CBSE Sample Paper 2026-27 · Section C · 3 marks

Q26: Find $PQ$ and $QP$ and the inverse of the product, if it exists

Sample 2026-27Ch 3 · Matrices Ch 4 · Determinants

The question

For two matrices,

P = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 1 & 2 \end{bmatrix}

and

Q = \begin{bmatrix} 1 & 0 \\ 0 & 2 \\ 1 & 0 \end{bmatrix},

find $PQ$ and $QP$ ⁠, whichever is defined. Also, find the inverse of the product matrix, if it exists.

Exam answer

P is of order $2 \times 3$ and Q is of order $3 \times 2$ ⁠, so both PQ and QP are defined. Let $A = PQ$ and $B = QP$ ⁠.

Why?

A product $XY$ is defined only when the number of columns of $X$ equals the number of rows of $Y$ ⁠. The answer then has the rows of $X$ and the columns of $Y$ ⁠:

\begin{aligned} (2 \times 3)(3 \times 2) &\to 2 \times 2 \\ (3 \times 2)(2 \times 3) &\to 3 \times 3 \end{aligned}

So PQ is $2 \times 2$ and QP is $3 \times 3$ ⁠. Both are defined, but they are different matrices.

\begin{aligned} A &= \begin{bmatrix} 2 & 0 & 0 \\ 0 & 1 & 2 \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 0 & 2 \\ 1 & 0 \end{bmatrix} \\ &= \begin{bmatrix} 2 + 0 + 0 & 0 + 0 + 0 \\ 0 + 0 + 2 & 0 + 2 + 0 \end{bmatrix} \\ &= \begin{bmatrix} 2 & 0 \\ 2 & 2 \end{bmatrix} \end{aligned}

Why?

Each entry is a row of P times a column of Q: multiply matching entries and add. For example, row 2 of P with column 1 of Q gives

0 \times 1 + 1 \times 0 + 2 \times 1 = 2

All four entries, row of P times column of Q:

\begin{aligned} &\text{R1, C1: } (2)(1) + (0)(0) + (0)(1) \\ &= 2 + 0 + 0 = 2 \\[4pt] &\text{R1, C2: } (2)(0) + (0)(2) + (0)(0) \\ &= 0 + 0 + 0 = 0 \\[4pt] &\text{R2, C1: } (0)(1) + (1)(0) + (2)(1) \\ &= 0 + 0 + 2 = 2 \\[4pt] &\text{R2, C2: } (0)(0) + (1)(2) + (2)(0) \\ &= 0 + 2 + 0 = 2 \end{aligned}

\begin{aligned} B &= \begin{bmatrix} 1 & 0 \\ 0 & 2 \\ 1 & 0 \end{bmatrix}\begin{bmatrix} 2 & 0 & 0 \\ 0 & 1 & 2 \end{bmatrix} \\ &= \begin{bmatrix} 2 + 0 & 0 + 0 & 0 + 0 \\ 0 + 0 & 0 + 2 & 0 + 4 \\ 2 + 0 & 0 + 0 & 0 + 0 \end{bmatrix} \\ &= \begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 4 \\ 2 & 0 & 0 \end{bmatrix} \end{aligned}

Why?

Now each row of Q has 2 entries and each column of P has 2 entries, so every entry is a sum of two products. For example, row 2 of Q with column 3 of P gives

0 \times 0 + 2 \times 2 = 4

All nine entries, row of Q times column of P. The columns of P are $(2, 0)$ ⁠, $(0, 1)$ and $(0, 2)$ ⁠:

\begin{aligned} &\text{Row 1 } (1, 0)\text{:} \\ &(1)(2) + (0)(0) = 2 \\ &(1)(0) + (0)(1) = 0 \\ &(1)(0) + (0)(2) = 0 \\[4pt] &\text{Row 2 } (0, 2)\text{:} \\ &(0)(2) + (2)(0) = 0 \\ &(0)(0) + (2)(1) = 2 \\ &(0)(0) + (2)(2) = 4 \\[4pt] &\text{Row 3 } (1, 0)\text{:} \\ &(1)(2) + (0)(0) = 2 \\ &(1)(0) + (0)(1) = 0 \\ &(1)(0) + (0)(2) = 0 \end{aligned}

$|A| = 4 - 0 = 4 \ne 0$ ⁠, so A is non-singular and $A^{-1}$ exists.

Why?

