Problem

Find the determinant of the matrix \n\[ A = \begin{bmatrix} 5 & 2 & 1 & 3 \\ 4 & 3 & 2 & 1 \\ 1 & 2 & 3 & 4 \\ 3 & 1 & 2 & 5 \end{bmatrix} \]

Answer

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Answer

The determinant of a triangular matrix is the product of its diagonal elements. So, the determinant of A is: \n\[ 5*1*1*-9 = -45 \]

Steps

Step 1 :First, we perform row operations to simplify the matrix: \n\[ \begin{bmatrix} 5 & 2 & 1 & 3 \\ 0 & 1 & 1 & -3 \\ 0 & 1 & 2 & 1 \\ 0 & -1 & 1 & 2 \end{bmatrix} \]

Step 2 :Next, we subtract row 2 from row 3 and add row 2 to row 4 to get: \n\[ \begin{bmatrix} 5 & 2 & 1 & 3 \\ 0 & 1 & 1 & -3 \\ 0 & 0 & 1 & 4 \\ 0 & 0 & 2 & -1 \end{bmatrix} \]

Step 3 :Finally, we subtract 2*row 3 from row 4 to get: \n\[ \begin{bmatrix} 5 & 2 & 1 & 3 \\ 0 & 1 & 1 & -3 \\ 0 & 0 & 1 & 4 \\ 0 & 0 & 0 & -9 \end{bmatrix} \]

Step 4 :The determinant of a triangular matrix is the product of its diagonal elements. So, the determinant of A is: \n\[ 5*1*1*-9 = -45 \]

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