystyle \textbf{Question 9. }\text{If }A\text{ is a skew-symmetric matrix of odd order }n,\text{ then}$
$\displaystyle \text{(a) }|A|=0\qquad \text{(b) }|A|=-1$
$\displaystyle \text{(c) }|A|=|A'|\qquad \text{(d) None of these}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For a skew-symmetric matrix, }A'=-A.$
$\displaystyle \therefore |A'|=|-A|=(-1)^n|A|$
$\displaystyle \text{Since }n\text{ is odd, }|A'|=-|A|$
$\displaystyle \text{But }|A'|=|A|$
$\displaystyle \therefore |A|=-|A|$
$\displaystyle \therefore 2|A|=0$
$\displaystyle \therefore |A|=0$
$\displaystyle \therefore \text{Correct option is (a).}$
$\\$

$\displaystyle \textbf{Question 10. }\text{Let }A\text{ be a square matrix of order }3\times3.\text{ Write the value of }|2A|, \\ \text{ where }|A|=4.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{If }A\text{ is a square matrix of order }n,\text{ then }|kA|=k^n|A|.$
$\displaystyle \text{Here, }n=3,\ k=2\text{ and }|A|=4.$
$\displaystyle \therefore |2A|=2^3|A|$
$\displaystyle =8\times4=32$
$\\$

$\displaystyle \textbf{Question 11. }\text{Evaluate }\begin{vmatrix}\cos15^\circ&\sin15^\circ\\ \sin75^\circ&\cos75^\circ\end{vmatrix}.$
$\displaystyle \text{Answer:}$
$\displaystyle \begin{vmatrix}\cos15^\circ&\sin15^\circ\\ \sin75^\circ&\cos75^\circ\end{vmatrix}$
$\displaystyle =\cos15^\circ\cos75^\circ-\sin15^\circ\sin75^\circ$
$\displaystyle =\cos(15^\circ+75^\circ)$
$\displaystyle =\cos90^\circ=0$
$\\$

$\displaystyle \textbf{Question 12. }\text{Evaluate }\begin{vmatrix}a+ib&c+id\\-c+id&a-ib\end{vmatrix}.$
$\displaystyle \text{Answer:}$
$\displaystyle \begin{vmatrix}a+ib&c+id\\-c+id&a-ib\end{vmatrix}$
$\displaystyle =(a+ib)(a-ib)-(c+id)(-c+id)$
$\displaystyle =a^2+b^2-\left[-c^2-d^2\right]$
$\displaystyle =a^2+b^2+c^2+d^2$
$\\$

$\displaystyle \textbf{Question 13. }\text{Find the area of triangle whose vertices are }(-2,-3),(3,2)\text{ and }(-1,-8).$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Area}=\frac{1}{2}\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|$
$\displaystyle =\frac{1}{2}\left|-2(2+8)+3(-8+3)+(-1)(-3-2)\right|$
$\displaystyle =\frac{1}{2}\left|-20-15+5\right|$
$\displaystyle =\frac{1}{2}|-30|=15$
$\displaystyle \therefore \text{Area of triangle}=15\text{ square units.}$
$\\$

$\displaystyle \textbf{Question 14. }\text{Find the equation of the line joining }(1,2)\text{ and }(3,6)\text{ using determinants.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Equation of line through }(1,2)\text{ and }(3,6)\text{ is}$
$\displaystyle \begin{vmatrix}x&y&1\\1&2&1\\3&6&1\end{vmatrix}=0$
$\displaystyle x(2-6)-y(1-3)+1(6-6)=0$
$\displaystyle -4x+2y=0$
$\displaystyle \therefore y=2x$
$\displaystyle \therefore \text{Required equation is }2x-y=0.$
$\\$

$\displaystyle \textbf{Question 15. }\text{Find the value of }k,\text{ if the points }(k+1,1),\ (2k+1,3)$
$\displaystyle \text{and }(2k+2,2k)\text{ are collinear.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For collinear points, area of triangle }=0.$
$\displaystyle \therefore \begin{vmatrix}k+1&1&1\\2k+1&3&1\\2k+2&2k&1\end{vmatrix}=0$
$\displaystyle (k-2)(2k+1)=0$
$\displaystyle \therefore k=2\text{ or }k=-\frac{1}{2}$
$\\$

