If `A=[(2,0,1),(2,1,3),(1,-1,0)]`, then find `(A^(2)-5A)`

Answer

To solve the problem, we need to find \( A^2 - 5A \) for the matrix \( A = \begin{pmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{pmatrix} \).

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

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

\[ A^2 = A \cdot A = \begin{pmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{pmatrix} \cdot \begin{pmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{pmatrix} \]

We will calculate each element of the resulting matrix:

- First row, first column:
\[ 2 \cdot 2 + 0 \cdot 2 + 1 \cdot 1 = 4 + 0 + 1 = 5 \]

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

- First row, third column:
\[ 2 \cdot 1 + 0 \cdot 3 + 1 \cdot 0 = 2 + 0 + 0 = 2 \]

- Second row, first column:
\[ 2 \cdot 2 + 1 \cdot 2 + 3 \cdot 1 = 4 + 2 + 3 = 9 \]

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

- Second row, third column:
\[ 2 \cdot 1 + 1 \cdot 3 + 3 \cdot 0 = 2 + 3 + 0 = 5 \]

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

- Third row, second column:
\[ 1 \cdot 0 + (-1) \cdot 1 + 0 \cdot (-1) = 0 - 1 + 0 = -1 \]

- Third row, third column:
\[ 1 \cdot 1 + (-1) \cdot 3 + 0 \cdot 0 = 1 - 3 + 0 = -2 \]

Thus, we have:
\[ A^2 = \begin{pmatrix} 5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2 \end{pmatrix} \]

### Step 2: Calculate \( 5A \)

Now, we need to calculate \( 5A \):
\[ 5A = 5 \cdot \begin{pmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{pmatrix} = \begin{pmatrix} 10 & 0 & 5 \\ 10 & 5 & 15 \\ 5 & -5 & 0 \end{pmatrix} \]

### Step 3: Calculate \( A^2 - 5A \)

Now, we will subtract \( 5A \) from \( A^2 \):
\[ A^2 - 5A = \begin{pmatrix} 5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2 \end{pmatrix} - \begin{pmatrix} 10 & 0 & 5 \\ 10 & 5 & 15 \\ 5 & -5 & 0 \end{pmatrix} \]

Calculating each element:
- First row:
\[ 5 - 10 = -5, \quad -1 - 0 = -1, \quad 2 - 5 = -3 \]

- Second row:
\[ 9 - 10 = -1, \quad -2 - 5 = -7, \quad 5 - 15 = -10 \]

- Third row:
\[ 0 - 5 = -5, \quad -1 - (-5) = 4, \quad -2 - 0 = -2 \]

Thus, we have:
\[ A^2 - 5A = \begin{pmatrix} -5 & -1 & -3 \\ -1 & -7 & -10 \\ -5 & 4 & -2 \end{pmatrix} \]

### Final Answer
\[ A^2 - 5A = \begin{pmatrix} -5 & -1 & -3 \\ -1 & -7 & -10 \\ -5 & 4 & -2 \end{pmatrix} \]