25 Solved Examples Across the Chapter

The problems below run in rough order of difficulty and cover every tool of the chapter: orders, construction and equality (Examples 1 to 4), the operations with their traps (5 to 9), transpose and the symmetric world (10 to 13), matrix equations and powers (14 to 17), and the closing classics — inverse-assisted equations, involutory and idempotent identities, counting and word problems (18 to 25).

Work each one before reading its solution — and in every multiplication, write the order bookkeeping first.

Examples 1 to 9 — Orders, Equality and Operations

Example 1 — Orders from an element count

A matrix has 18 elements. What are its possible orders? What if it has 5 elements?

Step 1 — factor pairs of 18: $(1,18), (2,9), (3,6), (6,3), (9,2), (18,1)$ — six orders.

Step 2 — 5 is prime: only $1 \times 5$ and $5 \times 1$.

Answer: six orders for 18 elements; two for 5.

Example 2 — Constructing from a rule

Construct the $3 \times 4$ matrix with $a_{ij} = \frac{1}{2}\vert -3i + j \vert$.

Step 1 — evaluate row by row: row 1 ($i=1$): $\vert -3+j \vert/2$ for $j = 1..4$: $1, \frac{1}{2}, 0, \frac{1}{2}$. Row 2: $\frac{5}{2}, 2, \frac{3}{2}, 1$. Row 3: $4, \frac{7}{2}, 3, \frac{5}{2}$.

Step 2 — assemble: $$A = \begin{pmatrix} 1 & \frac{1}{2} & 0 & \frac{1}{2} \\ \frac{5}{2} & 2 & \frac{3}{2} & 1 \\ 4 & \frac{7}{2} & 3 & \frac{5}{2} \end{pmatrix}$$

Answer: as above — evaluate with (row, column) discipline and the modulus handled per entry.

Example 3 — A linear combination

For $A = \begin{pmatrix} 1 & 2 & -3 \\ 5 & 0 & 2 \end{pmatrix}$ and $B = \begin{pmatrix} 3 & -1 & 2 \\ 4 & 2 & 5 \end{pmatrix}$, compute $A + B$ and $A - 2B$.

Step 1 — sum: $A + B = \begin{pmatrix} 4 & 1 & -1 \\ 9 & 2 & 7 \end{pmatrix}$.

Step 2 — combination: $A - 2B = \begin{pmatrix} 1-6 & 2+2 & -3-4 \\ 5-8 & -4 & 2-10 \end{pmatrix} = \begin{pmatrix} -5 & 4 & -7 \\ -3 & -4 & -8 \end{pmatrix}$.

Answer: entrywise throughout, signs distributed before adding.

Example 4 — Solving for an unknown matrix

Find $X$ if $X + \begin{pmatrix} 4 & 6 \\ -3 & 7 \end{pmatrix} = \begin{pmatrix} 3 & -6 \\ 5 & -8 \end{pmatrix}$.

Step 1 — subtract the known matrix from both sides: $$X = \begin{pmatrix} 3 - 4 & -6 - 6 \\ 5 + 3 & -8 - 7 \end{pmatrix} = \begin{pmatrix} -1 & -12 \\ 8 & -15 \end{pmatrix}$$

Answer: $X = \begin{pmatrix} -1 & -12 \\ 8 & -15 \end{pmatrix}$ — matrix equations at this level are entrywise arithmetic.

Example 5 — A rectangular product, both orders

For $A = \begin{pmatrix} 1 & -2 & 3 \\ -4 & 2 & 5 \end{pmatrix}$ and $B = \begin{pmatrix} 2 & 3 \\ 4 & 5 \\ 2 & 1 \end{pmatrix}$, find $AB$, and state the order of $BA$.

Step 1 — bookkeeping: $(2 \times 3)(3 \times 2) \to 2 \times 2$; $\ (3 \times 2)(2 \times 3) \to 3 \times 3$.

Step 2 — compute $AB$: $$AB = \begin{pmatrix} 2 - 8 + 6 & 3 - 10 + 3 \\ -8 + 8 + 10 & -12 + 10 + 5 \end{pmatrix} = \begin{pmatrix} 0 & -4 \\ 10 & 3 \end{pmatrix}$$

Answer: $AB$ as above; $BA$ exists too but is $3 \times 3$ — both defined, incomparably different.

Example 6 — Another zero-divisor pair

Show that $A = \begin{pmatrix} 1 & -1 \\ -1 & 1 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}$ satisfy $AB = O$.

Step 1 — compute: every entry of $AB$ is $1 - 1 = 0$ or $-1 + 1 = 0$: $$AB = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$$

Answer: two nonzero matrices annihilating each other — each row of $A$ is orthogonal to each column of $B$, which is the geometric anatomy of every zero-divisor pair.

