If $A = \begin{bmatrix}0 & -x\\ x & 0\end{bmatrix}$, $B = \begin{bmatrix}0 & 1\\ 1 & 0\end{bmatrix}$ and $x^2 = -1$, then show that $(A + B)^2 = A^2 + B^2$.

If A = 0 -xx 0 , B = 0 1\1 0 and x² = -1, then show that (A + B)² = A² + B².

NCERT Exemplar · Q34

A = \begin{bmatrix}0 & -x\\ x & 0\end{bmatrix}

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The key idea is that matrix multiplication is not generally commutative, but here $A$ and $B$ anti-commute ( $AB = -BA$ ), so the cross terms cancel. Using $x^2 = -1$ , we find $(A+B)^2 = A^2 + B^2$ .

We need to show that $(A+B)^2 = A^2 + B^2$ for the given matrices, where $x^2 = -1$ . The natural instinct is to expand $(A+B)^2 = A^2 + AB + BA + B^2$ . For this to equal $A^2 + B^2$ , we require $AB + BA = 0$ , i.e., $AB = -BA$ . So the problem reduces to checking whether $A$ and $B$ anti-commute.

Let’s verify this step by step.

Write down the matrices clearly.

$A = \begin{bmatrix}0 & -x \\ x & 0\end{bmatrix}$ , $B = \begin{bmatrix}0 & 1 \\ 1 & 0\end{bmatrix}$ , and we are given $x^2 = -1$ . Note that $x$ is not a real number — it behaves like the imaginary unit $i$ , but we treat it algebraically.
Compute $AB$ .

Multiply $A$ and $B$ :

AB = \begin{bmatrix}0 & -x \\ x & 0\end{bmatrix} \begin{bmatrix}0 & 1 \\ 1 & 0\end{bmatrix} = \begin{bmatrix} (0)(0) + (-x)(1) & (0)(1) + (-x)(0) \\ (x)(0) + (0)(1) & (x)(1) + (0)(0) \end{bmatrix} = \begin{bmatrix} -x & 0 \\ 0 & x \end{bmatrix}.

Compute $BA$ . Multiply in the reverse order:

BA = \begin{bmatrix}0 & 1 \\ 1 & 0\end{bmatrix} \begin{bmatrix}0 & -x \\ x & 0\end{bmatrix} = \begin{bmatrix} (0)(0) + (1)(x) & (0)(-x) + (1)(0) \\ (1)(0) + (0)(x) & (1)(-x) + (0)(0) \end{bmatrix} = \begin{bmatrix} x & 0 \\ 0 & -x \end{bmatrix}.

Observe the anti-commutation.

From steps 2 and 3, $AB = \begin{bmatrix} -x & 0 \\ 0 & x \end{bmatrix}$ and $BA = \begin{bmatrix} x & 0 \\ 0 & -x \end{bmatrix}$ . Clearly $AB = -BA$ , so $AB + BA = 0$ .

Tip

This anti-commutation property is the entire reason the cross terms vanish. In general, $(A+B)^2 = A^2 + B^2$ if and only if $AB = -BA$ . Always check this first.
Now compute $A^2$ and $B^2$ individually.

First, $A^2$ :

A^2 = \begin{bmatrix}0 & -x \\ x & 0\end{bmatrix} \begin{bmatrix}0 & -x \\ x & 0\end{bmatrix} = \begin{bmatrix} (0)(0) + (-x)(x) & (0)(-x) + (-x)(0) \\ (x)(0) + (0)(x) & (x)(-x) + (0)(0) \end{bmatrix} = \begin{bmatrix} -x^2 & 0 \\ 0 & -x^2 \end{bmatrix}.