For a $2 \times 2$ matrix, the determinant is (product of the leading diagonal) minus (product of the other diagonal): $2 \times 2 - 0 \times 2 = 4$ ⁠.

\begin{aligned} |A| &= \begin{vmatrix} 2 & 0 \\ 2 & 2 \end{vmatrix} \\ &= (2)(2) - (0)(2) \\ &= 4 - 0 = 4 \end{aligned}

A square matrix has an inverse exactly when its determinant is not zero. Always state this before finding $A^{-1}$ ⁠.

\begin{aligned} A^{-1} &= \frac{1}{|A|}\operatorname{adj}A \\ &= \frac14\begin{bmatrix} 2 & 0 \\ -2 & 2 \end{bmatrix} = \begin{bmatrix} \frac12 & 0 \\ -\frac12 & \frac12 \end{bmatrix} \end{aligned}

Why?

For a $2 \times 2$ matrix, the adjoint is found by swapping the two diagonal entries and changing the signs of the other two:

\begin{bmatrix} a & b \\ c & d \end{bmatrix} \to \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}

Here $a = 2$ ⁠, $b = 0$ ⁠, $c = 2$ ⁠, $d = 2$ ⁠:

\operatorname{adj}A = \begin{bmatrix} 2 & -0 \\ -2 & 2 \end{bmatrix} = \begin{bmatrix} 2 & 0 \\ -2 & 2 \end{bmatrix}

Then divide every entry by $|A| = 4$ ⁠:

\begin{aligned} \frac{2}{4} &= \frac{1}{2},\quad \frac{0}{4} = 0 \\ \frac{-2}{4} &= -\frac{1}{2},\quad \frac{2}{4} = \frac{1}{2} \end{aligned}

Check: $A \times A^{-1}$ gives the identity matrix $I$ ⁠.

\begin{aligned} &\text{R1, C1: } (2)\left(\tfrac12\right) + (0)\left(-\tfrac12\right) = 1 \\ &\text{R1, C2: } (2)(0) + (0)\left(\tfrac12\right) = 0 \\ &\text{R2, C1: } (2)\left(\tfrac12\right) + (2)\left(-\tfrac12\right) \\ &\quad = 1 - 1 = 0 \\ &\text{R2, C2: } (2)(0) + (2)\left(\tfrac12\right) = 1 \end{aligned}

\begin{aligned} |B| &= 2(2 \times 0 - 4 \times 0) \\ &\quad - 0 + 0 = 0 \end{aligned}

Why?

Expand along the first row. Only the entry 2 is non-zero, and its minor $2 \times 0 - 4 \times 0$ is 0.

\begin{aligned} |B| &= 2\begin{vmatrix} 2 & 4 \\ 0 & 0 \end{vmatrix} - 0(\ldots) \\ &\quad + 0(\ldots) \\[4pt] \begin{vmatrix} 2 & 4 \\ 0 & 0 \end{vmatrix} &= (2)(0) - (4)(0) \\ &= 0 - 0 = 0 \\[4pt] |B| &= 2(0) - 0 + 0 = 0 \end{aligned}

A quicker reason: a determinant with two equal rows is always 0, and rows 1 and 3 of B are both $2, 0, 0$ ⁠.

$|B| = 0$ ⁠, since its first and third rows are equal. So B is singular and $B^{-1}$ does not exist.

$\therefore\ (PQ)^{-1}$ is the matrix $A^{-1}$ found above, and $(QP)^{-1}$ does not exist.

Where this comes from

Chapter 3, Matrices: Matrix products; inverse of a product matrix.

Question type: Multiply two matrices + Inverse of a matrix (combines 2 types)

How to spot it: click to see

Ch 3, type 5. Multiply two matrices: row times column

Spot it: two matrices written side by side, or $AB$ ⁠, $BA$ or $A^2$ ⁠.

Here: First, write the orders: $P$ is $2 \times 3$ and $Q$ is $3 \times 2$ ⁠, so both $PQ$ ( $2 \times 2$ ⁠) and $QP$ ( $3 \times 3$ ⁠) exist. Work each out row by column.

Ch 4, type 6. Inverse of a matrix: $|A|$ ⁠, then the adjoint, then divide

Spot it: "find the inverse" or "find $A^{-1}$ ".

Here: Then the inverse of each product: $|PQ| = 4 \ne 0$ ⁠, so swap the diagonal, change the other two signs and divide by 4. $|QP| = 0$ (rows 1 and 3 are equal), so $QP$ has no inverse.

Quick lesson · Worked NCERT solutions · Interactive lesson

Also uses Chapter 4, Determinants.

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Q26: Find PQPQPQ and QPQPQP and the inverse of the product, if it exists

Q26: Find $PQ$ and $QP$ and the inverse of the product, if it exists