Q.If $A = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}$ and $B = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$, find $AB$ and $BA$, and show that $AB \neq BA$.

Showing the 12 most recent of 39 on this concept.

1. CBSE 2025Set 65/1/11 markMCQ
   Q.Let $A = \begin{bmatrix} 1 & -2 & -1 \\ 0 & 4 & -1 \\ -3 & 2 & 1 \end{bmatrix}$, $B = \begin{bmatrix} -2 \\ -5 \\ -7 \end{bmatrix}$, $C = [9 \ 8 \ 7]$, which of the following is defined ? (A) Only AB (B) Only AC (C) Only BA (D) All AB, AC and BA

   ›Reveal solutionSolution

   Matrix multiplication is defined only when the number of columns in the first matrix equals the number of rows in the second. Here, $A$ is $3 \times 3$, $B$ is $3 \times 1$, and $C$ is $1 \times 3$. So $AB$ ($3\times3$ times $3\times1$) is defined, $AC$ ($3\times3$ times $1\times3$) is defined, but $BA$ ($3\times1$ times $3\times3$) is not defined. The correct option is (B) Only AC.

   ---

   The key idea is simple: you can multiply two matrices only if the inner dimensions match. That is, if the first matrix has size $m \times n$ and the second has size $p \times q$, the product is defined iff $n = p$. The resulting matrix then has size $m \times q$.

   Let’s check each product one by one.

   1. Check $AB$

   $A$ is $3 \times 3$ (3 rows, 3 columns).

   $B$ is $3 \times 1$ (3 rows, 1 column).

   The inner dimensions: $3$ (columns of $A$) and $3$ (rows of $B$) are equal. So $AB$ is defined. The result will be a $3 \times 1$ matrix.

   2. Check $AC$

   $A$ is $3 \times 3$.

   $C$ is $1 \times 3$ (1 row, 3 columns).

   Inner dimensions: $3$ (columns of $A$) and $1$ (rows of $C$) — these are not equal. So $AC$ is not defined. …

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2. CBSE 20241 markMCQ
   Q.If for two non-zero square matrices A and B of the same order, $(\mathrm{A}+\mathrm{B})^{2}=\mathrm{A}^{2}+\mathrm{B}^{2}$, then : (A) $\mathrm{AB}=\mathrm{O}$ (B) $\mathrm{AB}=-\mathrm{BA}$ (C) $\mathrm{BA}=\mathrm{O}$ (D) $\mathrm{AB}=\mathrm{BA}$ Questions number 19 and 20 are Assertion and Reason based questions. Two statements are given, one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer to these questions from the codes (A), (B), (C) and (D) as given below. (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A). (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A). (C) Assertion (A) is true, but Reason (R) is false. (D) Assertion (A) is false, but Reason (R) is true.

   ›Reveal solutionSolution

   The given equation $(A+B)^2 = A^2 + B^2$ forces the cross terms to cancel, which means $AB = -BA$. The correct option is (B).

   Why This Works: The Core Idea

   Matrix multiplication is not commutative — $AB$ is generally not equal to $BA$. When you expand $(A+B)^2$, you get $A^2 + AB + BA + B^2$. The given condition says this equals $A^2 + B^2$, so the middle terms $AB + BA$ must vanish. That gives $AB = -BA$, a condition called anti-commutativity.

   Watch out

   A common mistake is to assume $AB = O$ (zero matrix) from $AB + BA = O$. But that’s only one possibility — the matrices could be non-zero and still satisfy $AB = -BA$. For example, take $A = \begin{pmatrix}0 & 1 \\ 0 & 0\end{pmatrix}$ and $B = \begin{pmatrix}0 & 0 \\ 1 & 0\end{pmatrix}$; then $AB = \begin{pmatrix}1 & 0 \\ 0 & 0\end{pmatrix}$ and $BA = \begin{pmatrix}0 & 0 \\ 0 & 1\end{pmatrix}$, so $AB = -BA$ holds but neither product is zero.

