Show that if A and B are square matrices such that AB = BA, then (A + B)² = A² + 2AB + B².

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Concept understanding — Matrix Addition

Matrix Addition: The Intuition

You run a fruit stall and record apples and bananas sold each morning and afternoon in a table:

Time	Apples	Bananas
Morning	10	5
Afternoon	8	12

That's a matrix — a rectangular array of numbers. Your friend's stall has its own table for the same day (morning: 6 apples, 7 bananas; afternoon: 4 apples, 9 bananas). To get the combined sales, you add the numbers in the same position: morning apples with morning apples, afternoon bananas with afternoon bananas, and so on.

That's matrix addition: you add corresponding entries — numbers in the same row and column.

The Precise Statement

(A + B)_{ij} = A_{ij} + B_{ij}

If $A$ and $B$ have the same size (same number of rows and columns), their sum $A + B$ is a matrix of that size where each entry is the sum of the corresponding entries.

Example:

A = \begin{bmatrix} 2 & 5 \\ 1 & -3 \end{bmatrix}, \quad B = \begin{bmatrix} 0 & 4 \\ 7 & 2 \end{bmatrix}

A + B = \begin{bmatrix} 2+0 & 5+4 \\ 1+7 & -3+2 \end{bmatrix} = \begin{bmatrix} 2 & 9 \\ 8 & -1 \end{bmatrix}

The One Rule You Cannot Break

Watch out

You can only add matrices with the exact same dimensions. A $2 \times 3$ matrix cannot be added to a $3 \times 2$ matrix — the positions don't match.

Properties That Feel Natural

Matrix addition behaves like ordinary number addition, inheriting these from the addition of individual entries:

Commutative: $A + B = B + A$
Associative: $(A + B) + C = A + (B + C)$
Zero matrix: there's a matrix $O$ (all zeros) with $A + O = A$

A Quick Check …

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Matrix multiplication is not commutative in general, but when $A$ and $B$ do commute ( $AB = BA$ ), the binomial expansion $(A+B)^2 = A^2 + 2AB + B^2$ holds exactly as in ordinary algebra — the cross terms $AB$ and $BA$ become identical and add to $2AB$ .

The key idea here is simple but often missed: the familiar formula $(x+y)^2 = x^2 + 2xy + y^2$ from school algebra relies on the fact that $x$ and $y$ are numbers, and numbers commute ( $xy = yx$ ). For matrices, multiplication is not commutative in general — $AB$ and $BA$ can be different. So the expansion of $(A+B)^2$ is actually $A^2 + AB + BA + B^2$ . The extra condition $AB = BA$ is what lets us combine $AB$ and $BA$ into $2AB$ .

Let’s walk through it step by step.

Write out the square. By definition, $(A+B)^2 = (A+B)(A+B)$ . Matrix multiplication is distributive, so we expand just like with numbers:

(A+B)(A+B) = A(A+B) + B(A+B) = A^2 + AB + BA + B^2.

No shortcuts yet — this is the raw expansion, valid for any square matrices $A$ and $B$ .

Apply the commuting condition. We are given $AB = BA$ . That means the two middle terms are equal:

AB = BA.

So in the sum $AB + BA$ , we can replace $BA$ with $AB$ (or vice versa), giving:

AB + BA = AB + AB = 2AB.

Substitute back. Putting this into the expansion from step 1:

(A+B)^2 = A^2 + (AB + BA) + B^2 = A^2 + 2AB + B^2.

That’s the whole proof — it’s just two lines of algebra once you know the expansion rule. …

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

Showing the 12 most recent of 13 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$A = \begin{bmatrix}1 & 2 & 3\\ 2 & 3 & 1\end{bmatrix}$
(a) $\begin{bmatrix}-1 & 5 & 3\\ 5 & 6 & 0\end{bmatrix}$
(b) $\begin{bmatrix}5 & -1 & 3\\ 5 & 6 & 0\end{bmatrix}$
(c) $\begin{bmatrix}2 & 4 & 6\\ 4 & 6 & 2\end{bmatrix}$
(d) $\begin{bmatrix}-3 & 1 & -3\\ 5 & 6 & 2\end{bmatrix}$
›Reveal solutionSolution

Scale $A$ by 2 and subtract $B$ entry-by-entry.