$\displaystyle \textbf{Question 16. }\text{Find the minors of the diagonal elements of the determinant}$
$\displaystyle \begin{vmatrix}1&i&-1\\-i&1&i\\1&-i&i\end{vmatrix}.$
$\displaystyle \text{Answer:}$
$\displaystyle M_{11}=\begin{vmatrix}1&i\\-i&i\end{vmatrix}=i-i(-i)=i-1$
$\displaystyle M_{22}=\begin{vmatrix}1&-1\\1&i\end{vmatrix}=i+1$
$\displaystyle M_{33}=\begin{vmatrix}1&i\\-i&1\end{vmatrix}=1-i(-i)=0$
$\displaystyle \therefore \text{Minors of diagonal elements are }i-1,\ i+1,\ 0.$
$\\$

$\displaystyle \textbf{Question 17. }\text{Using cofactors of elements of third row, evaluate}$
$\displaystyle \Delta=\begin{vmatrix}1&x&y+z\\1&y&z+x\\1&z&x+y\end{vmatrix}.$
$\displaystyle \text{Answer:}$
$\displaystyle \Delta=1\begin{vmatrix}x&y+z\\y&z+x\end{vmatrix}-z\begin{vmatrix}1&y+z\\1&z+x\end{vmatrix}+(x+y)\begin{vmatrix}1&x\\1&y\end{vmatrix}$
$\displaystyle =1[x(z+x)-y(y+z)]-z[(z+x)-(y+z)]+(x+y)(y-x)$
$\displaystyle =xz+x^2-y^2-yz-z(x-y)+y^2-x^2$
$\displaystyle =xz+x^2-y^2-yz-xz+yz+y^2-x^2$
$\displaystyle =0$
$\displaystyle \therefore \Delta=0$
$\\$

$\displaystyle \textbf{Question 18. }\text{Find the inverse of the matrix }\begin{bmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{bmatrix}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }A=\begin{bmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{bmatrix}$
$\displaystyle |A|=\cos^2\theta+\sin^2\theta=1$
$\displaystyle \text{adj}A=\begin{bmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{bmatrix}$
$\displaystyle \therefore A^{-1}=\frac{1}{|A|}\text{adj}A$
$\displaystyle =\begin{bmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{bmatrix}$
$\\$

$\displaystyle \textbf{Question 19. }\text{Solve the system of linear equations using matrix method.}$
$\displaystyle 2x-y=-2\text{ and }3x+4y=3$
$\displaystyle \text{Answer:}$
$\displaystyle \begin{bmatrix}2&-1\\3&4\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}-2\\3\end{bmatrix}$
$\displaystyle A=\begin{bmatrix}2&-1\\3&4\end{bmatrix},\quad B=\begin{bmatrix}-2\\3\end{bmatrix}$
$\displaystyle |A|=2\cdot4-(-1)\cdot3=11$
$\displaystyle A^{-1}=\frac{1}{11}\begin{bmatrix}4&1\\-3&2\end{bmatrix}$
$\displaystyle \therefore \begin{bmatrix}x\\y\end{bmatrix}=A^{-1}B$
$\displaystyle =\frac{1}{11}\begin{bmatrix}4&1\\-3&2\end{bmatrix}\begin{bmatrix}-2\\3\end{bmatrix}$
$\displaystyle =\frac{1}{11}\begin{bmatrix}-5\\12\end{bmatrix}$
$\displaystyle \therefore x=-\frac{5}{11},\ y=\frac{12}{11}$
$\\$

$\displaystyle \textbf{Question 20. }\text{If area of a triangle is }35\text{ sq units with vertices }(2,-6),\ (5,4)\text{ and } \\ (k,4),\text{ then find the values of }k.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Area}=\frac{1}{2}\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|$
$\displaystyle 35=\frac{1}{2}\left|2(4-4)+5(4+6)+k(-6-4)\right|$
$\displaystyle 35=\frac{1}{2}|50-10k|$
$\displaystyle 70=|50-10k|$
$\displaystyle |5-k|=7$
$\displaystyle \therefore 5-k=7\text{ or }5-k=-7$
$\displaystyle \therefore k=-2\text{ or }12$
$\\$

$\displaystyle \textbf{Question 21. }\text{Using determinants, find the area of the triangle whose vertices are }$
$\displaystyle (1,4),\ (2,3)\text{ and }(-5,-3). \text{ Are the given points collinear?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Area}=\frac{1}{2}\left|\begin{vmatrix}1&4&1\\2&3&1\\-5&-3&1\end{vmatrix}\right|$
$\displaystyle =\frac{1}{2}\left|1(3+3)-4(2+5)+1(-6+15)\right|$
$\displaystyle =\frac{1}{2}|6-28+9|$
$\displaystyle =\frac{1}{2}\times13$
$\displaystyle =\frac{13}{2}\text{ sq units}$
$\displaystyle \text{Since the area is non-zero, the points are not collinear.}$
$\\$

