If A `=[(1,2),(3,4)]` then `A^(2)-5A` equals

Options

A. O
B. I
C. 2I
D. none of these

Answer

To solve the problem, we need to compute \( A^2 - 5A \) for the matrix \( A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} \).

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

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

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

Calculating the elements of the resulting matrix:

- First row, first column:
\[ 1 \times 1 + 2 \times 3 = 1 + 6 = 7 \]

- First row, second column:
\[ 1 \times 2 + 2 \times 4 = 2 + 8 = 10 \]

- Second row, first column:
\[ 3 \times 1 + 4 \times 3 = 3 + 12 = 15 \]

- Second row, second column:
\[ 3 \times 2 + 4 \times 4 = 6 + 16 = 22 \]

Thus, we have:
\[ A^2 = \begin{pmatrix} 7 & 10 \\ 15 & 22 \end{pmatrix} \]

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

Next, we calculate \( 5A \) by multiplying each element of matrix \( A \) by 5:

\[ 5A = 5 \times \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} = \begin{pmatrix} 5 \times 1 & 5 \times 2 \\ 5 \times 3 & 5 \times 4 \end{pmatrix} = \begin{pmatrix} 5 & 10 \\ 15 & 20 \end{pmatrix} \]

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

Now we subtract \( 5A \) from \( A^2 \):

\[ A^2 - 5A = \begin{pmatrix} 7 & 10 \\ 15 & 22 \end{pmatrix} - \begin{pmatrix} 5 & 10 \\ 15 & 20 \end{pmatrix} \]

Calculating the elements of the resulting matrix:

- First row, first column:
\[ 7 - 5 = 2 \]

- First row, second column:
\[ 10 - 10 = 0 \]

- Second row, first column:
\[ 15 - 15 = 0 \]

- Second row, second column:
\[ 22 - 20 = 2 \]

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

### Step 4: Factor out the common term

We can factor out 2 from the resulting matrix:
\[ A^2 - 5A = 2 \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]

The matrix inside the parentheses is the identity matrix \( I \):
\[ A^2 - 5A = 2I \]

### Final Answer

Therefore, the final answer is:
\[ A^2 - 5A = 2I \]

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