If `{:A=[(0,1),(1,0)]:}`,I is the unit matrix of order 2 and a, b are arbitray constants, then `(aI +bA)^2` is equal to

Options

A. `a^2I-abA`
B. `a^2I+2abA`
C. `a^2I+b^2A`
D. none of these

Answer

To solve the problem, we need to calculate \((aI + bA)^2\) where \(A = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\) and \(I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\).

### Step 1: Write down the expression
We start with the expression:
\[ (aI + bA)^2 \]

### Step 2: Substitute the matrices
Substituting the values of \(I\) and \(A\):
\[ aI + bA = a \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} + b \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \]
This gives:
\[ = \begin{pmatrix} a & 0 \\ 0 & a \end{pmatrix} + \begin{pmatrix} 0 & b \\ b & 0 \end{pmatrix} \]
Now, we can add these two matrices:
\[ = \begin{pmatrix} a & b \\ b & a \end{pmatrix} \]

### Step 3: Square the resulting matrix
Now we need to square the resulting matrix:
\[ \left(\begin{pmatrix} a & b \\ b & a \end{pmatrix}\right)^2 \]
Using the formula for matrix multiplication:
\[ \begin{pmatrix} a & b \\ b & a \end{pmatrix} \begin{pmatrix} a & b \\ b & a \end{pmatrix} = \begin{pmatrix} a^2 + b^2 & ab + ab \\ ab + ab & b^2 + a^2 \end{pmatrix} \]
This simplifies to:
\[ = \begin{pmatrix} a^2 + b^2 & 2ab \\ 2ab & a^2 + b^2 \end{pmatrix} \]

### Final Result
Thus, we have:
\[ (aI + bA)^2 = \begin{pmatrix} a^2 + b^2 & 2ab \\ 2ab & a^2 + b^2 \end{pmatrix} \]

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