Compute Det of A where A = beginbmatrix 2 & -8 & 6 & 8 3 & -9 & 5 & 10 | Mathematics II 2080.1

Mathematics II2080.15 marksNumerical: matrix determinantProperties

Compute Det of A where $A = \begin{bmatrix} 2 & -8 & 6 & 8 \\ 3 & -9 & 5 & 10 \\ -3 & 0 & 1 & -2 \\ 1 & -4 & 0 & 6 \end{bmatrix}$ [5]

Compute Det of A where $A = \begin{bmatrix} 2 & -8 & 6 & 8 \ 3 & -9 & 5 & 10 \ -3 & 0 & 1 & -2 \ 1 & -4 & 0 & 6 \end{bmatrix}$ [5]

$A = \begin{bmatrix} 2 & -8 & 6 & 8 \\ 3 & -9 & 5 & 10 \\ -3 & 0 & 1 & -2 \\ 1 & -4 & 0 & 6 \end{bmatrix}$ Rules used: - Adding a multiple of one row to another does not change the determinant. - Factoring scalar $k$ out of a row means the original determinant is $k$ times the reduced one. --- ...

$$A = \begin{bmatrix} 2 & -8 & 6 & 8 \ 3 & -9 & 5 & 10 \ -3 & 0 & 1 & -2 \ 1 & -4 & 0 & 6 \end{bmatrix}$$ Rules used: - Adding a multiple of one row to another does not change the determinant. - Factoring scalar $k$ out of a row means the original determinant is $k$ times the reduced one. --- ...

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Compute Det of A where A = beginbmatrix 2 & -8 & 6 & 8 3 & -9 & 5 & 10 -3 & 0 & 1 & -2 1 & -4 & 0 & 6 endbmatrix