$\displaystyle \textbf{Question 22. }\text{Find the inverse of the matrix }\begin{bmatrix}1&0&0\\0&\cos\alpha&\sin\alpha\\0&\sin\alpha&-\cos\alpha\end{bmatrix}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }A=\begin{bmatrix}1&0&0\\0&\cos\alpha&\sin\alpha\\0&\sin\alpha&-\cos\alpha\end{bmatrix}$
$\displaystyle A^{2}=\begin{bmatrix}1&0&0\\0&\cos\alpha&\sin\alpha\\0&\sin\alpha&-\cos\alpha\end{bmatrix}\begin{bmatrix}1&0&0\\0&\cos\alpha&\sin\alpha\\0&\sin\alpha&-\cos\alpha\end{bmatrix}$
$\displaystyle =\begin{bmatrix}1&0&0\\0&\cos^{2}\alpha+\sin^{2}\alpha&0\\0&0&\sin^{2}\alpha+\cos^{2}\alpha\end{bmatrix}=I$
$\displaystyle \therefore A^{-1}=A$
$\displaystyle \therefore A^{-1}=\begin{bmatrix}1&0&0\\0&\cos\alpha&\sin\alpha\\0&\sin\alpha&-\cos\alpha\end{bmatrix}$
$\\$

$\displaystyle \textbf{Question 22. }\text{Show that }A=\begin{bmatrix}2&-3\\3&4\end{bmatrix}\text{ satisfies the equation } \\ x^{2}-6x+17=0.\text{ Hence, find }A^{-1}.$
$\displaystyle \text{Answer:}$
$\displaystyle A^{2}=\begin{bmatrix}2&-3\\3&4\end{bmatrix}\begin{bmatrix}2&-3\\3&4\end{bmatrix}=\begin{bmatrix}-5&-18\\18&7\end{bmatrix}$
$\displaystyle A^{2}-6A+17I$
$\displaystyle =\begin{bmatrix}-5&-18\\18&7\end{bmatrix}-6\begin{bmatrix}2&-3\\3&4\end{bmatrix}+17\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$\displaystyle =\begin{bmatrix}0&0\\0&0\end{bmatrix}$
$\displaystyle \therefore A\text{ satisfies }x^{2}-6x+17=0.$
$\displaystyle \text{Now, }A^{2}-6A+17I=0$
$\displaystyle \therefore A-6I+17A^{-1}=0$
$\displaystyle \therefore A^{-1}=\frac{1}{17}(6I-A)$
$\displaystyle =\frac{1}{17}\left[\begin{bmatrix}6&0\\0&6\end{bmatrix}-\begin{bmatrix}2&-3\\3&4\end{bmatrix}\right]$
$\displaystyle =\frac{1}{17}\begin{bmatrix}4&3\\-3&2\end{bmatrix}$
$\\$

$\displaystyle \textbf{Question 23. }\text{If }A=\begin{bmatrix}1&\tan x\\-\tan x&1\end{bmatrix},\text{ then show that}$
$\displaystyle A^{T}A^{-1}=\begin{bmatrix}\cos2x&-\sin2x\\\sin2x&\cos2x\end{bmatrix}.$
$\displaystyle \text{Answer:}$
$\displaystyle A^{T}=\begin{bmatrix}1&-\tan x\\\tan x&1\end{bmatrix}$
$\displaystyle A^{-1}=\frac{1}{1+\tan^{2}x}\begin{bmatrix}1&-\tan x\\\tan x&1\end{bmatrix}$
$\displaystyle \therefore A^{T}A^{-1}=\frac{1}{1+\tan^{2}x}\begin{bmatrix}1&-\tan x\\\tan x&1\end{bmatrix}\begin{bmatrix}1&-\tan x\\\tan x&1\end{bmatrix}$
$\displaystyle =\frac{1}{1+\tan^{2}x}\begin{bmatrix}1-\tan^{2}x&-2\tan x\\2\tan x&1-\tan^{2}x\end{bmatrix}$
$\displaystyle =\begin{bmatrix}\cos2x&-\sin2x\\\sin2x&\cos2x\end{bmatrix}$
$\displaystyle \text{Hence proved.}$
$\\$

