Class 12 Determinants · Long Answer3 Marksmatrix_inverse

Class 12 Maths · Determinants · Long Answer #53

For A=\(\begin{bmatrix}1 & -2 & 3\\0 & -1 & 4\\-2 & 2 & 1\end{bmatrix}\), find (Aᵀ)⁻¹.

✅ Step-by-Step Solution

Final Answer: \(\begin{bmatrix}-9 & -8 & -2\\8 & 7 & 2\\-5 & -4 & -1\end{bmatrix}\)

Expand the determinant: det(A)=1≠0, so A is invertible.
Compute the signed minors Cᵢⱼ=(−1)ⁱ⁺ʲMᵢⱼ to obtain the cofactor matrix \(\begin{bmatrix}-9 & 8 & -5\\-8 & 7 & -4\\-2 & 2 & -1\end{bmatrix}\).
Transpose the cofactor matrix to get adj(A)=\(\begin{bmatrix}-9 & -8 & -2\\8 & 7 & 2\\-5 & -4 & -1\end{bmatrix}\). Then A⁻¹=adj(A)/(1)=\(\begin{bmatrix}-9 & -8 & -2\\8 & 7 & 2\\-5 & -4 & -1\end{bmatrix}\).
Therefore (Aᵀ)⁻¹=\(\begin{bmatrix}-9 & -8 & -2\\8 & 7 & 2\\-5 & -4 & -1\end{bmatrix}\).

[0.5 marks] 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 \(\begin{bmatrix}-9 & 8 & -5\\-8 & 7 & -4\\-2 & 2 & -1\end{bmatrix}\).
[1 mark] Transpose the cofactor matrix to get adj(A)=\(\begin{bmatrix}-9 & -8 & -2\\8 & 7 & 2\\-5 & -4 & -1\end{bmatrix}\). Then A⁻¹=adj(A)/(1)=\(\begin{bmatrix}-9 & -8 & -2\\8 & 7 & 2\\-5 & -4 & -1\end{bmatrix}\).
[0.5 marks] Therefore (Aᵀ)⁻¹=\(\begin{bmatrix}-9 & -8 & -2\\8 & 7 & 2\\-5 & -4 & -1\end{bmatrix}\).

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