Fall 2026: Math 590 Quiz 5 Solutions

Problem 1

For \(A = \begin{pmatrix} 2 & 3 & -1\\4 & 0 & 6\\-2 & 2 & 1\end{pmatrix}\), find the \((2,2)\) entry and the \((3,1)\) entry of the classical adjoint of \(A\). (5 points)

Solution. Write \(C_{ij} = (-1)^{i+j}\det A_{ij}\) for the \((i,j)\) cofactor of \(A\), where \(A_{ij}\) is obtained from \(A\) by deleting row \(i\) and column \(j\). The classical adjoint \(A'\) is the transpose of the cofactor matrix, so \(A'_{ij} = C_{ji}\).

The \((2,2)\) entry of \(A'\) is

C_{22} = (-1)^{2+2}\det\begin{pmatrix} 2 & -1\\-2 & 1\end{pmatrix} = (2)(1)-(-1)(-2) = 2-2 = 0.

The \((3,1)\) entry of \(A'\) is

C_{13} = (-1)^{1+3}\det\begin{pmatrix} 4 & 0\\-2 & 2\end{pmatrix} = (4)(2)-(0)(-2) = 8.

Thus the \((2,2)\) entry is \(0\) and the \((3,1)\) entry is \(8\).

Problem 2

Use Cramer's rule to solve the system (5 points)

\begin{align*} 3x+4y &= 2\\ 2x+3y &= 1 \end{align*}

Solution. The coefficient matrix has determinant

\det\begin{pmatrix} 3 & 4\\2 & 3\end{pmatrix} = 9-8 = 1\neq 0,

so Cramer's rule applies. Replacing the first and second columns by \(\begin{pmatrix} 2\\1\end{pmatrix}\) in turn,

x = \frac{\det\begin{pmatrix} 2 & 4\\1 & 3\end{pmatrix}}{1} = 6-4 = 2, \qquad y = \frac{\det\begin{pmatrix} 3 & 2\\2 & 1\end{pmatrix}}{1} = 3-4 = -1.

Thus \(x = 2\), \(y = -1\). (Check: \(3(2)+4(-1) = 2\) and \(2(2)+3(-1) = 1\).)