Q.For two matrices $A = \begin{bmatrix} 3 & -6 & -1 \\ 2 & -5 & -1 \\ -2 & 4 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & -2 & -1 \\ 0 & -1 & -1 \\ 2 & 0 & 3 \end{bmatrix}$, find the product $AB$
and hence solve the system of equations:
$3x - 6y - z = 3$
$2x - 5y - z + 2 = 0$
$-2x + 4y + z = 5$

Showing the 12 most recent of 39 on this concept.

1. 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:
   

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

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

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

   This expansion relies only on the distributive property and the definition of matrix multiplication, both of which are always true for matrices. Notice that we cannot combine $AB$ and $BA$ into $2AB$ because $AB \neq BA$ in general.
   
   Therefore, statement (iii) is always true.
   

   4. Evaluate statement (iv): $AB = 0 \Rightarrow A = 0$ or $B = 0$

   This property holds for scalars: if the product of two numbers is zero, at least one of them must be zero. However, this is not always true for matrices. It is possible for the product of two non-zero matrices to be the zero matrix. Such matrices are called zero divisors.

   Consider:

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

   Neither $A$ nor $B$ is the zero matrix.
   Now, let's calculate their product:
   

   $AB = \begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ -1 & 0 \end{pmatrix} = \begin{pmatrix} 1 \cdot 1 + 1 \cdot (-1) & 1 \cdot 0 + 1 \cdot 0 \\ 0 \cdot 1 + 0 \cdot (-1) & 0 \cdot 0 + 0 \cdot 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$

   Here, $AB = 0$, but $A \neq 0$ and $B \neq 0$.
   
   Therefore, statement (iv) is not always true.
   

   Based on our analysis, only statement (iii) is always true.

   ✓Final answer

   Only statement (iii) is always true, so the correct option is (C).

2. 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 $=30$.

   $A = [1\ 2\ 3\ 4]$ is $1\times4$ and $B = \begin{bmatrix} 1 \\ 2 \\ 3 \\ 4 \end{bmatrix}$ is $4\times1$, so $AB$ is $1\times1$:

   $AB = 1(1) + 2(2) + 3(3) + 4(4) = 1 + 4 + 9 + 16 = 30.$

   ✓Final answer

   (a) $[30]$.

3. 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.

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

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

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

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

   $AB=\begin{bmatrix}0&-4\\10&3\end{bmatrix}$

   ✓Final answer

   $AB=\begin{bmatrix}0&-4\\10&3\end{bmatrix}$.

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

   So $A^2 = \begin{bmatrix} 8 & 20 \\ 5 & 13 \end{bmatrix}$. This does not match option (a), (b), or (c).

   ✓Final answer

   (d) None of these — the correct value is $A^2 = \begin{bmatrix} 8 & 20 \\ 5 & 13 \end{bmatrix}$.

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

   $AB = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$

   This is a useful reminder that for matrices, $AB=O$ does not force $A=O$ or $B=O$ (unlike ordinary numbers).

   ✓Final answer

   $AB = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$ (the zero matrix)

6. 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 $1\times1$.

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

   ✓Final answer

   Order of $AB=1\times1$; order of $BA=4\times4$.

7. 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.

   Watch out

   A common mistake is to think that because $C$ has 3 columns, it can multiply with $A$'s 3 rows. But the rule is about columns of the first matching rows of the second, not the other way around. $AC$ would require $C$ to have 3 rows, but it has only 1.

   3. Check $BA$

   $B$ is $3 \times 1$.

   $A$ is $3 \times 3$.

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

   Tip

   Notice that $BA$ would be $3 \times 1$ times $3 \times 3$ — the inner numbers (1 and 3) don’t match. But if you reversed the order to $AB$, they do match. Matrix multiplication is not commutative; order matters completely.

   So among the three, only $AB$ is defined. That matches option (A).

   ✓Final answer

   The correct option is (A) Only AB.

8. 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:

   • $(1,1): 5\cdot2 + (-1)\cdot3 = 10 - 3 = 7$
   • $(1,2): 5\cdot1 + (-1)\cdot4 = 5 - 4 = 1$
   • $(2,1): 6\cdot2 + 7\cdot3 = 12 + 21 = 33$
   • $(2,2): 6\cdot1 + 7\cdot4 = 6 + 28 = 34$

   $= \begin{bmatrix} 7 & 1 \\ 33 & 34 \end{bmatrix}.$

   ✓Final answer

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

9. 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

   $\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}.$

   ✓Final answer

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

10. 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:

    $\begin{bmatrix} 6 & 5 \end{bmatrix}\begin{bmatrix} -1 \\ 1 \end{bmatrix} = 6\cdot(-1) + 5\cdot1 = -6 + 5 = -1 = [-1].$

    ✓Final answer

    (C) $[-1]$.

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

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

    (Direct check: row 1 $= [1\cdot4+2\cdot0,\ 1\cdot0+2\cdot4] = [4,8]$; row 2 $= [12,16]$.)

    ✓Final answer

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

12. CBSE 2025Set E1 markMCQ
    Q.$\begin{bmatrix} 2 & 3 \\ 5 & 7 \end{bmatrix}\begin{bmatrix} 2 \\ 5 \end{bmatrix} =$

    (a) $\begin{bmatrix} 4 & 6 \\ 25 & 35 \end{bmatrix}$

    (b) $\begin{bmatrix} 4 & 15 \\ 10 & 35 \end{bmatrix}$

    (c) $\begin{bmatrix} 19 & 45 \end{bmatrix}$

    (d) $\begin{bmatrix} 19 \\ 45 \end{bmatrix}$

    ›Reveal solutionSolution

    $\begin{bmatrix} 2 & 3 \\ 5 & 7 \end{bmatrix}\begin{bmatrix} 2 \\ 5 \end{bmatrix} = \begin{bmatrix} 19 \\ 45 \end{bmatrix}$.

    Multiply each row by the column:

    • Row 1: $2\cdot2 + 3\cdot5 = 4 + 15 = 19$
    • Row 2: $5\cdot2 + 7\cdot5 = 10 + 35 = 45$

    A $2\times2$ times a $2\times1$ yields a $2\times1$ column:

    $\begin{bmatrix} 19 \\ 45 \end{bmatrix}.$

    ✓Final answer

    (D) $\begin{bmatrix} 19 \\ 45 \end{bmatrix}$.