Class 12 Determinants · Very Long Answer5 Marksadjugate

Class 12 Maths · Determinants · Very Long Answer #101

For A=\(\begin{bmatrix}-1 & -2 & -2\\2 & 1 & -2\\2 & -2 & 1\end{bmatrix}\), find adj(A) and verify A adj(A)=det(A)I₃.

✅ Step-by-Step Solution

Final Answer: \(\begin{bmatrix}-3 & 6 & 6\\-6 & 3 & -6\\-6 & -6 & 3\end{bmatrix}\)

Expand along row 1 to obtain det(A)=27.
The cofactor matrix is \(\begin{bmatrix}-3 & -6 & -6\\6 & 3 & -6\\6 & -6 & 3\end{bmatrix}\).
Transpose it: adj(A)=\(\begin{bmatrix}-3 & 6 & 6\\-6 & 3 & -6\\-6 & -6 & 3\end{bmatrix}\).
Direct multiplication gives A adj(A)=\(\begin{bmatrix}27 & 0 & 0\\0 & 27 & 0\\0 & 0 & 27\end{bmatrix}\).
This equals det(A)I₃=\(\begin{bmatrix}27 & 0 & 0\\0 & 27 & 0\\0 & 0 & 27\end{bmatrix}\), so the adjugate identity is verified.

[1 mark] Expand along row 1 to obtain det(A)=27.
[1 mark] The cofactor matrix is \(\begin{bmatrix}-3 & -6 & -6\\6 & 3 & -6\\6 & -6 & 3\end{bmatrix}\).
[1 mark] Transpose it: adj(A)=\(\begin{bmatrix}-3 & 6 & 6\\-6 & 3 & -6\\-6 & -6 & 3\end{bmatrix}\).
[1 mark] Direct multiplication gives A adj(A)=\(\begin{bmatrix}27 & 0 & 0\\0 & 27 & 0\\0 & 0 & 27\end{bmatrix}\).
[1 mark] This equals det(A)I₃=\(\begin{bmatrix}27 & 0 & 0\\0 & 27 & 0\\0 & 0 & 27\end{bmatrix}\), so the adjugate identity is verified.

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