$\displaystyle \textbf{Question 24. }\text{For positive numbers }x,\ y\text{ and }z,\text{ show that}$
$\displaystyle \begin{vmatrix}1&\log_x y&\log_x z\\ \log_y x&1&\log_y z\\ \log_z x&\log_z y&1\end{vmatrix}=0$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }\log_x y=a\text{ and }\log_y z=b.$
$\displaystyle \therefore \log_x z=ab,\quad \log_y x=\frac{1}{a},\quad \log_z y=\frac{1}{b},\quad \log_z x=\frac{1}{ab}$
$\displaystyle \therefore \begin{vmatrix}1&a&ab\\ \frac{1}{a}&1&b\\ \frac{1}{ab}&\frac{1}{b}&1\end{vmatrix}$
$\displaystyle =1(1-1)-a\left(\frac{1}{a}-\frac{1}{a}\right)+ab\left(\frac{1}{ab}-\frac{1}{ab}\right)$
$\displaystyle =0$
$\displaystyle \text{Hence proved.}$
$\\$

$\displaystyle \textbf{Question 25. }\text{Using the matrix method, solve the following system of equations}$
$\displaystyle x-y+2z=7,\ 3x+4y-5z=-5\text{ and }2x-y+3z=12.$
$\displaystyle \text{Answer:}$
$\displaystyle \begin{bmatrix}1&-1&2\\3&4&-5\\2&-1&3\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}7\\-5\\12\end{bmatrix}$
$\displaystyle \text{Solving by matrix method, we get}$
$\displaystyle \begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}2\\1\\3\end{bmatrix}$
$\displaystyle \therefore x=2,\ y=1,\ z=3$
$\\$

$\displaystyle \textbf{Question 26. }\text{Using the matrix method, solve the following system of equations}$
$\displaystyle 4x+3y+2z=60,\ x+2y+3z=45\text{ and }6x+2y+3z=70.$
$\displaystyle \text{Answer:}$
$\displaystyle \begin{bmatrix}4&3&2\\1&2&3\\6&2&3\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}60\\45\\70\end{bmatrix}$
$\displaystyle \text{Solving by matrix method, we get}$
$\displaystyle \begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}5\\8\\8\end{bmatrix}$
$\displaystyle \therefore x=5,\ y=8,\ z=8$
$\\$

$\displaystyle \textbf{Question 27. }\text{Use product }\begin{bmatrix}1&-1&2\\0&2&-3\\3&-2&4\end{bmatrix},\begin{bmatrix}-2&0&1\\9&2&-3\\6&1&-2\end{bmatrix}\text{ to solve the}$
$\displaystyle \text{system of equations }x-y+2z=1,\ 2y-3z=1\text{ and }3x-2y+4z=2.$
$\displaystyle \text{Answer:}$
$\displaystyle \begin{bmatrix}1&-1&2\\0&2&-3\\3&-2&4\end{bmatrix}\begin{bmatrix}-2&0&1\\9&2&-3\\6&1&-2\end{bmatrix}=I_3$
$\displaystyle \therefore A^{-1}=\begin{bmatrix}-2&0&1\\9&2&-3\\6&1&-2\end{bmatrix}$
$\displaystyle AX=B,\text{ where }B=\begin{bmatrix}1\\1\\2\end{bmatrix}$
$\displaystyle \therefore X=A^{-1}B$
$\displaystyle =\begin{bmatrix}-2&0&1\\9&2&-3\\6&1&-2\end{bmatrix}\begin{bmatrix}1\\1\\2\end{bmatrix}$
$\displaystyle =\begin{bmatrix}0\\5\\3\end{bmatrix}$
$\displaystyle \therefore x=0,\ y=5,\ z=3$
$\\$

$\displaystyle \textbf{Question 28. }\text{If }A=\begin{bmatrix}3&-4&2\\2&3&5\\1&0&1\end{bmatrix},\text{ then find }A^{-1}\text{ and hence solve the}$
$\displaystyle \text{system of equations }3x-4y+2z=-1,\ 2x+3y+5z=7\text{ and }x+z=2.$
$\displaystyle \text{Answer:}$
$\displaystyle A^{-1}=\begin{bmatrix}-\frac{1}{3}&-\frac{4}{9}&\frac{26}{9}\\-\frac{1}{3}&-\frac{1}{9}&\frac{11}{9}\\\frac{1}{3}&\frac{4}{9}&-\frac{17}{9}\end{bmatrix}$
$\displaystyle AX=B,\text{ where }B=\begin{bmatrix}-1\\7\\2\end{bmatrix}$
$\displaystyle \therefore X=A^{-1}B$
$\displaystyle =\begin{bmatrix}-\frac{1}{3}&-\frac{4}{9}&\frac{26}{9}\\-\frac{1}{3}&-\frac{1}{9}&\frac{11}{9}\\\frac{1}{3}&\frac{4}{9}&-\frac{17}{9}\end{bmatrix}\begin{bmatrix}-1\\7\\2\end{bmatrix}$
$\displaystyle =\begin{bmatrix}3\\2\\-1\end{bmatrix}$
$\displaystyle \therefore x=3,\ y=2,\ z=-1$
$\\$

