Board 2024 Class 12 Maths Q18 Solution: If (A + B)² = A² + B², then AB = −BA

The question

If A and B are two non-zero square matrices of same order such that $(A + B)^2 = A^2 + B^2$ ⁠, then :

(A) $AB = O$

(B) $AB = -BA$

(C) $BA = O$

(D) $AB = BA$

Exam answer

Matrix multiplication is not commutative in general, so

\begin{aligned} &(A + B)^2 \\ &= (A + B)(A + B) \\ &= A(A + B) + B(A + B) \\ &= A^2 + AB + BA + B^2 \end{aligned}

Why?

Multiply out using the distributive law, keeping each product in its order:

\begin{aligned} A(A + B) &= A^2 + AB \\ B(A + B) &= BA + B^2 \end{aligned}

Watch out: for numbers, $ab + ba = 2ab$ ⁠. For matrices, $AB$ and $BA$ are usually different, so the middle terms cannot be combined into $2AB$ ⁠.

Given $(A + B)^2 = A^2 + B^2$ ⁠:

\begin{aligned} A^2 + AB + BA + B^2 &= A^2 + B^2 \\ \Rightarrow AB + BA &= O \\ \Rightarrow AB &= -BA \end{aligned}

Why?

Subtract $A^2$ and $B^2$ from both sides. What is left must be the zero matrix $O$ ⁠.

For the other options, take

A = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix},\ B = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}

AB = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix},\ BA = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}

Here $AB = -BA$ ⁠, so the condition holds, but $AB \ne O$ ⁠, $BA \ne O$ and $AB \ne BA$ ⁠. So (A), (C) and (D) need not be true.

Why?

Each entry of $AB$ is a row of $A$ times a column of $B$ ⁠. For example, row 1 of $A$ times column 2 of $B$ is $0 \times 0 + 1 \times (-1) = -1$ ⁠.

All four entries of $AB$ (row of $A$ times column of $B$ ⁠):

\begin{aligned} (1,1)&: (0)(1) + (1)(0) = 0 \\ (1,2)&: (0)(0) + (1)(-1) = -1 \\ (2,1)&: (1)(1) + (0)(0) = 1 \\ (2,2)&: (1)(0) + (0)(-1) = 0 \end{aligned}

All four entries of $BA$ (row of $B$ times column of $A$ ⁠):

\begin{aligned} (1,1)&: (1)(0) + (0)(1) = 0 \\ (1,2)&: (1)(1) + (0)(0) = 1 \\ (2,1)&: (0)(0) + (-1)(1) = -1 \\ (2,2)&: (0)(1) + (-1)(0) = 0 \end{aligned}

Change every sign of $BA$ ⁠:

-BA = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} = AB

One pair of non-zero matrices that satisfies the condition but breaks an option is enough to rule that option out.

Answer: (B) $AB = -BA$

Where this comes from

Chapter 3, Matrices: Algebra of matrices: expanding (A + B)².

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Q18: If (A+B)2=A2+B2(A + B)^2 = A^2 + B^2(A+B)2=A2+B2⁠, then AB=−BAAB = -BAAB=−BA