If `A=[{:(3,-5),(-4,2):}]` then find `A^(2)-5A- 14I`. Hence, obtain `A^(3)`.

Options

A. I
B. 14I
C. 0
D. None of these

Answer

To solve the problem, we need to find \( A^2 - 5A - 14I \) where \( A = \begin{pmatrix} 3 & -5 \\ -4 & 2 \end{pmatrix} \) and \( I \) is the identity matrix of the same size. Then we will use the result to find \( A^3 \).

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

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

\[ A^2 = A \cdot A = \begin{pmatrix} 3 & -5 \\ -4 & 2 \end{pmatrix} \cdot \begin{pmatrix} 3 & -5 \\ -4 & 2 \end{pmatrix} \]

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

- First element: \( 3 \cdot 3 + (-5) \cdot (-4) = 9 + 20 = 29 \)
- Second element: \( 3 \cdot (-5) + (-5) \cdot 2 = -15 - 10 = -25 \)
- Third element: \( -4 \cdot 3 + 2 \cdot (-4) = -12 - 8 = -20 \)
- Fourth element: \( -4 \cdot (-5) + 2 \cdot 2 = 20 + 4 = 24 \)

Thus,

\[ A^2 = \begin{pmatrix} 29 & -25 \\ -20 & 24 \end{pmatrix} \]

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

Now, we calculate \( 5A \):

\[ 5A = 5 \cdot \begin{pmatrix} 3 & -5 \\ -4 & 2 \end{pmatrix} = \begin{pmatrix} 15 & -25 \\ -20 & 10 \end{pmatrix} \]

### Step 3: Calculate \( 14I \)

The identity matrix \( I \) for a \( 2 \times 2 \) matrix is:

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

Thus,

\[ 14I = 14 \cdot \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 14 & 0 \\ 0 & 14 \end{pmatrix} \]

### Step 4: Calculate \( A^2 - 5A - 14I \)

Now we can substitute \( A^2 \), \( 5A \), and \( 14I \) into the expression:

\[ A^2 - 5A - 14I = \begin{pmatrix} 29 & -25 \\ -20 & 24 \end{pmatrix} - \begin{pmatrix} 15 & -25 \\ -20 & 10 \end{pmatrix} - \begin{pmatrix} 14 & 0 \\ 0 & 14 \end{pmatrix} \]

Calculating this step by step:

1. First, subtract \( 5A \) from \( A^2 \):

\[ \begin{pmatrix} 29 & -25 \\ -20 & 24 \end{pmatrix} - \begin{pmatrix} 15 & -25 \\ -20 & 10 \end{pmatrix} = \begin{pmatrix} 29 - 15 & -25 - (-25) \\ -20 - (-20) & 24 - 10 \end{pmatrix} = \begin{pmatrix} 14 & 0 \\ 0 & 14 \end{pmatrix} \]

2. Now subtract \( 14I \):

\[ \begin{pmatrix} 14 & 0 \\ 0 & 14 \end{pmatrix} - \begin{pmatrix} 14 & 0 \\ 0 & 14 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} \]

Thus,

\[ A^2 - 5A - 14I = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} \]

### Step 5: Conclusion for \( A^2 - 5A - 14I \)

The result is the zero matrix, which corresponds to option 3.

### Step 6: Find \( A^3 \)

Since we know that \( A^2 - 5A - 14I = 0 \), we can rearrange this to find \( A^2 \):

\[ A^2 = 5A + 14I \]

Now, to find \( A^3 \):

\[ A^3 = A \cdot A^2 = A \cdot (5A + 14I) = 5A^2 + 14A \]

Substituting \( A^2 \):

\[ A^3 = 5 \cdot \begin{pmatrix} 29 & -25 \\ -20 & 24 \end{pmatrix} + 14 \cdot \begin{pmatrix} 3 & -5 \\ -4 & 2 \end{pmatrix} \]

Calculating \( 5A^2 \):

\[ 5A^2 = \begin{pmatrix} 145 & -125 \\ -100 & 120 \end{pmatrix} \]

Calculating \( 14A \):

\[ 14A = \begin{pmatrix} 42 & -70 \\ -56 & 28 \end{pmatrix} \]

Now, adding these two results:

\[ A^3 = \begin{pmatrix} 145 & -125 \\ -100 & 120 \end{pmatrix} + \begin{pmatrix} 42 & -70 \\ -56 & 28 \end{pmatrix} = \begin{pmatrix} 145 + 42 & -125 - 70 \\ -100 - 56 & 120 + 28 \end{pmatrix} = \begin{pmatrix} 187 & -195 \\ -156 & 148 \end{pmatrix} \]

### Final Answers

1. \( A^2 - 5A - 14I = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} \)
2. \( A^3 = \begin{pmatrix} 187 & -195 \\ -156 & 148 \end{pmatrix} \)