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Question
Find the adjoint of the following matrix:
\[\begin{bmatrix}- 3 & 5 \\ 2 & 4\end{bmatrix}\]
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Solution 1
Given below is the square matrix. Here, we will interchange the diagonal elements and change the signs of the off-diagonal elements.
\[\ A = \begin{bmatrix}- 3 & 5 \\ 2 & 4\end{bmatrix}\]
\[adjA = \begin{bmatrix}4 & - 5 \\ - 2 & - 3\end{bmatrix}\]
\[(adjA)A = \begin{bmatrix}- 22 & 0 \\ 0 & - 22\end{bmatrix}\]
\[\left| A \right| = - 22\]
\[\left| A \right|I = \begin{bmatrix}- 22 & 0 \\ 0 & - 22\end{bmatrix}\]
\[A(adjA) = \begin{bmatrix}- 22 & 0 \\ 0 & - 22\end{bmatrix}\]
\[ \therefore (adjA)A = \left| A \right|I = A(adjA)\]
Hence verified.
Solution 2
Given below is the square matrix. Here, we will interchange the diagonal elements and change the signs of the off-diagonal elements.
\[\ A = \begin{bmatrix}- 3 & 5 \\ 2 & 4\end{bmatrix}\]
\[adjA = \begin{bmatrix}4 & - 5 \\ - 2 & - 3\end{bmatrix}\]
\[(adjA)A = \begin{bmatrix}- 22 & 0 \\ 0 & - 22\end{bmatrix}\]
\[\left| A \right| = - 22\]
\[\left| A \right|I = \begin{bmatrix}- 22 & 0 \\ 0 & - 22\end{bmatrix}\]
\[A(adjA) = \begin{bmatrix}- 22 & 0 \\ 0 & - 22\end{bmatrix}\]
\[ \therefore (adjA)A = \left| A \right|I = A(adjA)\]
Hence verified.
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