Example 7 — A cost-aggregation word problem

A group plans contact of two types: type X costing ₹ 40 per contact and type Y costing ₹ 100 per contact. In city P they make 1000 X-contacts and 500 Y-contacts; in city Q, 3000 and 1000. Use matrix multiplication to find the total cost in each city.

Step 1 — quantities matrix times cost column: $$\begin{pmatrix} 1000 & 500 \\ 3000 & 1000 \end{pmatrix}\begin{pmatrix} 40 \\ 100 \end{pmatrix} = \begin{pmatrix} 40000 + 50000 \\ 120000 + 100000 \end{pmatrix} = \begin{pmatrix} 90000 \\ 220000 \end{pmatrix}$$

Answer: ₹ 90000 in city P and ₹ 220000 in city Q — one product answers both cities at once; that batching is the entire point of matrix multiplication.

Example 8 — A matrix polynomial

For $A = \begin{pmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{pmatrix}$, compute $A^2 - 5A + 6I$.

Step 1 — square: $$A^2 = \begin{pmatrix} 5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2 \end{pmatrix}$$

Step 2 — assemble the polynomial: $$A^2 - 5A + 6I = \begin{pmatrix} 5 - 10 + 6 & -1 - 0 & 2 - 5 \\ 9 - 10 & -2 - 5 + 6 & 5 - 15 \\ 0 - 5 & -1 + 5 & -2 - 0 + 6 \end{pmatrix} = \begin{pmatrix} 1 & -1 & -3 \\ -1 & -1 & -10 \\ -5 & 4 & 4 \end{pmatrix}$$

Answer: as above — $I$ contributes only to the diagonal, the commonest slip in matrix polynomials.

Example 9 — An entry-equation from a triple product

Find $x$ if $\begin{pmatrix} 1 & x & 1 \end{pmatrix}\begin{pmatrix} 1 & 3 & 2 \\ 2 & 5 & 1 \\ 15 & 3 & 2 \end{pmatrix}\begin{pmatrix} 1 \\ 2 \\ x \end{pmatrix} = O$.

Step 1 — multiply the row into the square matrix: $$\begin{pmatrix} 1 + 2x + 15 & 3 + 5x + 3 & 2 + x + 2 \end{pmatrix} = \begin{pmatrix} 2x + 16 & 5x + 6 & x + 4 \end{pmatrix}$$

Step 2 — multiply into the column and set to zero: $$(2x + 16) + 2(5x + 6) + x(x + 4) = x^2 + 16x + 28 = 0$$

Step 3 — solve: $(x + 2)(x + 14) = 0$.

Answer: $x = -2$ or $x = -14$ — a $1 \times 1$ product equation collapsing to a quadratic.

Examples 10 to 17 — Transpose, Symmetry and Powers

Example 10 — Reversal law on a column-row pair

For $A = \begin{pmatrix} 1 \\ -4 \\ 3 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & 3 & -6 \end{pmatrix}$, verify that $(AB)' = B'A'$.

Step 1 — the product $AB$ is $3 \times 3$: entries $a_i b_j$: $$AB = \begin{pmatrix} 1 & 3 & -6 \\ -4 & -12 & 24 \\ 3 & 9 & -18 \end{pmatrix} \implies (AB)' = \begin{pmatrix} 1 & -4 & 3 \\ 3 & -12 & 9 \\ -6 & 24 & -18 \end{pmatrix}$$

Step 2 — the reversed product: $B'A' = \begin{pmatrix} 1 \\ 3 \\ -6 \end{pmatrix}\begin{pmatrix} 1 & -4 & 3 \end{pmatrix}$ gives exactly the same $3 \times 3$ matrix ✓.

Answer: verified — and note $A'B'$ would be $(1 \times 3)(3 \times 1) = 1 \times 1$, a completely different object: the reversal is forced.

Example 11 — A quick decomposition

Express $M = \begin{pmatrix} 3 & 5 \\ 1 & -1 \end{pmatrix}$ as symmetric plus skew-symmetric.

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

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

Step 3 — check: $P + Q = M$ ✓, $P' = P$ ✓, $Q' = -Q$ ✓.

Answer: $M = P + Q$ as above.

Example 12 — Conjugation preserves symmetry

If $A$ is symmetric, show that $B'AB$ is symmetric (for any conformable $B$).

Step 1 — transpose with the reversal law (twice): $$(B'AB)' = B'A'(B')' = B'A'B$$

Step 2 — use $A' = A$: $\ (B'AB)' = B'AB$.

Answer: symmetric — and the same computation with $A' = -A$ shows $B'AB$ is skew when $A$ is skew: conjugation by any matrix preserves both symmetry types.

