Let `A=[{:(,4,-2),(,6,-3):}], B=[{:(,0,2),(,1,-1):}] and C=[{:(,-2,3),(,1,-1):}]`. Find `A^2-A+BC`.

Answer

To solve the problem, we need to calculate \( A^2 - A + BC \) where:

\[ A = \begin{pmatrix} 4 & -2 \\ 6 & -3 \end{pmatrix}, \quad B = \begin{pmatrix} 0 & 2 \\ 1 & -1 \end{pmatrix}, \quad C = \begin{pmatrix} -2 & 3 \\ 1 & -1 \end{pmatrix} \]

### Step 1: Calculate \( A^2 \)

To find \( A^2 \), we need to multiply matrix \( A \) by itself:

\[ A^2 = A \times A = \begin{pmatrix} 4 & -2 \\ 6 & -3 \end{pmatrix} \times \begin{pmatrix} 4 & -2 \\ 6 & -3 \end{pmatrix} \]

Calculating the elements of \( A^2 \):

- First row, first column:
\[ 4 \times 4 + (-2) \times 6 = 16 - 12 = 4 \]

- First row, second column:
\[ 4 \times (-2) + (-2) \times (-3) = -8 + 6 = -2 \]

- Second row, first column:
\[ 6 \times 4 + (-3) \times 6 = 24 - 18 = 6 \]

- Second row, second column:
\[ 6 \times (-2) + (-3) \times (-3) = -12 + 9 = -3 \]

Thus, we have:
\[ A^2 = \begin{pmatrix} 4 & -2 \\ 6 & -3 \end{pmatrix} \]

### Step 2: Calculate \( BC \)

Next, we calculate \( BC \):

\[ BC = B \times C = \begin{pmatrix} 0 & 2 \\ 1 & -1 \end{pmatrix} \times \begin{pmatrix} -2 & 3 \\ 1 & -1 \end{pmatrix} \]

Calculating the elements of \( BC \):

- First row, first column:
\[ 0 \times (-2) + 2 \times 1 = 0 + 2 = 2 \]

- First row, second column:
\[ 0 \times 3 + 2 \times (-1) = 0 - 2 = -2 \]

- Second row, first column:
\[ 1 \times (-2) + (-1) \times 1 = -2 - 1 = -3 \]

- Second row, second column:
\[ 1 \times 3 + (-1) \times (-1) = 3 + 1 = 4 \]

Thus, we have:
\[ BC = \begin{pmatrix} 2 & -2 \\ -3 & 4 \end{pmatrix} \]

### Step 3: Calculate \( A^2 - A + BC \)

Now we will combine all the results:

\[ A^2 - A + BC = \begin{pmatrix} 4 & -2 \\ 6 & -3 \end{pmatrix} - \begin{pmatrix} 4 & -2 \\ 6 & -3 \end{pmatrix} + \begin{pmatrix} 2 & -2 \\ -3 & 4 \end{pmatrix} \]

Calculating each element:

- First row, first column:
\[ 4 - 4 + 2 = 2 \]

- First row, second column:
\[ -2 - (-2) - 2 = -2 \]

- Second row, first column:
\[ 6 - 6 - 3 = -3 \]

- Second row, second column:
\[ -3 - (-3) + 4 = 4 \]

Thus, the final result is:
\[ A^2 - A + BC = \begin{pmatrix} 2 & -2 \\ -3 & 4 \end{pmatrix} \]

### Final Answer:
\[ \begin{pmatrix} 2 & -2 \\ -3 & 4 \end{pmatrix} \]