Since $x^2 = -1$ , we have $-x^2 = -(-1) = 1$ , so …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

TG EAPCET 2022Set eng-2022-07-18-AN1 markMCQ
$A+B = \begin{bmatrix} 2 & 1 & 2 \\ 1 & 2 & 0 \\ 0 & 2 & 2 \end{bmatrix},\ AB = \begin{bmatrix} 1 & 2 & 2 \\ 1 & 1 & 0 \\ 1 & 2 & 1 \end{bmatrix}$
›Reveal solutionSolution
The expression $A^2 + B(A+B)$ simplifies to $(A+B)^2 - AB$ by using the distributive property of matrix multiplication. We calculate $(A+B)^2$ and then subtract $AB$ to find the result $\boxed{\begin{bmatrix} 4 & 6 & 6 \\ 3 & 4 & 2 \\ 1 & 6 & 3 \end{bmatrix}}$ .

The core idea here is to simplify the given expression $A^2 + B(A+B)$ using the properties of matrix algebra, specifically the distributive property of matrix multiplication over addition. We are given $A+B$ and $AB$ , so we should try to express the target expression in terms of these known quantities.

First, let's expand $B(A+B)$ :

$B(A+B) = BA + B^2$

So, the expression we need to evaluate becomes:

$A^2 + B(A+B) = A^2 + BA + B^2$

Now, let's recall the expansion of $(A+B)^2$ for matrices:
$A$
Watch out

It is crucial to remember that matrix multiplication is generally not commutative, meaning $AB \neq BA$ . Therefore, $(A+B)^2$ is not equal to $A^2 + 2AB + B^2$ unless $AB=BA$ .

Comparing the expression we need, $A^2 + BA + B^2$ , with the expansion of $(A+B)^2$ , which is $A^2 + AB + BA + B^2$ , we can see a direct relationship.

If we subtract $AB$ from $(A+B)^2$ , we get:

$(A+B)^2 - AB = (A^2 + AB + BA + B^2) - AB$

$= A^2 + (AB - AB) + BA + B^2$

$= A^2 + 0 + BA + B^2$

$= A^2 + BA + B^2$

This shows that $A^2 + B(A+B) = (A+B)^2 - AB$ . This simplification is key because we are given $A+B$ and $AB$ directly.

Now, we can proceed with the calculations:

Calculate $(A+B)^2$ :

We are given $A+B = \begin{bmatrix} 2 & 1 & 2 \\ 1 & 2 & 0 \\ 0 & 2 & 2 \end{bmatrix}$ .

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

Let's perform the matrix multiplication:

First row:

$(2)(2) + (1)(1) + (2)(0) = 4 + 1 + 0 = 5$

$(2)(1) + (1)(2) + (2)(2) = 2 + 2 + 4 = 8$

$(2)(2) + (1)(0) + (2)(2) = 4 + 0 + 4 = 8$

Second row:

$(1)(2) + (2)(1) + (0)(0) = 2 + 2 + 0 = 4$

$(1)(1) + (2)(2) + (0)(2) = 1 + 4 + 0 = 5$ …
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Q.If $A = \begin{bmatrix}0 & -x\\ x & 0\end{bmatrix}$ , $B = \begin{bmatrix}0 & 1\\ 1 & 0\end{bmatrix}$ and $x^2 = -1$ , then show that $(A + B)^2 = A^2 + B^2$ .

Concept understanding — Matrix Multiplication Compatibility

Matrix Multiplication Compatibility: The Inner-Dimensions Rule

The Idea: A Row Meets a Column

The Precise Statement

Why the Rule Exists

Quick Check

Method: Checking $(A+B)^2=A^2+B^2$ — when do the cross terms vanish?

Steps

Common Mistakes

Q.If A=[0−xx0]A = \begin{bmatrix}0 & -x\\ x & 0\end{bmatrix}A=[0x​−x0​], B=[0110]B = \begin{bmatrix}0 & 1\\ 1 & 0\end{bmatrix}B=[01​10​] and x2=−1x^2 = -1x2=−1, then show that (A+B)2=A2+B2(A + B)^2 = A^2 + B^2(A+B)2=A2+B2.

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Q.If $A = \begin{bmatrix}0 & -x\\ x & 0\end{bmatrix}$ , $B = \begin{bmatrix}0 & 1\\ 1 & 0\end{bmatrix}$ and $x^2 = -1$ , then show that $(A + B)^2 = A^2 + B^2$ .

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