If `A=[(1,0),(0,1)]` then `7A^(3)+4A^(2)-11A=`………………..

Answer

To solve the expression \( 7A^3 + 4A^2 - 11A \) where \( A = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \), we will follow these steps:

### Step 1: Calculate \( A^2 \)
Since \( A \) is the identity matrix, we have:
\[ A^2 = A \cdot A = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \cdot \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]

### Step 2: Calculate \( A^3 \)
Next, we calculate \( A^3 \):
\[ A^3 = A^2 \cdot A = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \cdot \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]

### Step 3: Substitute \( A \), \( A^2 \), and \( A^3 \) into the expression
Now, substitute \( A \), \( A^2 \), and \( A^3 \) into the expression:
\[ 7A^3 + 4A^2 - 11A = 7 \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} + 4 \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} - 11 \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]

### Step 4: Calculate the coefficients
Calculating each term:
\[ 7A^3 = \begin{pmatrix} 7 & 0 \\ 0 & 7 \end{pmatrix} \]
\[ 4A^2 = \begin{pmatrix} 4 & 0 \\ 0 & 4 \end{pmatrix} \]
\[ -11A = \begin{pmatrix} -11 & 0 \\ 0 & -11 \end{pmatrix} \]

### Step 5: Combine the matrices
Now, combine these matrices:
\[ 7A^3 + 4A^2 - 11A = \begin{pmatrix} 7 & 0 \\ 0 & 7 \end{pmatrix} + \begin{pmatrix} 4 & 0 \\ 0 & 4 \end{pmatrix} + \begin{pmatrix} -11 & 0 \\ 0 & -11 \end{pmatrix} \]
\[ = \begin{pmatrix} 7 + 4 - 11 & 0 \\ 0 & 7 + 4 - 11 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} \]

### Final Answer
Thus, the final result is:
\[ \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} \]