$2A = \begin{bmatrix}2&4&6\\4&6&2\end{bmatrix}$ .

…

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CBSE 2026Set ANNUAL1 mark
$Q.If A = \begin{pmatrix}1 & 2 & 3\ 2 & 3 & 1\end{pmatrix} and B = \begin{pmatrix}3 & -1 & 3\ -1 & 0 & 2\end{pmatrix}, then find the value of 2A - B.$
›Reveal solutionSolution

Scale $A$ by $2$ (multiply every entry by 2), then subtract $B$ entry-by-entry.

Working:

$A = \begin{pmatrix}1 & 2 & 3\\ 2 & 3 & 1\end{pmatrix},\quad 2A = \begin{pmatrix}2 & 4 & 6\\ 4 & 6 & 2\end{pmatrix}$ …

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CBSE 2024Set ANNUAL1 markMCQ
$A = \begin{bmatrix}1 & 2 & 3\\ 2 & 3 & 1\end{bmatrix}$
(a) $\begin{bmatrix}1 & -5 & 2\\ 5 & 6 & 0\end{bmatrix}$
(b) $\begin{bmatrix}5 & 6 & 0\\ 1 & -5 & 3\end{bmatrix}$
(c) $\begin{bmatrix}-1 & 5 & 3\\ 5 & 6 & 0\end{bmatrix}$
(d) $\begin{bmatrix}-1 & 3 & 5\\ 5 & 6 & 0\end{bmatrix}$
›Reveal solutionSolution

Scale $A$ by 2 entrywise, then subtract $B$ entrywise.

$A = \begin{bmatrix}1 & 2 & 3\\ 2 & 3 & 1\end{bmatrix}$ , so $2A = \begin{bmatrix}2 & 4 & 6\\ 4 & 6 & 2\end{bmatrix}$ .

$B = \begin{bmatrix}3 & -1 & 3\\ -1 & 0 & 2\end{bmatrix}$ .

…

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CBSE 2024Set ANNUAL1 mark
$Q. If A = [[2, 4], [3, 2]], B = [[-2, 5], [3, 4]] then find 3A - B.$
›Reveal solutionSolution

Scale $A$ by 3, then subtract $B$ entrywise.

Given $A=\begin{pmatrix}2&4\\3&2\end{pmatrix}$ and $B=\begin{pmatrix}-2&5\\3&4\end{pmatrix}$ .

$3A=\begin{pmatrix}6&12\\9&6\end{pmatrix}$

…

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CBSE 2024Set ANNUAL1 markMCQ
$Q. Let [ILLEGIBLE — page defect obscures the matrix's own name/label] = [[2, 4], [3, 2]], C = [[-2, 5], [3, 4]], the value of [ILLEGIBLE — page defect obscures the operator between the two matrices] C is :$
(a) [[0, 9], [6, 6]]
(b) [[0, 6], [?, 6]] — one cell illegible (page defect)
(c) [[?, 7], [?, 2]] — two cells illegible (page defect)
(d) None of these
›Reveal solutionSolution

Although the matrix's own name/label and the operator symbol between the two matrices are obscured by a scan defect, the arithmetic can be recovered: treating the given matrix and C as being added, the sum matches option (a) exactly.

Note on the source scan: the label of the first matrix (e.g. 'A' or 'B') and the operator between the two matrices are lost to a page-defect blot. Given matrix (unnamed) = [[2, 4], [3, 2]] and C = [[−2, 5], [3, 4]].

…

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CBSE 2024Set ANNUAL1 markMCQ
$\begin{bmatrix}3 & -4\\ 1 & 2\end{bmatrix} + X = \begin{bmatrix}3 & 1\\ 5 & 0\end{bmatrix}$
(a) $\begin{bmatrix}6 & 5\\ 4 & 4\end{bmatrix}$
(b) $\begin{bmatrix}0 & 5\\ 4 & 2\end{bmatrix}$
(c) $\begin{bmatrix}0 & 5\\ 4 & -2\end{bmatrix}$
(d) $\begin{bmatrix}0 & 5\\ -4 & -2\end{bmatrix}$
›Reveal solutionSolution

Isolate X by subtracting the given matrix from the right-hand side matrix.