   Step-by-Step Reasoning

   1. Expand the square Since $A$ and $B$ are square matrices of the same order, we can multiply them. The distributive law holds for matrices, so:

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

   2. Apply the given condition The problem states:

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

   Substituting the expansion:

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

   3. Cancel the common terms Subtract $A^2 + B^2$ from both sides:

   $AB + BA = O$

   where $O$ is the zero matrix of the same order.

   4. Interpret the result The equation $AB + BA = O$ is equivalent to:

   $AB = -BA$

   This is the definition of anti-commuting matrices. It does not force $AB$ or $BA$ to be zero individually — only that they are negatives of each other. …

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3. CBSE 2026Set 65/2/11 markMCQ
   Q.If $A$ and $B$ are square matrices of same order, then which of the following statements is/are always true?

   (i) $(A + B)(A - B) = A^2 - B^2$

   (ii) $AB = BA$

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

   (iv) $AB = 0 \Rightarrow A = 0$ or $B = 0$ (A) Only

   (i) and

   (iii) (B) Only

   (ii) and

   (iii) (C) Only

   (iii) (D) Only

   (iii) and (iv)

   ›Reveal solutionSolution

   Matrix multiplication is generally not commutative ($AB \neq BA$), which means many algebraic identities from scalar arithmetic do not hold for matrices. Only statement (iii) is always true, making (C) the correct option.

   Concept and Intuition

   When we work with numbers (scalars), we are used to properties like $ab = ba$ (commutativity) and $ab = 0 \Rightarrow a = 0 \text{ or } b = 0$. However, matrices behave differently. The most crucial distinction is that matrix multiplication is generally not commutative. This means that for two matrices $A$ and $B$, $AB$ is usually not equal to $BA$. This single property is the root cause for why many familiar algebraic identities, which rely on terms like $AB$ and $BA$ cancelling or combining, do not hold true for matrices.

   Let's examine each statement with this fundamental understanding in mind.

   Step-by-step Evaluation

   1. Evaluate statement (i): $(A + B)(A - B) = A^2 - B^2$ To check if this is always true, we expand the left-hand side using the distributive property of matrix multiplication over addition, which does hold for matrices:

   $(A + B)(A - B) = A(A - B) + B(A - B)$

   $= A \cdot A - A \cdot B + B \cdot A - B \cdot B$

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

   For this expression to be equal to $A^2 - B^2$, we would need the terms $-AB + BA$ to be zero. This implies $BA = AB$. However, as discussed, matrix multiplication is generally not commutative, meaning $AB \neq BA$ in most cases.
   
   > [!WARNING]
   > This is a classic pitfall! The identity $(x+y)(x-y) = x^2 - y^2$ is true for scalars because $xy = yx$. For matrices, this is only true if $A$ and $B$ commute.
   
   Therefore, statement (i) is not always true.
   

   2. Evaluate statement (ii): $AB = BA$

   This statement claims that matrix multiplication is always commutative. This is false. Matrix multiplication is generally not commutative. We can easily find counterexamples.

   Consider:

   $A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}, \quad B = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}$

   Then:
   

   $AB = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} 1 \cdot 1 + 1 \cdot 1 & 1 \cdot 0 + 1 \cdot 1 \\ 0 \cdot 1 + 1 \cdot 1 & 0 \cdot 0 + 1 \cdot 1 \end{pmatrix} = \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix}$

   And:
   

   $BA = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 \cdot 1 + 0 \cdot 0 & 1 \cdot 1 + 0 \cdot 1 \\ 1 \cdot 1 + 1 \cdot 0 & 1 \cdot 1 + 1 \cdot 1 \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix}$

   Since $AB \neq BA$, statement (ii) is not always true.
   

