How Matrices Appears in the Board Exam

A dependable scorer in the CBSE paper:

Marks are awarded for the frame: order bookkeeping before a product, entrywise equations written out from equality, and verification lines ($P' = P$, $Q' = -Q$, $P + Q = A$) in decompositions.

Below are 12 board-style written questions with complete solutions, organised by marks, followed by a 15-question MCQ quiz.

2-Mark Questions

Question 1 — Construction from a rule

Write the $2 \times 2$ matrix $A = [a_{ij}]$ whose elements are given by $a_{ij} = i + 2j$.

Step 1 — evaluate: $a_{11} = 3, \ a_{12} = 5, \ a_{21} = 4, \ a_{22} = 6$.

Answer: $A = \begin{pmatrix} 3 & 5 \\ 4 & 6 \end{pmatrix}$.

Question 2 — Unknowns from a matrix equation

Find $x$ and $y$ if $2\begin{pmatrix} 1 & 3 \\ 0 & x \end{pmatrix} + \begin{pmatrix} y & 0 \\ 1 & 2 \end{pmatrix} = \begin{pmatrix} 5 & 6 \\ 1 & 8 \end{pmatrix}$.

Step 1 — combine the left side: $\begin{pmatrix} 2 + y & 6 \\ 1 & 2x + 2 \end{pmatrix}$.

Step 2 — equate entries: $2 + y = 5 \Rightarrow y = 3$; $\ 2x + 2 = 8 \Rightarrow x = 3$ (spare entries $6 = 6$, $1 = 1$ ✓).

Answer: $x = 3, \ y = 3$.

Question 3 — Order bookkeeping

$A$ has order $3 \times 4$ and $B$ has order $4 \times 3$. State the orders of $AB$ and $BA$.

Step 1 — $AB$: $(3 \times 4)(4 \times 3) \to 3 \times 3$.

Step 2 — $BA$: $(4 \times 3)(3 \times 4) \to 4 \times 4$.

Answer: $AB$ is $3 \times 3$ and $BA$ is $4 \times 4$ — both defined, never comparable.

Question 4 — Recognising symmetry types

Classify $P = \begin{pmatrix} 1 & 5 \\ 5 & 2 \end{pmatrix}$ and $Q = \begin{pmatrix} 0 & 3 \\ -3 & 0 \end{pmatrix}$ as symmetric or skew-symmetric, with reason.

Step 1 — test $P$: $P' = P$ (mirror entries $5, 5$ agree) — symmetric.

Step 2 — test $Q$: $Q' = \begin{pmatrix} 0 & -3 \\ 3 & 0 \end{pmatrix} = -Q$, with zero diagonal — skew-symmetric.

Answer: $P$ symmetric, $Q$ skew-symmetric.

3-Mark Questions

Question 5 — Squaring a matrix

For $A = \begin{pmatrix} 1 & 2 \\ 4 & -3 \end{pmatrix}$, compute $A^2$.

Step 1 — row-by-column: $$A^2 = \begin{pmatrix} 1 + 8 & 2 - 6 \\ 4 - 12 & 8 + 9 \end{pmatrix} = \begin{pmatrix} 9 & -4 \\ -8 & 17 \end{pmatrix}$$

Answer: $\begin{pmatrix} 9 & -4 \\ -8 & 17 \end{pmatrix}$ — note $A^2$ is NOT the matrix of squared entries.

Question 6 — Verifying distributivity

For $A = \begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix}$, $B = \begin{pmatrix} 1 & 1 \\ 0 & 2 \end{pmatrix}$, $C = \begin{pmatrix} 2 & 0 \\ 1 & 1 \end{pmatrix}$, verify $A(B + C) = AB + AC$.

Step 1 — left side: $B + C = \begin{pmatrix} 3 & 1 \\ 1 & 3 \end{pmatrix}$, so $A(B + C) = \begin{pmatrix} 3 & 1 \\ 7 & 5 \end{pmatrix}$.

Step 2 — right side: $AB = \begin{pmatrix} 1 & 1 \\ 2 & 4 \end{pmatrix}$, $\ AC = \begin{pmatrix} 2 & 0 \\ 5 & 1 \end{pmatrix}$; sum $= \begin{pmatrix} 3 & 1 \\ 7 & 5 \end{pmatrix}$.

Answer: both sides agree ✓ — distributivity verified numerically, as boards ask.

Question 7 — Simultaneous matrix equations

Solve for $X$ and $Y$: $\ 2X + 3Y = \begin{pmatrix} 2 & 3 \\ 4 & 0 \end{pmatrix}$ and $3X + 2Y = \begin{pmatrix} -2 & 2 \\ 1 & -5 \end{pmatrix}$.

Step 1 — eliminate $X$: $3 \times$(first)$- 2 \times$(second): $$5Y = \begin{pmatrix} 6 + 4 & 9 - 4 \\ 12 - 2 & 0 + 10 \end{pmatrix} = \begin{pmatrix} 10 & 5 \\ 10 & 10 \end{pmatrix} \implies Y = \begin{pmatrix} 2 & 1 \\ 2 & 2 \end{pmatrix}$$

Step 2 — back-substitute: $2X = \begin{pmatrix} 2 & 3 \\ 4 & 0 \end{pmatrix} - 3Y = \begin{pmatrix} -4 & 0 \\ -2 & -6 \end{pmatrix}$, so $X = \begin{pmatrix} -2 & 0 \\ -1 & -3 \end{pmatrix}$.

