1.

Find the (adj A), ifA = \(\rm \begin{bmatrix} 3& 7& 1 \\ 2& 1& 8\\ 4& 5& 0\end{bmatrix}\)1. \(\rm \begin{bmatrix} -40& 32& 55 \\ 5& -4& 13\\ 55& -22& -11\end{bmatrix}\)2. \(\rm \begin{bmatrix} 3& 7& 1 \\ 2& 1& 8\\ 4& 5& 0\end{bmatrix}\)3. \(\rm \begin{bmatrix} -40& 32& 6 \\ 5& -4& 13\\ 55& -22& -11\end{bmatrix}\)4. \(\rm \begin{bmatrix} -40& 5& 55\\ 32& -4& -22\\ 6& 13& -11\end{bmatrix}\)

Answer» Correct Answer - Option 4 : \(\rm \begin{bmatrix} -40& 5& 55\\ 32& -4& -22\\ 6& 13& -11\end{bmatrix}\)

Concept:

If matrix A = [aij]

The adj A matrix is the transpose of the matrix [Aij] is the cofactor of the element aij.

Calculation:

A = \(\rm \begin{bmatrix} 3& 7& 1 \\ 2& 1& 8\\ 4& 5& 0\end{bmatrix}\) 

Cof(a11) = 0 - 40 = -40

Cof(a12) = -(0 - 32) = 32

Cof(a13) = 10 - 4 = 6

Cof(a21) = -(0 - 5) = 5

Cof(a22) = 0 - 4 = -4

Cof(a23) = -(15 - 28) = 13

Cof(a31) = 56 - 1 = 55

Cof(a32) = -(24 - 2) = -22

Cof(a33) = 3 - 14 = -11

cof A = \(\rm \begin{bmatrix} -40& 32& 6 \\ 5& -4& 13\\ 55& -22& -11\end{bmatrix}\)

Adj A = (cof A)T = \(\rm \begin{bmatrix} -40& 32& 6 \\ 5& -4& 13\\ 55& -22& -11\end{bmatrix}^T\)

⇒ Adj A = \(\rm \begin{bmatrix} -40& 5& 55\\ 32& -4& -22\\ 6& 13& -11\end{bmatrix}\)



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