CBSE Class 12 Mathematics 2024: question 32(a)

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Step-by-step solution

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1. $|A|=1(0+6)-2(0+3)-3(4-0)=6-6-12=-12\ne 0$, so $A^{-1}$ exists.

   1 mark

2. Cofactors: $A_{11}=6,A_{12}=-3,A_{13}=4$; $A_{21}=-6,A_{22}=3,A_{23}=0$; $A_{31}=-6,A_{32}=-3,A_{33}=-4$. So $\text{adj}A=\left[\begin{array}{ccc}6& -6& -6\\ -3& 3& -3\\ 4& 0& -4\end{array}\right]$.

   2 marks

3. $A^{-1}=\frac{1}{|A|}\text{adj}A=-\frac{1}{12}\left[\begin{array}{ccc}6& -6& -6\\ -3& 3& -3\\ 4& 0& -4\end{array}\right]$.

   ½ mark

4. The system is $AX=B$ with $X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]$, $B=\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right]$. $X=A^{-1}B=-\frac{1}{12}\left[\begin{array}{c}6-12-18\\ -3+6-9\\ 4+0-12\end{array}\right]=-\frac{1}{12}\left[\begin{array}{c}-24\\ -6\\ -8\end{array}\right]=\left[\begin{array}{c}2\\ \frac{1}{2}\\ \frac{2}{3}\end{array}\right]$. Hence $x=2$, $y=\frac{1}{2}$, $z=\frac{2}{3}$.

   1½ marks

Answer: $A^{-1}=-\frac{1}{12}\left[\begin{array}{ccc}6& -6& -6\\ -3& 3& -3\\ 4& 0& -4\end{array}\right]$; $x=2$, $y=\frac{1}{2}$, $z=\frac{2}{3}$

Revise the chapter

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