If $A=\begin{bmatrix}1&-1\ 2&-1\end{bmatrix}$ and $B=\begin{bmatrix}a&1\ b&-1\end{bmatrix}$ such that $A^{2}+B^{2}=(A+B)^{2},$ then :

Q: If $A=\begin{bmatrix}1&-1\\ 2&-1\end{bmatrix}$ and $B=\begin{bmatrix}a&1\\ b&-1\end{bmatrix}$ such that $A^{2}+B^{2}=(A+B)^{2},$ then :

If $A=\begin{bmatrix}1&-1\\ 2&-1\end{bmatrix}$ and $B=\begin{bmatrix}a&1\\ b&-1\end{bmatrix}$ such that $A^{2}+B^{2}=(A+B)^{2},$ then :

Question

Accepted Answer

To solve the problem, we need to verify the condition $ A^2 + B^2 = (A + B)^2 $ for the given matrices $ A $ and $ B $. Let's start by calculating each component.

First, calculate $ A^2 $:

$$
A = \begin{bmatrix} 1 & -1 \ 2 & -1 \end{bmatrix}
$$

$$
A^2 = A \cdot A = \begin{bmatrix} 1 & -1 \ 2 & -1 \end{bmatrix} \cdot \begin{bmatrix} 1 & -1 \ 2 & -1 \end{bmatrix} = \begin{bmatrix} 1 \cdot 1 + (-1) \cdot 2 & 1 \cdot (-1) + (-1) \cdot (-1) \ 2 \cdot 1 + (-1) \cdot 2 & 2 \cdot (-1) + (-1) \cdot (-1) \end{bmatrix}
$$

$$
= \begin{bmatrix} 1 - 2 & -1 + 1 \ 2 - 2 & -2 + 1 \end{bmatrix} = \begin{bmatrix} -1 & 0 \ 0 & -1 \end{bmatrix}
$$

Next, calculate $ B^2 $:

$$
B = \begin{bmatrix} a & 1 \ b & -1 \end{bmatrix}
$$

$$
B^2 = B \cdot B = \begin{bmatrix} a & 1 \ b & -1 \end{bmatrix} \cdot \begin{bmatrix} a & 1 \ b & -1 \end{bmatrix} = \begin{bmatrix} a^2 + 1 \cdot b & a \cdot 1 + 1 \cdot (-1) \ b \cdot a + (-1) \cdot b & b \cdot 1 + (-1) \cdot (-1) \end{bmatrix}
$$

$$
= \begin{bmatrix} a^2 + b & a - 1 \ ab - b & b + 1 \end{bmatrix}
$$

Now, calculate $ A + B $:

$$
A + B = \begin{bmatrix} 1 & -1 \ 2 & -1 \end{bmatrix} + \begin{bmatrix} a & 1 \ b & -1 \end{bmatrix} = \begin{bmatrix} 1 + a & -1 + 1 \ 2 + b & -1 - 1 \end{bmatrix}
$$

$$
= \begin{bmatrix} 1 + a & 0 \ 2 + b & -2 \end{bmatrix}
$$

Calculate $ (A + B)^2 $:

$$
(A + B)^2 = \begin{bmatrix} 1 + a & 0 \ 2 + b & -2 \end{bmatrix} \cdot \begin{bmatrix} 1 + a & 0 \ 2 + b & -2 \end{bmatrix}
$$

$$
= \begin{bmatrix} (1 + a)^2 & 0 \ (2 + b)(1 + a) & (2 + b) \cdot 0 + (-2)(-2) \end{bmatrix}
$$

$$
= \begin{bmatrix} 1 + 2a + a^2 & 0 \ (2 + b)(1 + a) & 4 \end{bmatrix}
$$

Now, equate $ A^2 + B^2 $ and $ (A + B)^2 $:

$$
A^2 + B^2 = \begin{bmatrix} -1 & 0 \ 0 & -1 \end{bmatrix} + \begin{bmatrix} a^2 + b & a - 1 \ ab - b & b + 1 \end{bmatrix} = \begin{bmatrix} a^2 + b - 1 & a - 1 \ ab - b & b \end{bmatrix}
$$

Equating with $ (A + B)^2 $:

$$
\begin{bmatrix} a^2 + b - 1 & a - 1 \ ab - b & b \end{bmatrix} = \begin{bmatrix} 1 + 2a + a^2 & 0 \ (2 + b)(1 + a) & 4 \end{bmatrix}
$$

From the first element of the matrices:

$$ a^2 + b - 1 = 1 + 2a + a^2 $$

$$ b - 1 = 1 + 2a $$

$$ b = 2a + 2 $$

From the second element of the matrices:

$$ a - 1 = 0 $$

$$ a = 1 $$

Substitute $ a = 1 $ into $ b = 2a + 2 $:

$$ b = 2(1) + 2 = 4 $$

Thus, $ a = 1 $ and $ b = 4 $, which corresponds to Option 2.

Now, let's verify why the other options are incorrect:

1. **Option 1: $ a = 4, b = 1 $**

Substituting these values into $ b = 2a + 2 $ gives $ 1 = 2(4) + 2 = 10 $, which is incorrect.

3. **Option 3: $ a = 0, b = 4 $**

Substituting these values into $ b = 2a + 2 $ gives $ 4 = 2(0) + 2 = 2 $, which is incorrect.

4. **Option 4: $ a = 2, b = 4 $**

Substituting these values into $ b = 2a + 2 $ gives $ 4 = 2(2) + 2 = 6 $, which is incorrect.

Therefore, Option 2 is the only correct choice.