The problem in the image provides two matrices $A$ and $B$, and asks to find the matrix $C$, where:

$C = A^2 - (AB) + 3E$

Here:

• $A^2$ is the square of matrix $A$ (i.e., $A \times A$).
• $AB$ is the product of matrices $A$ and $B$.
• $E$ is the identity matrix, and multiplying it by 3 gives $3E$.

The matrices provided are: $A = \begin{pmatrix} 5 & -6 \\ 0 & -1 \end{pmatrix}, \quad B = \begin{pmatrix} -1 & 1 \\ 3 & 2 \end{pmatrix}$

Step-by-step solution:

1. Calculate $A^2$:

$A^2 = A \times A = \begin{pmatrix} 5 & -6 \\ 0 & -1 \end{pmatrix} \times \begin{pmatrix} 5 & -6 \\ 0 & -1 \end{pmatrix}$

Perform the matrix multiplication:

$A^2 = \begin{pmatrix} (5 \times 5 + (-6) \times 0) & (5 \times (-6) + (-6) \times (-1)) \\ (0 \times 5 + (-1) \times 0) & (0 \times (-6) + (-1) \times (-1)) \end{pmatrix}$ $A^2 = \begin{pmatrix} 25 & -30 + 6 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 25 & -24 \\ 0 & 1 \end{pmatrix}$

2. Calculate $AB$:

$AB = \begin{pmatrix} 5 & -6 \\ 0 & -1 \end{pmatrix} \times \begin{pmatrix} -1 & 1 \\ 3 & 2 \end{pmatrix}$

Perform the matrix multiplication:

$AB = \begin{pmatrix} (5 \times (-1) + (-6) \times 3) & (5 \times 1 + (-6) \times 2) \\ (0 \times (-1) + (-1) \times 3) & (0 \times 1 + (-1) \times 2) \end{pmatrix}$ $AB = \begin{pmatrix} -5 - 18 & 5 - 12 \\ 0 - 3 & 0 - 2 \end{pmatrix} = \begin{pmatrix} -23 & -7 \\ -3 & -2 \end{pmatrix}$

3. Find $3E$:

The identity matrix $E$ is a 2x2 matrix:

$E = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$

Multiplying by 3:

$3E = 3 \times \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix}$

4. Calculate $C = A^2 - (AB) + 3E$:

Now, subtract $AB$ from $A^2$ and add $3E$:

$C = \begin{pmatrix} 25 & -24 \\ 0 & 1 \end{pmatrix} - \begin{pmatrix} -23 & -7 \\ -3 & -2 \end{pmatrix} + \begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix}$

First, subtract:

$A^2 - AB = \begin{pmatrix} 25 - (-23) & -24 - (-7) \\ 0 - (-3) & 1 - (-2) \end{pmatrix} = \begin{pmatrix} 48 & -17 \\ 3 & 3 \end{pmatrix}$

Now, add $3E$:

$C = \begin{pmatrix} 48 & -17 \\ 3 & 3 \end{pmatrix} + \begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix} = \begin{pmatrix} 48 + 3 & -17 + 0 \\ 3 + 0 & 3 + 3 \end{pmatrix}$

$C = \begin{pmatrix} 51 & -17 \\ 3 & 6 \end{pmatrix}$

Final Answer:

$C = \begin{pmatrix} 51 & -17 \\ 3 & 6 \end{pmatrix}$

Let me know if you need any further details or clarifications!

Here are 5 related questions for practice:

1. How would you calculate $A^3$ using the matrix $A$ provided?
2. Can you compute the determinant of matrix $A$?
3. How would the solution change if $B$ was a 3x2 matrix?
4. What properties do the identity matrix $E$ and its multiples have?
5. What is the inverse of matrix $A$ (if it exists)?

Tip: Always double-check your matrix multiplications row by column to avoid simple errors.