### Instruction: Let $\mathbf{P}$ be the matrix for projecting onto the vector $\begin{pmatrix} -3 \\ -2 \end{pmatrix}.$  Find $\mathbf{P}^{-1}.$

If the inverse does not exist, then enter the zero matrix.### Response: I know that the projection matrix onto a vector $\mathbf{v}$ is given by $\mathbf{P} = \frac{\mathbf{v}\mathbf{v}^T}{\mathbf{v}^T\mathbf{v}}.$
So in this case, I have $\mathbf{v} = \begin{pmatrix} -3 \\ -2 \end{pmatrix}$ and $\mathbf{v}^T = \begin{pmatrix} -3 & -2 \end{pmatrix}.$
I can compute $\mathbf{v}^T\mathbf{v} = (-3)^2 + (-2)^2 = 13.$
I can also compute $\mathbf{v}\mathbf{v}^T = \begin{pmatrix} -3 \\ -2 \end{pmatrix}\begin{pmatrix} -3 & -2 \end{pmatrix} = \begin{pmatrix} 9 & 6 \\ 6 & 4 \end{pmatrix}.$
Therefore, $\mathbf{P} = \frac{1}{13}\begin{pmatrix} 9 & 6 \\ 6 & 4 \end{pmatrix}.$
To find the inverse of $\mathbf{P},$ I need to solve the equation $\mathbf{P}\mathbf{X} = \mathbf{I},$ where $\mathbf{I}$ is the identity matrix and $\mathbf{X}$ is the unknown matrix.
I can write this equation as $\frac{1}{13}\begin{pmatrix} 9 & 6 \\ 6 & 4 \end{pmatrix}\mathbf{X} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}.$
To simplify the calculation, I can multiply both sides by $13$ and get $\begin{pmatrix} 9 & 6 \\ 6 & 4 \end{pmatrix}\mathbf{X} = \begin{pmatrix} 13 & 0 \\ 0 & 13 \end{pmatrix}.$
Now, I can use the formula for the inverse of a 2x2 matrix, which is $\begin{pmatrix} a & b \\ c & d \end{pmatrix}^{-1} = \frac{1}{ad - bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix},$ as long as $ad - bc \neq 0.$
In this case, I have $a = 9, b = 6, c = 6, d = 4,$ and $ad - bc = 9 \times 4 - 6 \times 6 = 0.$
This means that the inverse does not exist, because the denominator is zero.
Therefore, I enter the zero matrix as the answer.