   3. Evaluate statement (iii): $(A + B)^2 = A^2 + AB + BA + B^2$

   Let's expand the left-hand side:

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

   Again, using the distributive property:
   

   $= A(A + B) + B(A + B)$

   $= A \cdot A + A \cdot B + B \cdot A + B \cdot B$

   $$= A^2 + AB + BA + B^2$$ …
   

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4. CBSE 2026Set A1 markMCQ
   Q.If $A = [1\ 2\ 3\ 4]$ and $B = \begin{bmatrix} 1 \\ 2 \\ 3 \\ 4 \end{bmatrix}$ then $AB =$

   (a) $[30]$

   (b) $[10]$

   (c) $[20]$

   (d) $[40]$

   ›Reveal solutionSolution

   A $1\times4$ row times a $4\times1$ column is the dot product $=30$.

   …

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5. CBSE 2026Set ANNUAL1 mark
   Q.If $A = \begin{bmatrix}1 & -2 & 3\\ -4 & 2 & 5\end{bmatrix}$ and $B = \begin{bmatrix}2 & 3\\ 4 & 5\\ 2 & 1\end{bmatrix}$, then find AB.

   ›Reveal solutionSolution

   Multiply the $2\times3$ matrix $A$ by the $3\times2$ matrix $B$ row-by-column.

   $AB_{11}=1(2)+(-2)(4)+3(2)=2-8+6=0$

   $AB_{12}=1(3)+(-2)(5)+3(1)=3-10+3=-4$

   $AB_{21}=-4(2)+2(4)+5(2)=-8+8+10=10$

   $AB_{22}=-4(3)+2(5)+5(1)=-12+10+5=3$ …

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6. CBSE 2026Set ANNUAL1 markMCQ
   Q.If $A = \begin{bmatrix} 2 & 4 \\ 1 & 3 \end{bmatrix}$ then $A^2 =$

   (a) $\begin{bmatrix} 4 & 16 \\ 9 & 1 \end{bmatrix}$

   (b) $\begin{bmatrix} 2 & 12 \\ 6 & 2 \end{bmatrix}$

   (c) $\begin{bmatrix} 16 & 13 \\ 9 & 12 \end{bmatrix}$

   (d) None of these

   ›Reveal solutionSolution

   Multiply $A$ by itself using row-by-column matrix multiplication; the result matches none of the printed options.

   Given $A = \begin{bmatrix} 2 & 4 \\ 1 & 3 \end{bmatrix}$.

   $A^2 = A \cdot A = \begin{bmatrix} 2 & 4 \\ 1 & 3 \end{bmatrix}\begin{bmatrix} 2 & 4 \\ 1 & 3 \end{bmatrix}$

   Entry $(1,1)$: $2(2)+4(1) = 4+4=8$

   Entry $(1,2)$: $2(4)+4(3) = 8+12=20$

   Entry $(2,1)$: $1(2)+3(1) = 2+3=5$

   Entry $(2,2)$: $1(4)+3(3) = 4+9=13$

   …

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7. CBSE 2026Set ANNUAL1 mark
   Q.Find $AB$, if $A = \begin{bmatrix} 0 & -1 \\ 0 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} 3 & 5 \\ 0 & 0 \end{bmatrix}$.

   ›Reveal solutionSolution

   Multiply the two $2\times2$ matrices row-by-column; every entry of the product turns out to be $0$.

   Given $A = \begin{bmatrix} 0 & -1 \\ 0 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} 3 & 5 \\ 0 & 0 \end{bmatrix}$.

   $AB = \begin{bmatrix} 0 & -1 \\ 0 & 2 \end{bmatrix}\begin{bmatrix} 3 & 5 \\ 0 & 0 \end{bmatrix}$

   Entry $(1,1)$: $0(3)+(-1)(0) = 0$

   Entry $(1,2)$: $0(5)+(-1)(0) = 0$

   Entry $(2,1)$: $0(3)+2(0) = 0$

   Entry $(2,2)$: $0(5)+2(0) = 0$

   …

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8. CBSE 2026Set ANNUAL1 mark
   Q.If $A=[1\ \ 2\ \ 5\ \ 7]$ and $B=\begin{bmatrix} 6 \\ 2 \\ 4 \\ 8 \end{bmatrix}$, write the orders of $AB$ and $BA$.

