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CBSE Class 12 Math 2025 All Sets Solved Paper

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Question : 5 of 20
Marks: +1, -0
If A=[−10001000−1],A=\begin{bmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{bmatrix}, then A−1A^{-1} is
Solution:  
A=[−100010001]A=\begin{bmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} as ∣A∣=−1≠0⇒A−1\left| A \right| = -1 \neq 0 \Rightarrow A^{-1} exists
A−1=  adj⁡A∣A∣A^{-1}= \; \frac{\operatorname{adj} A}{\left| A \right|}
adj⁡A=[C11C12C13C21C22C23C31C32C33]\operatorname{adj} A = \begin{bmatrix} C_{11} & C_{12} & C_{13} \\ C_{21} & C_{22} & C_{23} \\ C_{31} & C_{32} & C_{33} \end{bmatrix}
where Cij=(−1)i+jMijC_{ij}=(-1)^{i+j} M_{ij}
where MijM_{ij} is the minor
C11=(−1)1+1∣1001∣=1C_{11}=(-1)^{1+1} \begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix} = 1
C12=(−1)1+2∣0001∣=0C_{12}=(-1)^{1+2} \begin{vmatrix} 0 & 0 \\ 0 & 1 \end{vmatrix} = 0
C13=(−1)1+3∣0100∣=0C_{13}=(-1)^{1+3} \begin{vmatrix} 0 & 1 \\ 0 & 0 \end{vmatrix} = 0
C21=(−1)2+1∣0001∣=0C_{21}=(-1)^{2+1} \begin{vmatrix} 0 & 0 \\ 0 & 1 \end{vmatrix} = 0
C22=(−1)2+2∣−1001∣=−1C_{22}=(-1)^{2+2} \begin{vmatrix} -1 & 0 \\ 0 & 1 \end{vmatrix} = -1
C23=(−1)2+3∣−1000∣=0C_{23}=(-1)^{2+3} \begin{vmatrix} -1 & 0 \\ 0 & 0 \end{vmatrix} = 0
C31=(−1)3+1∣0010∣=0C_{31}=(-1)^{3+1} \begin{vmatrix} 0 & 0 \\ 1 & 0 \end{vmatrix} = 0
C32=(−1)3+2∣−1000∣=0C_{32}=(-1)^{3+2} \begin{vmatrix} -1 & 0 \\ 0 & 0 \end{vmatrix} = 0
C33=(−1)3+3∣−1001∣=−1C_{33}=(-1)^{3+3} \begin{vmatrix} -1 & 0 \\ 0 & 1 \end{vmatrix} = -1
adj⁡A=[1000−1000−1]\operatorname{adj} A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{bmatrix}
A−1=  adj⁡A∣A∣A^{-1}= \; \frac{\operatorname{adj} A}{\left| A \right|}
A−1=−1[1000−1000−1]A^{-1} = -1 \begin{bmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{bmatrix}
A−1=[−100010001]A^{-1} = \begin{bmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}
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