Answer: $X = \begin{pmatrix} -2 & 0 \\ -1 & -3 \end{pmatrix}$, $\ Y = \begin{pmatrix} 2 & 1 \\ 2 & 2 \end{pmatrix}$ (both original equations check ✓) — ordinary elimination, with matrices as the unknowns.

Question 8 — Transpose properties in numbers

For $A = \begin{pmatrix} 2 & -1 \\ 3 & 0 \end{pmatrix}$, verify that $(A')' = A$ and $(3A)' = 3A'$.

Step 1 — double transpose: $A' = \begin{pmatrix} 2 & 3 \\ -1 & 0 \end{pmatrix}$, and transposing again returns $\begin{pmatrix} 2 & -1 \\ 3 & 0 \end{pmatrix} = A$ ✓.

Step 2 — scalar law: $3A = \begin{pmatrix} 6 & -3 \\ 9 & 0 \end{pmatrix}$, so $(3A)' = \begin{pmatrix} 6 & 9 \\ -3 & 0 \end{pmatrix} = 3A'$ ✓.

Answer: both laws verified.

5-Mark Questions

Question 9 — The decomposition, full marks version

Express $A = \begin{pmatrix} 1 & 3 & 5 \\ -6 & 8 & 3 \\ -4 & 6 & 5 \end{pmatrix}$ as the sum of a symmetric and a skew-symmetric matrix.

Step 1 — symmetric part: $$P = \frac{1}{2}(A + A') = \begin{pmatrix} 1 & -\frac{3}{2} & \frac{1}{2} \\ -\frac{3}{2} & 8 & \frac{9}{2} \\ \frac{1}{2} & \frac{9}{2} & 5 \end{pmatrix}$$

Step 2 — skew part: $$Q = \frac{1}{2}(A - A') = \begin{pmatrix} 0 & \frac{9}{2} & \frac{9}{2} \\ -\frac{9}{2} & 0 & -\frac{3}{2} \\ -\frac{9}{2} & \frac{3}{2} & 0 \end{pmatrix}$$

Step 3 — the three verification lines: $P' = P$ ✓, $\ Q' = -Q$ (zero diagonal ✓), $\ P + Q = A$ ✓.

Answer: $A = P + Q$ as above — the verifications are where the last mark lives.

Question 10 — A cubic matrix identity

For $A = \begin{pmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{pmatrix}$, show that $A^3 - 6A^2 + 7A + 2I = O$.

Step 1 — compute $A^2$: $$A^2 = \begin{pmatrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{pmatrix}$$

Step 2 — compute $A^3 = A^2 \cdot A$: $$A^3 = \begin{pmatrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{pmatrix}$$

Step 3 — assemble the combination: every entry of $A^3 - 6A^2 + 7A + 2I$ cancels; for instance the $(1,1)$ entry is $21 - 30 + 7 + 2 = 0$ and the $(3,3)$ entry is $55 - 78 + 21 + 2 = 0$.

Answer: $A^3 - 6A^2 + 7A + 2I = O$ — a matrix satisfying its own cubic; keep the arithmetic in a neat grid, entry by entry.

Question 11 — A quadratic matrix identity

For $A = \begin{pmatrix} 3 & 1 \\ -1 & 2 \end{pmatrix}$, show that $A^2 - 5A + 7I = O$.

Step 1 — square: $$A^2 = \begin{pmatrix} 9 - 1 & 3 + 2 \\ -3 - 2 & -1 + 4 \end{pmatrix} = \begin{pmatrix} 8 & 5 \\ -5 & 3 \end{pmatrix}$$

Step 2 — combine: $$A^2 - 5A + 7I = \begin{pmatrix} 8 - 15 + 7 & 5 - 5 \\ -5 + 5 & 3 - 10 + 7 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$$

Answer: $O$ — the chapter's most famous identity (it also yields $A^{-1} = \frac{1}{7}(5I - A)$, a preview of the JEE Corner's matrix-equation technique).

Question 12 — An applied product

A campaign uses three contact modes costing ₹ 40, ₹ 100 and ₹ 50 per contact. City X makes 400, 300 and 100 contacts of the three modes; city Y makes 300, 100 and 50. Using matrix multiplication, find the total campaign cost in each city.

Step 1 — quantities matrix times cost column: $$\begin{pmatrix} 400 & 300 & 100 \\ 300 & 100 & 50 \end{pmatrix}\begin{pmatrix} 40 \\ 100 \\ 50 \end{pmatrix} = \begin{pmatrix} 16000 + 30000 + 5000 \\ 12000 + 10000 + 2500 \end{pmatrix} = \begin{pmatrix} 51000 \\ 24500 \end{pmatrix}$$

Answer: ₹ 51000 for city X and ₹ 24500 for city Y — state the row/column conventions, show the product, interpret the result in words.

Presentation tip: in every 5-mark answer above, one explicit verification or interpretation line closes the solution — that closing line is a mark by itself.