InterviewSolution
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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. \(\rm \begin{bmatrix} -40& 5& 55\\ 32& -4& -22\\ 6& 13& -11\end{bmatrix}\) |
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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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