To solve this problem, we need to compute the inverse of matrix $M$, verify the result by performing matrix multiplication, and match it with the identity matrix $I$.

The given matrix $M$ is: $M = \begin{bmatrix} -5 & 4 \\ -1 & 1 \end{bmatrix}$

Step 1: Finding $M^{-1}$

For a $2 \times 2$ matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$, the inverse, if it exists, is given by: $M^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$ where $ad - bc$ is the determinant of $M$.

Let's calculate each component for $M$:

1. Determinant $\det(M) = (-5)(1) - (4)(-1) = -5 + 4 = -1$.
2. Since the determinant is non-zero, $M$ is invertible.
3. Applying the formula, we get: $M^{-1} = \frac{1}{-1} \begin{bmatrix} 1 & -4 \\ 1 & -5 \end{bmatrix} = \begin{bmatrix} -1 & 4 \\ -1 & 5 \end{bmatrix}$

Thus, $M^{-1} = \begin{bmatrix} -1 & 4 \\ -1 & 5 \end{bmatrix}$.

Step 2: Verifying that $M \cdot M^{-1} = I$

Now, we calculate $M \cdot M^{-1}$ to see if we obtain the identity matrix $I$.

Matrix Multiplication

$M \cdot M^{-1} = \begin{bmatrix} -5 & 4 \\ -1 & 1 \end{bmatrix} \begin{bmatrix} -1 & 4 \\ -1 & 5 \end{bmatrix}$

1. First row, first column: $(-5)(-1) + (4)(-1) = 5 - 4 = 1$

2. First row, second column: $(-5)(4) + (4)(5) = -20 + 20 = 0$

3. Second row, first column: $(-1)(-1) + (1)(-1) = 1 - 1 = 0$

4. Second row, second column: $(-1)(4) + (1)(5) = -4 + 5 = 1$

Thus, $M \cdot M^{-1} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I$, confirming that $M^{-1}$ is correct.

Answers for Multiple-Choice Options

From the choices in the image, let's match these calculations to the corresponding expressions:

1. First row, first column: The correct answer is $A: (-1)(-5) + (4)(-1) = 1$.
2. First row, second column: The correct answer is $D: (-5)(4) + (4)(5) = 0$.
3. Second row, first column: The correct answer is $C: (-1)(-1) + (1)(-1) = 0$.
4. Second row, second column: The correct answer is $B: (-1)(4) + (1)(5) = 1$.

The matrix formed by these values is indeed $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$, showing that $M \cdot M^{-1} = I$, as required.

Do you want further details or have any questions about this solution?

Further Related Questions:

1. How do you find the inverse of a $3 \times 3$ matrix?
2. What conditions make a matrix non-invertible?
3. How does matrix multiplication work for general $n \times n$ matrices?
4. What are the properties of the identity matrix in linear algebra?
5. How can we verify matrix inverses using determinants?

Tip:

Always double-check the determinant before attempting to find an inverse. A zero determinant means the matrix is non-invertible.