Problem 5
Find the determinant of the \(2 \times 2\) matrix. \(\left[\begin{array}{ll}8 & 10 \\ 2 & -3\end{array}\right]\)
Problem 5
Matrices \(A\) and \(\vec{b}\) are given. (a) Give \(\operatorname{det}(A)\) and \(\operatorname{det}\left(A_{i}\right)\) for all \(i\). (b) Use Cramer's Rule to solve \(A \vec{x}=\vec{b}\). If Cramer's Rule cannot be used to find the solution, then state whether or not a solution exists. \(A=\left[\begin{array}{cc}2 & 10 \\ -1 & 3\end{array}\right], \quad \vec{b}=\left[\begin{array}{l}42 \\ 19\end{array}\right]\)
Problem 5
Find the trace of the given matrix. \(\left[\begin{array}{ccc}-4 & 1 & 1 \\ -2 & 0 & 0 \\ -1 & -2 & -5\end{array}\right]\)
Problem 5
Find the determinant of the given matrix using cofactor expansion along any row or column you choose. \(\left[\begin{array}{ccc}-2 & -3 & 5 \\ 5 & 2 & 0 \\ -1 & 0 & 0\end{array}\right]\)
Problem 5
Find \(A^{T} ;\) make note if \(A\) is upper/lower triangular, diagonal, symmetric and/or skew symmetric. \(\left[\begin{array}{cc}-5 & -9 \\ 3 & 1 \\ -10 & -8\end{array}\right]\)
Problem 6
Find the determinant of the \(2 \times 2\) matrix. \(\left[\begin{array}{cc}10 & -10 \\ -10 & 0\end{array}\right]\)
Problem 6
Find \(A^{T} ;\) make note if \(A\) is upper/lower triangular, diagonal, symmetric and/or skew symmetric. \(\left[\begin{array}{cc}-2 & 10 \\ 1 & -7 \\ 9 & -2\end{array}\right]\)
Problem 6
Find the determinant of the given matrix using cofactor expansion along any row or column you choose. \(\left[\begin{array}{ccc}-2 & -2 & 0 \\ 2 & -5 & -3 \\ -5 & 1 & 0\end{array}\right]\)
Problem 6
Find the trace of the given matrix. \(\left[\begin{array}{ccc}0 & -3 & 1 \\ 5 & -5 & 5 \\ -4 & 1 & 0\end{array}\right]\)
Problem 6
Matrices \(A\) and \(\vec{b}\) are given. (a) Give \(\operatorname{det}(A)\) and \(\operatorname{det}\left(A_{i}\right)\) for all \(i\). (b) Use Cramer's Rule to solve \(A \vec{x}=\vec{b}\). If Cramer's Rule cannot be used to find the solution, then state whether or not a solution exists. \(A=\left[\begin{array}{cc}7 & 14 \\ -2 & -4\end{array}\right], \quad \vec{b}=\left[\begin{array}{c}-1 \\ 4\end{array}\right]\)