   ›Reveal solutionSolution

   $A_{1\times4}B_{4\times1}\to 1\times1$; $B_{4\times1}A_{1\times4}\to 4\times4$.

   $A=[1\ 2\ 5\ 7]$ has order $1\times4$. $B=\begin{bmatrix}6\\2\\4\\8\end{bmatrix}$ has order $4\times1$.

   • $AB$: $(1\times4)(4\times1)$ — inner dimensions ($4$) agree, result order $1\times1$. …

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9. CBSE 2025Set E1 markMCQ
   Q.$\begin{bmatrix} 5 & -1 \\ 6 & 7 \end{bmatrix} \cdot \begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix} =$

   (a) $\begin{bmatrix} 7 & 11 \\ 33 & 34 \end{bmatrix}$

   (b) $\begin{bmatrix} 7 & 1 \\ 33 & 34 \end{bmatrix}$

   (c) $\begin{bmatrix} 7 & 1 \\ 34 & 33 \end{bmatrix}$

   (d) $\begin{bmatrix} 16 & 5 \\ 39 & 25 \end{bmatrix}$

   ›Reveal solutionSolution

   Row-by-column multiplication gives $\begin{bmatrix} 7 & 1 \\ 33 & 34 \end{bmatrix}$.

   Multiply $\begin{bmatrix} 5 & -1 \\ 6 & 7 \end{bmatrix}\begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix}$ entry by entry:

   • $(1,1): 5\cdot2 + (-1)\cdot3 = 10 - 3 = 7$
   • $(1,2): 5\cdot1 + (-1)\cdot4 = 5 - 4 = 1$
   • $(2,1): 6\cdot2 + 7\cdot3 = 12 + 21 = 33$ …

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10. CBSE 2025Set E1 markMCQ
    Q.$\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} =$

    (a) $\begin{bmatrix} 1 & 0 \\ 0 & 4 \end{bmatrix}$

    (b) $\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$

    (c) $\begin{bmatrix} 1 & 2 \\ 0 & 4 \end{bmatrix}$

    (d) $\begin{bmatrix} 1 & 2 \\ 3 & 0 \end{bmatrix}$

    ›Reveal solutionSolution

    $AI = A$, so the product is the original matrix.

    The second matrix $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ is the $2\times2$ identity $I$. For any matrix $A$, $AI = A$. Hence …

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11. CBSE 2025Set E1 markMCQ
    Q.$\begin{bmatrix} 6 & 5 \end{bmatrix}\begin{bmatrix} -1 \\ 1 \end{bmatrix} =$

    (a) $\begin{bmatrix} -6 & 5 \end{bmatrix}$

    (b) $\begin{bmatrix} -6 \\ 5 \end{bmatrix}$

    (c) $[-1]$

    (d) $[1]$

    ›Reveal solutionSolution

    $[6\ 5]\begin{bmatrix} -1 \\ 1 \end{bmatrix} = 6(-1)+5(1) = -1$, a $1\times1$ matrix.

    A $1\times2$ matrix times a $2\times1$ matrix gives a $1\times1$ matrix: …

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12. CBSE 2025Set E1 markMCQ
    Q.$\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}\begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix} =$

    (a) $\begin{bmatrix} 4 & 8 \\ 0 & 16 \end{bmatrix}$

    (b) $\begin{bmatrix} 5 & 2 \\ 3 & 8 \end{bmatrix}$

    (c) $\begin{bmatrix} 4 & 8 \\ 12 & 16 \end{bmatrix}$

    (d) $\begin{bmatrix} 4 & 12 \\ 8 & 16 \end{bmatrix}$

    ›Reveal solutionSolution

    $\begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix} = 4I$, so the product is $4\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$.

    Since $\begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix} = 4I$,

    $\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}(4I) = 4\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 4 & 8 \\ 12 & 16 \end{bmatrix}.$ …

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