Class 12 Determinants · Very Long Answer5 Marksmatrix_inverse

Class 12 Maths · Determinants · Very Long Answer #102

Find A⁻¹ for A=\(\begin{bmatrix}3 & -1 & 1\\-15 & 6 & -5\\5 & -2 & 2\end{bmatrix}\) and verify A⁻¹A=I.

✅ Step-by-Step Solution

Final Answer: \(\begin{bmatrix}2 & 0 & -1\\5 & 1 & 0\\0 & 1 & 3\end{bmatrix}\)

Expand the determinant: det(A)=1≠0, so A is invertible.
Compute the signed minors Cᵢⱼ=(−1)ⁱ⁺ʲMᵢⱼ to obtain the cofactor matrix \(\begin{bmatrix}2 & 5 & 0\\0 & 1 & 1\\-1 & 0 & 3\end{bmatrix}\).
Transpose the cofactor matrix to get adj(A)=\(\begin{bmatrix}2 & 0 & -1\\5 & 1 & 0\\0 & 1 & 3\end{bmatrix}\). Then A⁻¹=adj(A)/(1)=\(\begin{bmatrix}2 & 0 & -1\\5 & 1 & 0\\0 & 1 & 3\end{bmatrix}\).
Direct multiplication of the displayed inverse by A gives \(\begin{bmatrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{bmatrix}\).
Hence A⁻¹=\(\begin{bmatrix}2 & 0 & -1\\5 & 1 & 0\\0 & 1 & 3\end{bmatrix}\) and A⁻¹A=I₃.

[1 mark] Expand the determinant: det(A)=1≠0, so A is invertible.
[1 mark] Compute the signed minors Cᵢⱼ=(−1)ⁱ⁺ʲMᵢⱼ to obtain the cofactor matrix \(\begin{bmatrix}2 & 5 & 0\\0 & 1 & 1\\-1 & 0 & 3\end{bmatrix}\).
[1 mark] Transpose the cofactor matrix to get adj(A)=\(\begin{bmatrix}2 & 0 & -1\\5 & 1 & 0\\0 & 1 & 3\end{bmatrix}\). Then A⁻¹=adj(A)/(1)=\(\begin{bmatrix}2 & 0 & -1\\5 & 1 & 0\\0 & 1 & 3\end{bmatrix}\).
[1 mark] Direct multiplication of the displayed inverse by A gives \(\begin{bmatrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{bmatrix}\).
[1 mark] Hence A⁻¹=\(\begin{bmatrix}2 & 0 & -1\\5 & 1 & 0\\0 & 1 & 3\end{bmatrix}\) and A⁻¹A=I₃.

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