If `f(x)=x^(2)+4x-5andA={:[(1,2),(4,-3)]:}`, then f(A) is equal to

Options

A. `{:[(0,-4),(8,8)]:}`
B. `{:[(2,1),(2,0)]:}`
C. `{:[(1,1),(1,0)]:}`
D. `{:[(8,4),(8,0)]:}`

Answer

To solve the problem, we need to find \( f(A) \) where \( f(x) = x^2 + 4x - 5 \) and \( A = \begin{pmatrix} 1 & 2 \\ 4 & -3 \end{pmatrix} \).

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

We first need to find \( A^2 \) by multiplying matrix \( A \) by itself.

\[ A^2 = A \cdot A = \begin{pmatrix} 1 & 2 \\ 4 & -3 \end{pmatrix} \cdot \begin{pmatrix} 1 & 2 \\ 4 & -3 \end{pmatrix} \]

Calculating the elements:

- First row, first column: \( 1 \cdot 1 + 2 \cdot 4 = 1 + 8 = 9 \)
- First row, second column: \( 1 \cdot 2 + 2 \cdot (-3) = 2 - 6 = -4 \)
- Second row, first column: \( 4 \cdot 1 + (-3) \cdot 4 = 4 - 12 = -8 \)
- Second row, second column: \( 4 \cdot 2 + (-3) \cdot (-3) = 8 + 9 = 17 \)

Thus,

\[ A^2 = \begin{pmatrix} 9 & -4 \\ -8 & 17 \end{pmatrix} \]

### Step 2: Calculate \( 4A \)

Next, we calculate \( 4A \):

\[ 4A = 4 \cdot \begin{pmatrix} 1 & 2 \\ 4 & -3 \end{pmatrix} = \begin{pmatrix} 4 \cdot 1 & 4 \cdot 2 \\ 4 \cdot 4 & 4 \cdot (-3) \end{pmatrix} = \begin{pmatrix} 4 & 8 \\ 16 & -12 \end{pmatrix} \]

### Step 3: Calculate \( 5I \)

Now, we calculate \( 5I \) where \( I \) is the identity matrix:

\[ I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \implies 5I = 5 \cdot \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 5 & 0 \\ 0 & 5 \end{pmatrix} \]

### Step 4: Calculate \( f(A) = A^2 + 4A - 5I \)

Now we can compute \( f(A) \):

\[ f(A) = A^2 + 4A - 5I = \begin{pmatrix} 9 & -4 \\ -8 & 17 \end{pmatrix} + \begin{pmatrix} 4 & 8 \\ 16 & -12 \end{pmatrix} - \begin{pmatrix} 5 & 0 \\ 0 & 5 \end{pmatrix} \]

Calculating the sum:

1. First, add \( A^2 \) and \( 4A \):

\[ \begin{pmatrix} 9 & -4 \\ -8 & 17 \end{pmatrix} + \begin{pmatrix} 4 & 8 \\ 16 & -12 \end{pmatrix} = \begin{pmatrix} 9 + 4 & -4 + 8 \\ -8 + 16 & 17 - 12 \end{pmatrix} = \begin{pmatrix} 13 & 4 \\ 8 & 5 \end{pmatrix} \]

2. Now subtract \( 5I \):

\[ \begin{pmatrix} 13 & 4 \\ 8 & 5 \end{pmatrix} - \begin{pmatrix} 5 & 0 \\ 0 & 5 \end{pmatrix} = \begin{pmatrix} 13 - 5 & 4 - 0 \\ 8 - 0 & 5 - 5 \end{pmatrix} = \begin{pmatrix} 8 & 4 \\ 8 & 0 \end{pmatrix} \]

### Final Result

Thus,

\[ f(A) = \begin{pmatrix} 8 & 4 \\ 8 & 0 \end{pmatrix} \]