For `A=[{:(4,2i),(i,1):}],(A-2l)(A-3l)` is a

Options

A. null matrix
B. Hermition matrix
C. unit matrix
D. none of these

Answer

To solve the problem, we need to compute the expression \((A - 2I)(A - 3I)\), where \(A\) is given as the matrix:

\[ A = \begin{pmatrix} 4 & 2i \\ i & 1 \end{pmatrix} \]

and \(I\) is the identity matrix of the same size:

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

### Step 1: Compute \(A - 2I\)

To find \(A - 2I\), we first calculate \(2I\):

\[ 2I = 2 \cdot \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} \]

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

\[ A - 2I = \begin{pmatrix} 4 & 2i \\ i & 1 \end{pmatrix} - \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} = \begin{pmatrix} 4 - 2 & 2i - 0 \\ i - 0 & 1 - 2 \end{pmatrix} = \begin{pmatrix} 2 & 2i \\ i & -1 \end{pmatrix} \]

### Step 2: Compute \(A - 3I\)

Next, we calculate \(3I\):

\[ 3I = 3 \cdot \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix} \]

Now, we subtract \(3I\) from \(A\):

\[ A - 3I = \begin{pmatrix} 4 & 2i \\ i & 1 \end{pmatrix} - \begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix} = \begin{pmatrix} 4 - 3 & 2i - 0 \\ i - 0 & 1 - 3 \end{pmatrix} = \begin{pmatrix} 1 & 2i \\ i & -2 \end{pmatrix} \]

### Step 3: Compute the product \((A - 2I)(A - 3I)\)

Now we need to multiply the two matrices we obtained:

\[ (A - 2I)(A - 3I) = \begin{pmatrix} 2 & 2i \\ i & -1 \end{pmatrix} \begin{pmatrix} 1 & 2i \\ i & -2 \end{pmatrix} \]

Using the row-column multiplication rule, we compute each element of the resulting matrix:

1. First row, first column:
\[ 2 \cdot 1 + 2i \cdot i = 2 + 2i^2 = 2 - 2 = 0 \]

2. First row, second column:
\[ 2 \cdot 2i + 2i \cdot (-2) = 4i - 4i = 0 \]

3. Second row, first column:
\[ i \cdot 1 + (-1) \cdot i = i - i = 0 \]

4. Second row, second column:
\[ i \cdot 2i + (-1) \cdot (-2) = 2i^2 + 2 = -2 + 2 = 0 \]

Putting it all together, we have:

\[ (A - 2I)(A - 3I) = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} \]

This is the null matrix.

### Final Answer

The result of the expression \((A - 2I)(A - 3I)\) is the null matrix.

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