If `A=[{:(1,-1),(2,-1):}],B=[{:(a,-1),(b,-1):}]" and "(A+B)^(2)=(A^(2)+B^(2))` then find the values of a and b.

Options

A. 4
B. 5
C. 6
D. 7

Answer

To solve the equation \((A + B)^2 = A^2 + B^2\) for the matrices \(A\) and \(B\), we start with the given matrices:

\[ A = \begin{pmatrix} 1 & -1 \\ 2 & -1 \end{pmatrix}, \quad B = \begin{pmatrix} a & -1 \\ b & -1 \end{pmatrix} \]

### Step 1: Expand the Left Side
We need to expand \((A + B)^2\):

\[ A + B = \begin{pmatrix} 1 & -1 \\ 2 & -1 \end{pmatrix} + \begin{pmatrix} a & -1 \\ b & -1 \end{pmatrix} = \begin{pmatrix} 1 + a & -1 - 1 \\ 2 + b & -1 - 1 \end{pmatrix} = \begin{pmatrix} 1 + a & -2 \\ 2 + b & -2 \end{pmatrix} \]

Now, we square this matrix:

\[ (A + B)^2 = \begin{pmatrix} 1 + a & -2 \\ 2 + b & -2 \end{pmatrix} \cdot \begin{pmatrix} 1 + a & -2 \\ 2 + b & -2 \end{pmatrix} \]

Using matrix multiplication:

\[ = \begin{pmatrix} (1 + a)(1 + a) + (-2)(2 + b) & (1 + a)(-2) + (-2)(-2) \\ (2 + b)(1 + a) + (-2)(2 + b) & (2 + b)(-2) + (-2)(-2) \end{pmatrix} \]

Calculating each element:

1. First row, first column:
\[ (1 + a)^2 - 4 - 2b = (1 + 2a + a^2 - 4 - 2b) = a^2 + 2a - 2b - 3 \]

2. First row, second column:
\[ -2(1 + a) + 4 = -2 - 2a + 4 = 2 - 2a \]

3. Second row, first column:
\[ (2 + b)(1 + a) - 4 - 2b = (2 + 2a + b + ab - 4 - 2b) = 2a + ab - b - 2 \]

4. Second row, second column:
\[ -2(2 + b) + 4 = -4 - 2b + 4 = -2b \]

Thus, we have:

\[ (A + B)^2 = \begin{pmatrix} a^2 + 2a - 2b - 3 & 2 - 2a \\ 2a + ab - b - 2 & -2b \end{pmatrix} \]

### Step 2: Calculate the Right Side
Now we calculate \(A^2 + B^2\):

1. Calculate \(A^2\):
\[ A^2 = \begin{pmatrix} 1 & -1 \\ 2 & -1 \end{pmatrix} \cdot \begin{pmatrix} 1 & -1 \\ 2 & -1 \end{pmatrix} = \begin{pmatrix} 1 - 2 & -1 + 1 \\ 2 - 2 & -2 + 1 \end{pmatrix} = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} \]

2. Calculate \(B^2\):
\[ B^2 = \begin{pmatrix} a & -1 \\ b & -1 \end{pmatrix} \cdot \begin{pmatrix} a & -1 \\ b & -1 \end{pmatrix} = \begin{pmatrix} a^2 - b & -a + 1 \\ ab - b & -b - 1 \end{pmatrix} \]

Now, we add \(A^2\) and \(B^2\):

\[ A^2 + B^2 = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} + \begin{pmatrix} a^2 - b & -a + 1 \\ ab - b & -b - 1 \end{pmatrix} = \begin{pmatrix} a^2 - b - 1 & -a + 1 \\ ab - b & -b - 2 \end{pmatrix} \]

### Step 3: Set the Matrices Equal
Now we set the two matrices equal:

\[ \begin{pmatrix} a^2 + 2a - 2b - 3 & 2 - 2a \\ 2a + ab - b - 2 & -2b \end{pmatrix} = \begin{pmatrix} a^2 - b - 1 & -a + 1 \\ ab - b & -b - 2 \end{pmatrix} \]

From this, we can equate the corresponding elements:

1. \(a^2 + 2a - 2b - 3 = a^2 - b - 1\)
2. \(2 - 2a = -a + 1\)
3. \(2a + ab - b - 2 = ab - b\)
4. \(-2b = -b - 2\)

### Step 4: Solve the Equations
1. From \(a^2 + 2a - 2b - 3 = a^2 - b - 1\):
\[ 2a - b - 2 = 0 \implies b = 2a - 2 \]

2. From \(2 - 2a = -a + 1\):
\[ 1 - 2 = -a + 2a \implies a = 1 \]

3. Substitute \(a = 1\) into \(b = 2a - 2\):
\[ b = 2(1) - 2 = 0 \]

### Final Values
Thus, the values of \(a\) and \(b\) are:

\[ \boxed{a = 1, b = 0} \]