CBSE Class 12 Mathematics 2026: question 33(a)

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Step-by-step solution

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1. $|A|=0(0-2)-2(0+2)+1(2+1)=-4+3=-1\ne 0$, so $A^{-1}$ exists.

   1 mark

2. Cofactors: $C_{11}=-2,C_{12}=-2,C_{13}=3;C_{21}=-1,C_{22}=-1,C_{23}=2;C_{31}=-3,C_{32}=-2,C_{33}=4$.

   $\text{adj}A=\left[\begin{array}{ccc}-2& -1& -3\\ -2& -1& -2\\ 3& 2& 4\end{array}\right]$

   1½ marks

3. $A^{-1}=\frac{1}{|A|}\text{adj}A=\left[\begin{array}{ccc}2& 1& 3\\ 2& 1& 2\\ -3& -2& -4\end{array}\right]$

   ½ mark

4. The coefficient matrix of the system is $\left[\begin{array}{ccc}0& -2& 1\\ 2& -1& -1\\ 1& -2& 0\end{array}\right]=A^T$. So the system is $A^{T}X=B$ with $X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]$, $B=\left[\begin{array}{c}7\\ 8\\ 10\end{array}\right]$, giving $X=(A^T{)}^{-1}B=(A^{-1}{)}^{T}B$.

   ½ mark

5. $\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{ccc}2& 2& -3\\ 1& 1& -2\\ 3& 2& -4\end{array}\right]\left[\begin{array}{c}7\\ 8\\ 10\end{array}\right]=\left[\begin{array}{c}14+16-30\\ 7+8-20\\ 21+16-40\end{array}\right]=\left[\begin{array}{c}0\\ -5\\ -3\end{array}\right]$

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

   1½ marks

Answer: $A^{-1}=\left[\begin{array}{ccc}2& 1& 3\\ 2& 1& 2\\ -3& -2& -4\end{array}\right]$; $x=0,y=-5,z=-3$

Revise the chapter

• Chapter: Matrices (notes, formulas, where marks are lost and every past question)