\(\left( AB \right) ^{ -1 }=\left[ \begin{matrix} 9 & -3 & 5 \\ -2 & 1 & 0 \\ 1 & 0 & 2 \end{matrix} \right] \)

\(A=\left[\begin{array}{ll} 3 & 2 \\ 1 & 1 \end{array}\right]\)
\(\therefore A^{2}=\left[\begin{array}{ll} 3 & 2 \\ 1 & 1 \end{array}\right]\left[\begin{array}{ll} 3 & 2 \\ 1 & 1 \end{array}\right]=\left[\begin{array}{ll} 9+2 & 6+2 \\ 3+1 & 2+1 \end{array}\right]=\left[\begin{array}{ll} 11 & 8 \\ 4 & 3 \end{array}\right]\)
\(\text { Now, }\)
\(A^{2}+a A+b I=O\)
\(\Rightarrow(A A) A^{-1}+a A A^{-1}+b I A^{-1}=O \quad\left[\text { Post-multiplying by } A^{-1} \text { as }|A| \neq 0\right]\)
\(\Rightarrow A\left(A A^{-1}\right)+a I+b\left(I A^{-1}\right)=O\)
\(\Rightarrow A I+a I+b A^{-1}=O\)
\(\Rightarrow A+a I=-b A^{-1} \)
\(\Rightarrow A^{-1}=-\frac{1}{b}(A+a I)\)
\(\text {Now, }A^{-1}=\frac{1}{|A|} a d j A=\frac{1}{1}\left[\begin{array}{cc} 1 & -2 \\ -1 & 3 \end{array}\right]=\left[\begin{array}{cc} 1 & -2 \\ -1 & 3 \end{array}\right]\)
\(\text {We have: }\left[\begin{array}{cc} 1 & -2 \\ -1 & 3 \end{array}\right]=-\frac{1}{b}\left(\left[\begin{array}{ll} 3 & 2 \\ 1 & 1 \end{array}\right]+\left[\begin{array}{cc} a & 0 \\ 0 & a \end{array}\right]\right)=-\frac{1}{b}\left[\begin{array}{cc} 3+a & 2 \\ 1 & 1+a \end{array}\right]=\left[\begin{array}{cc} \frac{-3-a}{b} & -\frac{2}{b} \\ -\frac{1}{b} & \frac{-1-a}{b} \end{array}\right]\)
Comparing the corresponding elements of the two matrices, we have:
\(-\frac{1}{b}=-1 \Rightarrow b=1 \)
\(\frac{-3-a}{b}=1 \Rightarrow-3-a=1 \Rightarrow a=-4 \)

 Hence, −4 and 1 are the required values of a and b respectively.

0

x=500, y=2000, z=3500

\( \left|\begin{array}{ccc} x & \sin \theta & \cos \theta \\ -\sin \theta & -x & 1 \\ \cos \theta & 1 & x \end{array}\right| \)
\( =x\left(-x^2-1\right)-\sin \theta(-x \sin \theta-\cos \theta)+\cos \theta(-\sin \theta+x \cos \theta) \)
\( =-x^3-x+x \sin ^2 \theta+\sin \theta \cos \theta-\sin \theta \cos \theta+x \cos ^2 \theta \)
\( =-x^3-x+x\left(\sin ^2 \theta+\cos ^2 \theta\right) \)
\(=-x^3-x+x \)
\(=-x^3 \text { (Which is Independent of } \theta \text { ) }\)

Consider the product \(\left[\begin{array}{ccc} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{array}\right]\left[\begin{array}{ccc} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{array}\right]\)
\(=\left[\begin{array}{ccc} -2-9+12 & 0-2+2 & 1+3-4 \\ 0+18-18 & 0+4-3 & 0-6+6 \\ -6-18+24 & 0-4+4 & 3+6-8 \end{array}\right]=\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]\)
Hence, \(\left[\begin{array}{ccc} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{array}\right]^{-1}=\left[\begin{array}{ccc} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{array}\right]\)
Now, given system of equations can be written, in matrix form, as follows
\(\left[\begin{array}{ccc} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{l} 1 \\ 1 \\ 2 \end{array}\right]\)
\(\begin{aligned} &x \\ &y \\ &z \end{aligned}=\left[\begin{array}{rrr} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{array}\right]^{-1}\left[\begin{array}{l} 1 \\ 1 \\ 2 \end{array}\right]=\left[\begin{array}{lll} 2 & 0 & 1 \\ 9 & 2 & 3 \\ 6 & 1 & 2 \end{array}\right]\left[\begin{array}{l} 1 \\ 1 \\ 2 \end{array}\right]\)
\(=\left[\begin{array}{r} -2+0+2 \\ 9+2-6 \\ 6+1-4 \end{array}\right]=\left[\begin{array}{l} 0 \\ 5 \\ 3 \end{array}\right]\)
Hence, x = 0, y = 5, z = 3

\(\text { Given }\left|\begin{array}{ccc} x+a & x & x \\ x & x+a & x \\ x & x & x+a \end{array}\right|=0, a \neq 0\)
\(\Rightarrow\left|\begin{array}{rrc} a & 0 & x \\ -a & a & x \\ 0 & -a & x+a \end{array}\right|=0\)
\(\left[\text { by performing } C_{1} \rightarrow C_{1}-C_{2} \text { and } C_{2} \rightarrow C_{2}-C_{3}\right]\)
\(\Rightarrow\left|\begin{array}{ccc} a & 0 & x \\ 0 & a & 2 x \\ 0 & -a & x+a \end{array}\right|=0\left[\text { by performing } R_{2} \rightarrow R_{2}+R_{1}\right]\)
\(\Rightarrow a\left(a x+a^{2}+2 a x\right)=0 \text { [on expanding along } \left.C_{1}\right] \)
\(\Rightarrow 3 a x+a^{2}=0 \Rightarrow x=-\frac{a}{3}, a \neq 0\)

\(\left|\begin{array}{ccc} 3 a-x & 3 a-x & 3 a-x \\ a-x & a+x & a-x \\ a-x & a-x & a+x \end{array}\right|=0\)
\(\text { [by performing } \left.R_{1} \rightarrow R_{1}+\left(R_{2}+R_{3}\right)\right]\)
\(\Rightarrow(3 a-x)\left|\begin{array}{ccc} 1 & 1 & 1 \\ a-x & a+x & a-x \\ a-x & a-x & a+x \end{array}\right|=0\)
\(\text { [by taking } \left.(3 a-x) \text { common from } R_{1}\right]\)
\(\Rightarrow(3 a-x)\left|\begin{array}{crc} 0 & 0 & 1 \\ -2 x & 2 x & a-x \\ 0 & -2 x & a+x \end{array}\right|=0\)
\(\text { [by performing } \left.C_{1} \rightarrow C_{1}-C_{2} \text { and } C_{2} \rightarrow C_{2}-C_{3}\right]\)
\(\Rightarrow 4 x^{2}(3 a-x)=0 \)
\(\Rightarrow x=0 \text { or } x=3 a \)