Determinants
- Compute the determinant by cofactor expansion. solution
\( \begin{vmatrix}\begin{array}{rrrr}{1} & {-2} & {5} & {2} \\
{0} & {0} & {3} & {0} \\ {2} & {-4} & {-3}
& {5} \\ {2} & {0} & {3} & {5}\end{array}\end{vmatrix} \)
- Compute det(A) where A is given below. solution
\(A=\left[\begin{array}{rrrr}{2} & {-8} & {6} & {8} \\
{3} & {-9} & {5} & {10} \\ {-3} & {0} & {1}
& {-2} \\ {1} & {-4} & {0} &
{6}\end{array}\right]\)
- Show that if \( \mathrm{det}(A^T) = \mathrm{det}(A^{-1})\)
then \(\mathrm{det}(A)=\pm1. \) proof
- Let A and B be \(4 \times 4\) matrices, with
\(\operatorname{det}(A) = -3\) and \(\operatorname{det} (B) = -1
\). Compute: solution
a. \(\operatorname{det} (AB)\) b.
\(\operatorname{det} (B^{5})\) c.
\(\operatorname{det} (2 A)\) d.
\(\operatorname{det} (A^{T} B A)\) e.
\(\operatorname{det} (B^{-1} A B)\)