Example 13 — Completing a skew-symmetric matrix

Find $a, b, c$ if $A = \begin{pmatrix} 0 & a & 3 \\ 2 & b & -1 \\ c & 1 & 0 \end{pmatrix}$ is skew-symmetric.

Step 1 — diagonal must vanish: $b = 0$.

Step 2 — mirror pairs must negate: $a = -a_{21} = -2$; $\ c = -a_{13} = -3$ (and the pair $(-1, 1)$ checks: $a_{32} = 1 = -a_{23}$ ✓).

Answer: $a = -2, \ b = 0, \ c = -3$.

Example 14 — Fitting a matrix equation

For $A = \begin{pmatrix} 3 & -2 \\ 4 & -2 \end{pmatrix}$, find $k$ such that $A^2 = kA - 2I$.

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

Step 2 — compute $kA - 2I$ and match: $$kA - 2I = \begin{pmatrix} 3k - 2 & -2k \\ 4k & -2k - 2 \end{pmatrix}$$ The $(1,2)$ entries give $-2k = -2$, so $k = 1$; the remaining entries confirm ($3 - 2 = 1$, $4 = 4$, $-4 = -4$) ✓.

Answer: $k = 1$ — solve from one convenient entry, verify with the rest.

Example 15 — The rotation matrix composes angles

For $F(x) = \begin{pmatrix} \cos x & \sin x \\ -\sin x & \cos x \end{pmatrix}$, show that $F(x)F(y) = F(x + y)$.

Step 1 — multiply: $$F(x)F(y) = \begin{pmatrix} \cos x\cos y - \sin x \sin y & \cos x \sin y + \sin x \cos y \\ -\sin x \cos y - \cos x \sin y & -\sin x \sin y + \cos x \cos y \end{pmatrix}$$

Step 2 — recognise the addition formulas: the entries are $\cos(x+y)$, $\sin(x+y)$, $-\sin(x+y)$, $\cos(x+y)$.

Answer: $F(x)F(y) = F(x+y)$ — rotations compose by adding angles, and in particular all $F$'s commute with each other.

Example 16 — Powers by induction (rotation form)

If $A = \begin{pmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{pmatrix}$, prove $A^n = \begin{pmatrix} \cos n\theta & \sin n\theta \\ -\sin n\theta & \cos n\theta \end{pmatrix}$ for all $n \in \mathbb{N}$.

Step 1 — base case $n = 1$: trivially true.

Step 2 — inductive step: assuming the formula for $n$, use Example 15's composition law: $$A^{n+1} = A^n \cdot A = F(n\theta)F(\theta) = F\big((n+1)\theta\big)$$

Answer: by induction the formula holds for every $n$ — the model "prove $A^n = \ldots$" solution: verify the base, multiply once, invoke the pattern.

Example 17 — Powers by induction (shear form)

If $A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$, prove $A^n = \begin{pmatrix} 1 & n \\ 0 & 1 \end{pmatrix}$.

Step 1 — base: $n = 1$ ✓.

Step 2 — step: $$A^{n+1} = A^n A = \begin{pmatrix} 1 & n \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & n + 1 \\ 0 & 1 \end{pmatrix}$$

Answer: proved — shears accumulate linearly ($A^3 = \begin{pmatrix} 1 & 3 \\ 0 & 1 \end{pmatrix}$, a quick sanity check by direct multiplication ✓).

Examples 18 to 25 — Closing Classics

Example 18 — Sandwich products of symmetric matrices

If $A$ and $B$ are symmetric matrices of the same order, show that $ABA$ is symmetric.

Step 1 — transpose with the reversal law: $$(ABA)' = A'B'A' = ABA$$ using $A' = A$ and $B' = B$.

Answer: symmetric — sandwiching $B$ between two copies of a symmetric $A$ preserves symmetry even though the plain product $AB$ generally does not.

Example 19 — Solving $CD - AB = O$

Let $A = \begin{pmatrix} 2 & -1 \\ 3 & 4 \end{pmatrix}$, $B = \begin{pmatrix} 5 & 2 \\ 7 & 4 \end{pmatrix}$, $C = \begin{pmatrix} 2 & 5 \\ 3 & 8 \end{pmatrix}$. Find a matrix $D$ such that $CD - AB = O$.

Step 1 — the target: $CD = AB = \begin{pmatrix} 10 - 7 & 4 - 4 \\ 15 + 28 & 6 + 16 \end{pmatrix} = \begin{pmatrix} 3 & 0 \\ 43 & 22 \end{pmatrix}$.

