If `A=[(5,3),(12,7)]`, show that `A^(2)-12A-I=0`. Hence find `A^(-1)`.

Answer

To solve the problem, we need to show that \( A^2 - 12A - I = 0 \) for the matrix \( A = \begin{pmatrix} 5 & 3 \\ 12 & 7 \end{pmatrix} \) and then find the inverse of \( A \).

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

First, we need to compute \( A^2 \):

\[ A^2 = A \cdot A = \begin{pmatrix} 5 & 3 \\ 12 & 7 \end{pmatrix} \cdot \begin{pmatrix} 5 & 3 \\ 12 & 7 \end{pmatrix} \]

Calculating the product:

\[ A^2 = \begin{pmatrix} (5 \cdot 5 + 3 \cdot 12) & (5 \cdot 3 + 3 \cdot 7) \\ (12 \cdot 5 + 7 \cdot 12) & (12 \cdot 3 + 7 \cdot 7) \end{pmatrix} \]

Calculating each element:

- First row, first column: \( 5 \cdot 5 + 3 \cdot 12 = 25 + 36 = 61 \)
- First row, second column: \( 5 \cdot 3 + 3 \cdot 7 = 15 + 21 = 36 \)
- Second row, first column: \( 12 \cdot 5 + 7 \cdot 12 = 60 + 84 = 144 \)
- Second row, second column: \( 12 \cdot 3 + 7 \cdot 7 = 36 + 49 = 85 \)

Thus,

\[ A^2 = \begin{pmatrix} 61 & 36 \\ 144 & 85 \end{pmatrix} \]

### Step 2: Calculate \( 12A \)

Next, we calculate \( 12A \):

\[ 12A = 12 \cdot \begin{pmatrix} 5 & 3 \\ 12 & 7 \end{pmatrix} = \begin{pmatrix} 60 & 36 \\ 144 & 84 \end{pmatrix} \]

### Step 3: Calculate \( A^2 - 12A - I \)

Now we need to compute \( A^2 - 12A - I \):

First, we need the identity matrix \( I \):

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

Now, we subtract \( 12A \) and \( I \) from \( A^2 \):

\[ A^2 - 12A - I = \begin{pmatrix} 61 & 36 \\ 144 & 85 \end{pmatrix} - \begin{pmatrix} 60 & 36 \\ 144 & 84 \end{pmatrix} - \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]

Calculating the subtraction:

\[ = \begin{pmatrix} 61 - 60 - 1 & 36 - 36 - 0 \\ 144 - 144 - 0 & 85 - 84 - 1 \end{pmatrix} \]

This simplifies to:

\[ = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} \]

Thus, we have shown that:

\[ A^2 - 12A - I = 0 \]

### Step 4: Find \( A^{-1} \)

From the equation \( A^2 - 12A - I = 0 \), we can rearrange it to find \( A^{-1} \):

\[ A^2 - 12A = I \]

Multiplying both sides by \( A^{-1} \):

\[ A - 12I = A^{-1} \]

Now, we calculate \( 12I \):

\[ 12I = 12 \cdot \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 12 & 0 \\ 0 & 12 \end{pmatrix} \]

Now, substituting \( A \) and \( 12I \):

\[ A^{-1} = A - 12I = \begin{pmatrix} 5 & 3 \\ 12 & 7 \end{pmatrix} - \begin{pmatrix} 12 & 0 \\ 0 & 12 \end{pmatrix} \]

Calculating the subtraction:

\[ A^{-1} = \begin{pmatrix} 5 - 12 & 3 - 0 \\ 12 - 0 & 7 - 12 \end{pmatrix} = \begin{pmatrix} -7 & 3 \\ 12 & -5 \end{pmatrix} \]

Thus, the inverse of \( A \) is:

\[ A^{-1} = \begin{pmatrix} -7 & 3 \\ 12 & -5 \end{pmatrix} \]

### Summary of Steps:

1. Calculate \( A^2 \).
2. Calculate \( 12A \).
3. Show that \( A^2 - 12A - I = 0 \).
4. Rearrange to find \( A^{-1} \).