Let A =`[(1,1,0),(0,1,0),(0,0,1)]` and let I denote the `3xx3` identity matrix . Then `2A^(2) -A^(3)` =

Options

A. A+I
B. A-I
C. I-A
D. A

Answer

To solve the problem, we need to compute \(2A^2 - A^3\) where \(A = \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}\).

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

We start by calculating \(A^2\) which is \(A \times A\).

\[ A^2 = A \times A = \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \times \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

Calculating the elements:

- First row, first column: \(1 \cdot 1 + 1 \cdot 0 + 0 \cdot 0 = 1\)
- First row, second column: \(1 \cdot 1 + 1 \cdot 1 + 0 \cdot 0 = 2\)
- First row, third column: \(1 \cdot 0 + 1 \cdot 0 + 0 \cdot 1 = 0\)

- Second row, first column: \(0 \cdot 1 + 1 \cdot 0 + 0 \cdot 0 = 0\)
- Second row, second column: \(0 \cdot 1 + 1 \cdot 1 + 0 \cdot 0 = 1\)
- Second row, third column: \(0 \cdot 0 + 1 \cdot 0 + 0 \cdot 1 = 0\)

- Third row, first column: \(0 \cdot 1 + 0 \cdot 0 + 1 \cdot 0 = 0\)
- Third row, second column: \(0 \cdot 1 + 0 \cdot 1 + 1 \cdot 0 = 0\)
- Third row, third column: \(0 \cdot 0 + 0 \cdot 0 + 1 \cdot 1 = 1\)

Thus, we have:

\[ A^2 = \begin{pmatrix} 1 & 2 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

### Step 2: Calculate \(A^3\)

Next, we calculate \(A^3\) which is \(A^2 \times A\).

\[ A^3 = A^2 \times A = \begin{pmatrix} 1 & 2 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \times \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

Calculating the elements:

- First row, first column: \(1 \cdot 1 + 2 \cdot 0 + 0 \cdot 0 = 1\)
- First row, second column: \(1 \cdot 1 + 2 \cdot 1 + 0 \cdot 0 = 3\)
- First row, third column: \(1 \cdot 0 + 2 \cdot 0 + 0 \cdot 1 = 0\)

- Second row, first column: \(0 \cdot 1 + 1 \cdot 0 + 0 \cdot 0 = 0\)
- Second row, second column: \(0 \cdot 1 + 1 \cdot 1 + 0 \cdot 0 = 1\)
- Second row, third column: \(0 \cdot 0 + 1 \cdot 0 + 0 \cdot 1 = 0\)

- Third row, first column: \(0 \cdot 1 + 0 \cdot 0 + 1 \cdot 0 = 0\)
- Third row, second column: \(0 \cdot 1 + 0 \cdot 1 + 1 \cdot 0 = 0\)
- Third row, third column: \(0 \cdot 0 + 0 \cdot 0 + 1 \cdot 1 = 1\)

Thus, we have:

\[ A^3 = \begin{pmatrix} 1 & 3 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

### Step 3: Calculate \(2A^2\)

Now we calculate \(2A^2\):

\[ 2A^2 = 2 \times \begin{pmatrix} 1 & 2 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 2 & 4 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{pmatrix} \]

### Step 4: Calculate \(2A^2 - A^3\)

Finally, we compute \(2A^2 - A^3\):

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

Calculating the elements:

- First row, first column: \(2 - 1 = 1\)
- First row, second column: \(4 - 3 = 1\)
- First row, third column: \(0 - 0 = 0\)

- Second row, first column: \(0 - 0 = 0\)
- Second row, second column: \(2 - 1 = 1\)
- Second row, third column: \(0 - 0 = 0\)

- Third row, first column: \(0 - 0 = 0\)
- Third row, second column: \(0 - 0 = 0\)
- Third row, third column: \(2 - 1 = 1\)

Thus, we have:

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

### Final Answer

\[ 2A^2 - A^3 = A \]