In some cases, we may want to use an alternative method to find the inverse of a nonsingular square matrix. This method involves using determinants and the adjoint (adjugate). To use this process, we will state a formula that tells us that the inverse of an invertible square matrix can be found by multiplying 1/det times the adjoint (adjugate) of the matrix.

Test Objectives
• Demonstrate the ability to find the determinant of a matrix
• Demonstrate the ability to find the adjoint of a matrix
• Demonstrate the ability to find the inverse of a matrix
Inverse Adjoint and Determinant Method Practice Test:

#1:

Instructions: find the inverse.

$$a)\hspace{.2em}$$ $$\left[ \begin{array}{cc}1 & -12\\ 1 & -8\end{array}\right]$$

$$b)\hspace{.2em}$$ $$\left[ \begin{array}{cc}0 & 8\\ 0 & -9\end{array}\right]$$

#2:

Instructions: find the inverse.

$$a)\hspace{.2em}$$ $$\left[ \begin{array}{cc}-8 & 10\\ -3 & 3\end{array}\right]$$

$$b)\hspace{.2em}$$ $$\left[ \begin{array}{cc}6 & -10\\ 2 & -4\end{array}\right]$$

#3:

Instructions: find the inverse.

$$a)\hspace{.2em}$$ $$\left[ \begin{array}{cc}0 & -2\\ 4 & -8\end{array}\right]$$

$$b)\hspace{.2em}$$ $$\left[ \begin{array}{ccc}-3 & -3 & -1\\ -2 & 2 & -3 \\ 1 & -6 & 5\end{array}\right]$$

#4:

Instructions: find the inverse.

$$a)\hspace{.2em}$$ $$\left[ \begin{array}{ccc}-1 & 3 & -6\\ -4 & -4 & 6 \\ 4 & 0 & 2\end{array}\right]$$

$$b)\hspace{.2em}$$ $$\left[ \begin{array}{ccc}0 & 2 & 4\\ -1 & 0 & 4 \\ 1 & 3 & 0\end{array}\right]$$

#5:

Instructions: find the inverse.

$$a)\hspace{.2em}$$ $$\left[ \begin{array}{ccc}1 & 3 & -6\\ 5 & 6 & 0 \\ -2 & -6 & 12\end{array}\right]$$

$$b)\hspace{.2em}$$ $$\left[ \begin{array}{cccc}1 & 0 & 0 & 0\\ 4 & 1 & 2 & -2 \\ 1 & -1 & 0 & 0 \\0 & 1 & -4 & 3\end{array}\right]$$

Written Solutions:

#1:

Solutions:

$$a)\hspace{.2em}$$ $$\left[ \begin{array}{cc}-2 & 3\\ -\frac{1}{4}& \frac{1}{4}\end{array}\right]$$

$$b)\hspace{.2em}Singular$$

#2:

Solutions:

$$a)\hspace{.2em}$$ $$\large{\left[ \begin{array}{cc}\frac{1}{2}& -\frac{5}{3}\\ \frac{1}{2}& -\frac{4}{3}\end{array}\right]}$$

$$b)\hspace{.2em}$$ $$\large{\left[ \begin{array}{cc}1 & -\frac{5}{2}\\ \frac{1}{2}& -\frac{3}{2}\end{array}\right]}$$

#3:

Solutions:

$$a)\hspace{.2em}$$ $$\large{\left[ \begin{array}{cc}-1 & \frac{1}{4}\\ -\frac{1}{2}& 0 \end{array}\right]}$$

$$b)\hspace{.2em}$$ $$\left[ \begin{array}{ccc}\frac{8}{7}& -3 & -\frac{11}{7}\\ -1 & 2 & 1 \\ -\frac{10}{7}& 3 & \frac{12}{7}\end{array}\right]$$

#4:

Solutions:

$$a)\hspace{.2em}$$ $$\left[ \begin{array}{ccc}-1 & -\frac{3}{4}& -\frac{3}{4}\\ 4 & \frac{11}{4}& \frac{15}{4}\\ 2 & \frac{3}{2}& 2\end{array}\right]$$

$$b)\hspace{.2em}$$ $$\left[ \begin{array}{ccc}3 & -3 & -2\\ -1 & 1 & 1 \\ \frac{3}{4}& -\frac{1}{2}& -\frac{1}{2}\end{array}\right]$$

#5:

Solutions:

$$a)\hspace{.2em}Singular$$

$$b)\hspace{.2em}$$ $$\left[ \begin{array}{cccc}1 & 0 & 0 & 0\\ 1 & 0 & -1 & 0 \\ \frac{17}{2}& -\frac{3}{2}& -\frac{5}{2}& -1 \\ 11 & -2 & -3 & -1\end{array}\right]$$