We are given the matrix $A = \begin{bmatrix} -2 & 1 \\ -5 & 3 \end{bmatrix}$ and are tasked to:

1. Evaluate $A^2 - A$
2. Express $A^{-1}$ in terms of $A$ and the identity matrix $I$.

Step 1: Compute $A^2$

To compute $A^2$, we multiply $A$ by itself:

$A^2 = A \cdot A = \begin{bmatrix} -2 & 1 \\ -5 & 3 \end{bmatrix} \cdot \begin{bmatrix} -2 & 1 \\ -5 & 3 \end{bmatrix}$

Performing matrix multiplication:

$A^2 = \begin{bmatrix} (-2)(-2) + (1)(-5) & (-2)(1) + (1)(3) \\ (-5)(-2) + (3)(-5) & (-5)(1) + (3)(3) \end{bmatrix}$

$A^2 = \begin{bmatrix} 4 - 5 & -2 + 3 \\ 10 - 15 & -5 + 9 \end{bmatrix} = \begin{bmatrix} -1 & 1 \\ -5 & 4 \end{bmatrix}$

Step 2: Compute $A^2 - A$

Now, subtract $A$ from $A^2$:

$A^2 - A = \begin{bmatrix} -1 & 1 \\ -5 & 4 \end{bmatrix} - \begin{bmatrix} -2 & 1 \\ -5 & 3 \end{bmatrix}$

$A^2 - A = \begin{bmatrix} -1 + 2 & 1 - 1 \\ -5 + 5 & 4 - 3 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

Thus, $A^2 - A = I$, the identity matrix.

Step 3: Express $A^{-1}$

From the previous result, we have:

$A^2 - A = I$

This can be rewritten as:

$A(A - I) = 0$

Multiplying both sides by $A^{-1}$, we get:

$A^{-1} = A - I$

Thus, the inverse of $A$ is:

$A^{-1} = A - I = \begin{bmatrix} -2 & 1 \\ -5 & 3 \end{bmatrix} - \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} -3 & 1 \\ -5 & 2 \end{bmatrix}$

Summary:

• $A^2 - A = I$
• $A^{-1} = A - I = \begin{bmatrix} -3 & 1 \\ -5 & 2 \end{bmatrix}$

Would you like more details on any part of the solution?

Here are some related questions:

1. What are the properties of the inverse of a matrix?
2. How can we verify if a matrix is invertible?
3. What other methods can be used to find the inverse of a matrix?
4. How does the determinant of a matrix relate to its inverse?
5. Can a non-square matrix have an inverse?

Tip: Always check if the determinant of a matrix is non-zero before attempting to find its inverse.