Given $\begin{bmatrix}3 & -4\\ 1 & 2\end{bmatrix} + X = \begin{bmatrix}3 & 1\\ 5 & 0\end{bmatrix}$

…

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CBSE 2023Set ANNUAL1 mark
$Q. If A = [[1, 3], [-2, 5]] and B = [[2, 4], [3, 2]] then find A + B.$
›Reveal solutionSolution

Add corresponding entries of the two matrices.

Matrix addition is defined entry-by-entry: if $A=[a_{ij}]$ and $B=[b_{ij}]$ are matrices of the same order, then $A+B=[a_{ij}+b_{ij}]$ .

Here $A = [[1, 3], [-2, 5]]$ and $B = [[2, 4], [3, 2]]$ .

…

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CBSE 2022Set ANNUAL1 markMCQ
$A=\begin{bmatrix}1&2\\0&1\end{bmatrix}$
(a) $\begin{bmatrix}6&5\\0&2\end{bmatrix}$
(b) $\begin{bmatrix}8&7\\-3&2\end{bmatrix}$
(c) $\begin{bmatrix}7&7\\-3&2\end{bmatrix}$
(d) $\begin{bmatrix}8&7\\-2&3\end{bmatrix}$
›Reveal solutionSolution

Scale each matrix entrywise, then add corresponding entries.

$A=\begin{bmatrix}1&2\\0&1\end{bmatrix}$ , $B=\begin{bmatrix}2&1\\-1&0\end{bmatrix}$ .

$2A=\begin{bmatrix}2&4\\0&2\end{bmatrix}$ , $3B=\begin{bmatrix}6&3\\-3&0\end{bmatrix}$ .

…

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CBSE 2022Set ANNUAL1 mark
$A = \begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix}$
›Reveal solutionSolution

Subtract matrices entry-by-entry.

$A=\begin{bmatrix}1 & 3\\ -2 & 5\end{bmatrix}$ , $B=\begin{bmatrix}2 & 4\\ 3 & 2\end{bmatrix}$ .

…

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CBSE 2022Set ANNUAL1 mark
$A = \begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix}$
›Reveal solutionSolution

Scale $A$ by 3 entrywise, then subtract $B$ entrywise.

Given $A = \begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix}$ , $B = \begin{bmatrix} -2 & 5 \\ 3 & 4 \end{bmatrix}$ .

Step 1: Compute $3A$ by multiplying every entry by 3:

$3A = \begin{bmatrix} 6 & 12 \\ 9 & 6 \end{bmatrix}$

Step 2: Subtract $B$ entrywise (matrices must have the same order to add/subtract, which they do here, both $2\times2$ ): …

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CBSE 2021Set I1 markMCQ
$\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} =$
(a) $\begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}$
(b) $\begin{bmatrix} 0 & 2 \\ 2 & 0 \end{bmatrix}$
(c) $\begin{bmatrix} 2 & 2 \\ 2 & 2 \end{bmatrix}$
(d) $\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$
›Reveal solutionSolution

Matrix addition is component-wise.

Add corresponding entries:

…

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CBSE 2020Set ANNUAL1 mark
$Q. If A = [[1, 3], [-2, 5]] and B = [[-2, 5], [3, 4]], find the value of A - B.$
›Reveal solutionSolution

Subtract corresponding entries of the two matrices.

…

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Q.Show that if $A$ and $B$ are square matrices such that $AB = BA$ , then $(A + B)^2 = A^2 + 2AB + B^2$ .

Concept understanding — Matrix Addition

Matrix Addition: The Intuition

The Precise Statement

The One Rule You Cannot Break

Properties That Feel Natural

A Quick Check …

Method: Expanding a matrix binomial using a commutativity condition

Steps

Common Mistakes

Q.Show that if AAA and BBB are square matrices such that AB=BAAB = BAAB=BA, then (A+B)2=A2+2AB+B2(A + B)^2 = A^2 + 2AB + B^2(A+B)2=A2+2AB+B2.

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Q.Show that if $A$ and $B$ are square matrices such that $AB = BA$ , then $(A + B)^2 = A^2 + 2AB + B^2$ .

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