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1. Find the adjoint of the matrix \( \left[\begin{array}{cc}2 & -4 \\ -1 & 2\end{array}\right] \)

Ask by Bird Reeves. in Nepal
Nov 11,2024

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The adjoint of the matrix \( \begin{bmatrix}2 & -4 \\ -1 & 2\end{bmatrix} \) is \( \begin{bmatrix} 2 & 4 \\ 1 & 2 \end{bmatrix} \).

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To find the adjoint of the matrix \( \left[\begin{array}{cc}2 & -4 \\ -1 & 2\end{array}\right] \), we first calculate the matrix of cofactors. The cofactor of an element \( a_{ij} \) is given by \( (-1)^{i+j} \) times the determinant of the minor matrix obtained by removing the \( i \)-th row and \( j \)-th column. The matrix given is 2x2, so its adjoint (which is the transpose of the cofactor matrix) can be computed directly: 1. The cofactor for the element \( a_{11} = 2 \) is \( \text{det}(\begin{bmatrix}2\end{bmatrix}) = 2 \), so the cofactor is \( 2 \). 2. The cofactor for the element \( a_{12} = -4 \) is \( \text{det}(\begin{bmatrix}-1\end{bmatrix}) = -1 \), the cofactor is \( -(-1) = 1 \). 3. The cofactor for the element \( a_{21} = -1 \) is \( \text{det}(\begin{bmatrix}-4\end{bmatrix}) = -4 \), the cofactor is \( -(-4) = 4 \). 4. The cofactor for the element \( a_{22} = 2 \) is \( \text{det}(\begin{bmatrix}2\end{bmatrix}) = 2 \), the cofactor is \( 2 \). Putting this all together, the cofactor matrix is: \[ \begin{bmatrix} 2 & 1 \\ 4 & 2 \end{bmatrix} \] Now, we take the transpose of this cofactor matrix to find the adjoint: \[ \text{adjoint} = \begin{bmatrix} 2 & 4 \\ 1 & 2 \end{bmatrix} \] So the adjoint of the matrix \( \left[\begin{array}{cc}2 & -4 \\ -1 & 2\end{array}\right] \) is \( \begin{bmatrix} 2 & 4 \\ 1 & 2\end{bmatrix} \).

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