$\displaystyle \textbf{Question 29. }\text{Find the inverse of the matrix }A=\begin{bmatrix}a&b\\c&\frac{1+bc}{a}\end{bmatrix}\text{ and}$
$\displaystyle \text{show that }aA^{-1}=(a^2+bc+1)I-aA.$
$\displaystyle \text{Answer:}$
$\displaystyle |A|=a\cdot\frac{1+bc}{a}-bc=1+bc-bc=1$
$\displaystyle \therefore A^{-1}=\begin{bmatrix}\frac{1+bc}{a}&-b\\-c&a\end{bmatrix}$
$\displaystyle \therefore aA^{-1}=a\begin{bmatrix}\frac{1+bc}{a}&-b\\-c&a\end{bmatrix}=\begin{bmatrix}1+bc&-ab\\-ac&a^2\end{bmatrix}$
$\displaystyle \text{Now, }(a^2+bc+1)I-aA$
$\displaystyle =(a^2+bc+1)\begin{bmatrix}1&0\\0&1\end{bmatrix}-a\begin{bmatrix}a&b\\c&\frac{1+bc}{a}\end{bmatrix}$
$\displaystyle =\begin{bmatrix}a^2+bc+1&0\\0&a^2+bc+1\end{bmatrix}-\begin{bmatrix}a^2&ab\\ac&1+bc\end{bmatrix}$
$\displaystyle =\begin{bmatrix}1+bc&-ab\\-ac&a^2\end{bmatrix}$
$\displaystyle \therefore aA^{-1}=(a^2+bc+1)I-aA$
$\displaystyle \text{Hence proved.}$
$\\$

$\displaystyle \textbf{Question 30. }\text{If }A=\begin{bmatrix}2&3&1\\1&2&2\\-3&1&-1\end{bmatrix},\text{ find }A^{-1}\text{ and hence solve the}$
$\displaystyle \text{system of equations }2x+y-3z=13,\ 3x+2y+z=4\text{ and }x+2y-z=8.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }A=\begin{bmatrix}2&3&1\\1&2&2\\-3&1&-1\end{bmatrix}$
$\displaystyle |A|=2\begin{vmatrix}2&2\\1&-1\end{vmatrix}-3\begin{vmatrix}1&2\\-3&-1\end{vmatrix}+\begin{vmatrix}1&2\\-3&1\end{vmatrix}$
$\displaystyle =2(-4)-3(5)+7=-16$
$\displaystyle \text{Cofactor matrix of }A=\begin{bmatrix}-4&-5&7\\4&1&-11\\4&-3&7\end{bmatrix}$
$\displaystyle \therefore \ \text{adj}(A)=\begin{bmatrix}-4&4&4\\-5&1&-3\\7&-11&7\end{bmatrix}$
$\displaystyle A^{-1}=\frac{1}{|A|}\text{adj}(A)=\frac{-1}{16}\begin{bmatrix}-4&4&4\\-5&1&-3\\7&-11&7\end{bmatrix}$
$\displaystyle A^{-1}=\begin{bmatrix}\frac14&-\frac14&-\frac14\\\frac5{16}&-\frac1{16}&\frac3{16}\\-\frac7{16}&\frac{11}{16}&-\frac7{16}\end{bmatrix}$
$\displaystyle \text{The given system is }A^T\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}13\\4\\8\end{bmatrix}$
$\displaystyle \therefore \begin{bmatrix}x\\y\\z\end{bmatrix}=(A^T)^{-1}\begin{bmatrix}13\\4\\8\end{bmatrix}=(A^{-1})^T\begin{bmatrix}13\\4\\8\end{bmatrix}$
$\displaystyle =\begin{bmatrix}\frac14&\frac5{16}&-\frac7{16}\\-\frac14&-\frac1{16}&\frac{11}{16}\\-\frac14&\frac3{16}&-\frac7{16}\end{bmatrix}\begin{bmatrix}13\\4\\8\end{bmatrix}$
$\displaystyle =\begin{bmatrix}1\\2\\-3\end{bmatrix}$
$\displaystyle \text{Hence, }x=1,\ y=2,\ z=-3.$