### About Finding the Inverse of a Matrix:

When a square matrix has an inverse, it is known as invertible or nonsingular. When we want to find the inverse of a nonsingular square matrix A, we set up the following augmented matrix:
[A | I]
Where A is on the left and the identity matrix of the same order is on the right. Basically, we can use row operations to transform the augmented matrix into the following form:
[I | A(-1)]
Now we will have the identity matrix on the left and the inverse of A on the right.

Test Objectives
• Demonstrate the ability to set up an augmented matrix
• Demonstrate the ability to achieve reduced-row echelon form
• Demonstrate the ability to find the inverse of a matrix
Finding the Inverse of a Matrix Practice Test:

#1:

Instructions: find the inverse.

$$a)\hspace{.2em}$$ $$\left[ \begin{array}{cc}8 & -1 \\ 0 & 0\end{array}\right]$$

$$b)\hspace{.2em}$$ $$\left[ \begin{array}{cc}-1 & 7 \\ 1 & -5\end{array}\right]$$

#2:

Instructions: find the inverse.

$$a)\hspace{.2em}$$ $$\left[ \begin{array}{cc}9 & -1 \\ -3 & 0\end{array}\right]$$

$$b)\hspace{.2em}$$ $$\left[ \begin{array}{cc}0 & 1 \\ -9 & -8\end{array}\right]$$

#3:

Instructions: find the inverse.

$$a)\hspace{.2em}$$ $$\left[ \begin{array}{cc}6 & -9 \\ 1 & -1\end{array}\right]$$

$$b)\hspace{.2em}$$ $$\left[ \begin{array}{ccc}4 & -5 & -3 \\ -1 & -1 & 0 \\4 & 6 & 0\end{array}\right]$$

#4:

Instructions: find the inverse.

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

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

#5:

Instructions: find the inverse.

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

$$b)\hspace{.2em}$$ $$\left[ \begin{array}{cccc}3 & 0 & 1 & 0 \\-1 & 3 & 5 & -2 \\ 1 & -1 & 0 & 0\\-2 & 5 & -4 &2\end{array}\right]$$

Written Solutions:

#1:

Solutions:

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

$$b)\hspace{.2em}$$ $$\large{\left[ \begin{array}{cc}\frac{5}{2}& \frac{7}{2}\\ \frac{1}{2}& \frac{1}{2}\end{array}\right]}$$

#2:

Solutions:

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

$$b)\hspace{.2em}$$ $$\left[ \begin{array}{cc}-\frac{8}{9}& -\frac{1}{9}\\ 1 & 0\end{array}\right]$$

#3:

Solutions:

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

$$b)\hspace{.2em}$$ $$\left[ \begin{array}{ccc}0 & -3 & -\frac{1}{2}\\ 0 & 2 & \frac{1}{2}\\ -\frac{1}{3}& -\frac{22}{3}& -\frac{3}{2}\end{array}\right]$$

#4:

Solutions:

$$a)\hspace{.2em}$$ $$\left[ \begin{array}{ccc}-\frac{10}{3}& 3 & -\frac{14}{3}\\ -\frac{17}{6}& \frac{5}{2}& -\frac{11}{3}\\ -\frac{3}{2}& \frac{3}{2}& -2\end{array}\right]$$

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

#5:

Solutions:

$$a)\hspace{.2em}$$ $$\left[ \begin{array}{ccc}0 & \frac{1}{2}& \frac{3}{4}\\ 0 & -1 & -1 \\ -\frac{1}{2}& 2 & \frac{9}{4}\end{array}\right]$$

$$b)\hspace{.2em}$$ $$\left[ \begin{array}{cccc}-\frac{1}{2}& \frac{1}{2}& 4 & \frac{1}{2}\\ -\frac{1}{2}& \frac{1}{2}& 3 & \frac{1}{2}\\ \frac{5}{2}& -\frac{3}{2}& -12 & -\frac{3}{2}\\ \frac{23}{4}& -\frac{15}{4}& -\frac{55}{2}& -\frac{13}{4}\end{array}\right]$$