If `A=[(2,3),(5,7)], B=[(0,4),(-1,7)]andC=[(1,0),(-1,4)]`
find `AC+B^(2)-10C`.

Answer

To solve the problem, we need to calculate the expression \( AC + B^2 - 10C \) using the given matrices \( A \), \( B \), and \( C \).

### Step 1: Define the Matrices
We have:
\[ A = \begin{pmatrix} 2 & 3 \\ 5 & 7 \end{pmatrix}, \quad B = \begin{pmatrix} 0 & 4 \\ -1 & 7 \end{pmatrix}, \quad C = \begin{pmatrix} 1 & 0 \\ -1 & 4 \end{pmatrix} \]

### Step 2: Calculate \( AC \)
To find \( AC \), we multiply matrix \( A \) with matrix \( C \):
\[ AC = A \cdot C = \begin{pmatrix} 2 & 3 \\ 5 & 7 \end{pmatrix} \cdot \begin{pmatrix} 1 & 0 \\ -1 & 4 \end{pmatrix} \]
Calculating the elements:
- First row, first column: \( 2 \cdot 1 + 3 \cdot (-1) = 2 - 3 = -1 \)
- First row, second column: \( 2 \cdot 0 + 3 \cdot 4 = 0 + 12 = 12 \)
- Second row, first column: \( 5 \cdot 1 + 7 \cdot (-1) = 5 - 7 = -2 \)
- Second row, second column: \( 5 \cdot 0 + 7 \cdot 4 = 0 + 28 = 28 \)

Thus,
\[ AC = \begin{pmatrix} -1 & 12 \\ -2 & 28 \end{pmatrix} \]

### Step 3: Calculate \( B^2 \)
Now, we find \( B^2 \) by multiplying \( B \) with itself:
\[ B^2 = B \cdot B = \begin{pmatrix} 0 & 4 \\ -1 & 7 \end{pmatrix} \cdot \begin{pmatrix} 0 & 4 \\ -1 & 7 \end{pmatrix} \]
Calculating the elements:
- First row, first column: \( 0 \cdot 0 + 4 \cdot (-1) = 0 - 4 = -4 \)
- First row, second column: \( 0 \cdot 4 + 4 \cdot 7 = 0 + 28 = 28 \)
- Second row, first column: \( -1 \cdot 0 + 7 \cdot (-1) = 0 - 7 = -7 \)
- Second row, second column: \( -1 \cdot 4 + 7 \cdot 7 = -4 + 49 = 45 \)

Thus,
\[ B^2 = \begin{pmatrix} -4 & 28 \\ -7 & 45 \end{pmatrix} \]

### Step 4: Calculate \( 10C \)
Next, we calculate \( 10C \):
\[ 10C = 10 \cdot \begin{pmatrix} 1 & 0 \\ -1 & 4 \end{pmatrix} = \begin{pmatrix} 10 & 0 \\ -10 & 40 \end{pmatrix} \]

### Step 5: Combine the Results
Now we need to compute \( AC + B^2 - 10C \):
\[ AC + B^2 - 10C = \begin{pmatrix} -1 & 12 \\ -2 & 28 \end{pmatrix} + \begin{pmatrix} -4 & 28 \\ -7 & 45 \end{pmatrix} - \begin{pmatrix} 10 & 0 \\ -10 & 40 \end{pmatrix} \]

Calculating the sum:
- First row, first column: \( -1 - 4 - 10 = -15 \)
- First row, second column: \( 12 + 28 - 0 = 40 \)
- Second row, first column: \( -2 - 7 + 10 = 1 \)
- Second row, second column: \( 28 + 45 - 40 = 33 \)

Thus, the final result is:
\[ AC + B^2 - 10C = \begin{pmatrix} -15 & 40 \\ 1 & 33 \end{pmatrix} \]