Step 2 — write $D = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ and unpack $CD = AB$: $$2a + 5c = 3, \quad 2b + 5d = 0, \quad 3a + 8c = 43, \quad 3b + 8d = 22$$

Step 3 — solve the two independent pairs: from the first and third: $a = -191, c = 77$; from the second and fourth: $b = -110, d = 44$.

Answer: $D = \begin{pmatrix} -191 & -110 \\ 77 & 44 \end{pmatrix}$ (verify: $CD = \begin{pmatrix} 3 & 0 \\ 43 & 22 \end{pmatrix}$ ✓).

Example 20 — Finding a left factor

Find the matrix $X$ such that $X\begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{pmatrix} = \begin{pmatrix} -7 & -8 & -9 \\ 2 & 4 & 6 \end{pmatrix}$.

Step 1 — order bookkeeping: $X (2 \times 3) = (2 \times 3)$ forces $X$ to be $2 \times 2$: $X = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$.

Step 2 — unpack columns: first column: $a + 4b = -7$, $c + 4d = 2$; second: $2a + 5b = -8$, $2c + 5d = 4$.

Step 3 — solve: $a = 1, b = -2$; $\ c = 2, d = 0$ — and the third column checks: $3(1) + 6(-2) = -9$ ✓, $3(2) + 0 = 6$ ✓.

Answer: $X = \begin{pmatrix} 1 & -2 \\ 2 & 0 \end{pmatrix}$ — the spare column is the built-in verification.

Example 21 — The involutory template

For $A = \begin{pmatrix} \alpha & \beta \\ \gamma & -\alpha \end{pmatrix}$, compute $A^2$ and state the condition for $A^2 = I$.

Step 1 — square: $$A^2 = \begin{pmatrix} \alpha^2 + \beta\gamma & 0 \\ 0 & \alpha^2 + \beta\gamma \end{pmatrix} = (\alpha^2 + \beta\gamma)\, I$$

Step 2 — impose $A^2 = I$: $\ \alpha^2 + \beta\gamma = 1$, i.e. $1 - \alpha^2 - \beta\gamma = 0$.

Answer: $A^2 = I \iff 1 - \alpha^2 - \beta\gamma = 0$ — the trace-zero $2 \times 2$ shape always squares to a scalar matrix, a fact worth keeping for the JEE Corner.

Example 22 — The idempotent identity

If $A^2 = A$, show that $(I + A)^3 - 7A = I$.

Step 1 — expand $(I + A)^3$ (legal since $I$ commutes with everything): $$(I + A)^3 = I + 3A + 3A^2 + A^3$$

Step 2 — collapse the powers with $A^2 = A$: $A^3 = A^2 \cdot A = A \cdot A = A$, so $$(I + A)^3 = I + 3A + 3A + A = I + 7A$$

Step 3 — subtract: $(I + A)^3 - 7A = I$.

Answer: $I$ — idempotency telescopes every power of $A$ down to $A$ itself.

Example 23 — Classifying combination matrices

For any square matrix $A$, classify (i) $AA'$ and (ii) $A'A$ as symmetric or skew-symmetric.

Step 1 — transpose (i): $(AA')' = (A')'A' = AA'$ — symmetric.

Step 2 — transpose (ii): $(A'A)' = A'(A')' = A'A$ — symmetric.

Answer: both $AA'$ and $A'A$ are always symmetric (though generally different from each other) — two more members of the always-symmetric family alongside $A + A'$.

Example 24 — Counting symmetric 0-1 matrices

How many $2 \times 2$ matrices with every entry 0 or 1 are symmetric?

Step 1 — symmetry ties the mirror pair: $a_{12} = a_{21}$, so the free choices are $a_{11}, a_{22}, a_{12}$ — three independent binary entries.

Step 2 — count: $2^3 = 8$.

Answer: 8 (out of the $2^4 = 16$ total) — symmetry halves the free off-diagonal entries, a Ch1-style count inside Ch3.

Example 25 — The trust-fund word problem

A trust fund has ₹ 30000 to invest in two bonds paying 5% and 7% simple interest per year. Using matrix multiplication, determine how to split the money to obtain an annual interest of ₹ 1800.

Step 1 — set up the product: invest $x$ at 5% and $30000 - x$ at 7%: $$\begin{pmatrix} x & 30000 - x \end{pmatrix}\begin{pmatrix} 0.05 \\ 0.07 \end{pmatrix} = [\,0.05x + 0.07(30000 - x)\,] = [1800]$$

Step 2 — solve: $2100 - 0.02x = 1800$, so $0.02x = 300$ and $x = 15000$.

Answer: ₹ 15000 in each bond (check: $750 + 1050 = 1800$ ✓) — the row-times-column product once again encoding "amounts times rates, summed".