CBSE Class 12 Mathematics 2023: question 33(a)

Hints

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Step-by-step solution

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1. Use the reversal law: $(AB{)}^{-1}=B^{-1}{A}^{-1}$.

   ½ mark

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

   1 mark

3. Cofactors of $A$:

   • $A_{11}=3,A_{12}=1,A_{13}=2$
   • $A_{21}=2,A_{22}=1,A_{23}=2$
   • $A_{31}=6,A_{32}=2,A_{33}=5$

   So $\text{adj}A=\left[\begin{array}{ccc}3& 2& 6\\ 1& 1& 2\\ 2& 2& 5\end{array}\right]$.

   2 marks

4. $A^{-1}=\frac{1}{|A|}\text{adj}A=\left[\begin{array}{ccc}3& 2& 6\\ 1& 1& 2\\ 2& 2& 5\end{array}\right]$

   ½ mark

5. $(AB{)}^{-1}=B^{-1}{A}^{-1}=\left[\begin{array}{ccc}3& -1& 1\\ -15& 6& -5\\ 5& -2& 2\end{array}\right]\left[\begin{array}{ccc}3& 2& 6\\ 1& 1& 2\\ 2& 2& 5\end{array}\right]=\left[\begin{array}{ccc}10& 7& 21\\ -49& -34& -103\\ 17& 12& 36\end{array}\right]$

   1 mark

Answer: $(AB{)}^{-1}=\left[\begin{array}{ccc}10& 7& 21\\ -49& -34& -103\\ 17& 12& 36\end{array}\right]$

Revise the chapter

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