Consider the following matrix. A=[} 1&-1 -3&2 ] Choose the correct description of A. Find A^-1 if it exists. A is nonsingular That is, it has an inverse. A^-1=square A is singular. That is, its inverse doesn't exist.

To determine whether the matrix \( A \) is singular or nonsingular, we need to calculate its determinant. If the determinant is nonzero, then the matrix is nonsingular and has an inverse. If the determinant is zero, then the matrix is singular and does not have an inverse.<br /><br />The matrix \( A \) is given by:<br />\[ A = \begin{bmatrix} 1 & -1 \\ -3 & 2 \end{bmatrix} \]<br /><br />The determinant of \( A \) is calculated as follows:<br />\[ \text{det}(A) = (1)(2) - (-1)(-3) = 2 - 3 = -1 \]<br /><br />Since the determinant of \( A \) is \(-1\), which is nonzero, the matrix \( A \) is nonsingular and has an inverse.<br /><br />To find the inverse of \( A \), we use the formula for the inverse of a 2x2 matrix:<br />\[ A^{-1} = \frac{1}{\text{det}(A)} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \]<br /><br />For the matrix \( A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \), the elements are:<br />\[ a = 1, \quad b = -1, \quad c = -3, \quad d = 2 \]<br /><br />Substituting these values into the formula, we get:<br />\[ A^{-1} = \frac{1}{-1} \begin{bmatrix} 2 & 1 \\ 3 & 1 \end{bmatrix} = \begin{bmatrix} -2 & -1 \\ -3 & -1 \end{bmatrix} \]<br /><br />Therefore, the inverse of \( A \) is:<br />\[ A^{-1} = \begin{bmatrix} -2 & -1 \\ -3 & -1 \end{bmatrix} \]<br /><br />So, the correct description of \( A \) is:<br />\[ A \text{ is nonsingular. That is, it has an inverse.} \]<br /><br />And the inverse of \( A \) is:<br />\[ A^{-1} = \begin{bmatrix} -2 & -1 \\ -3 & -1 \end{bmatrix} \]