Block A — Evaluating Determinants and Areas

Example 1 — A full $3 \times 3$ evaluation

Evaluate $\begin{vmatrix} 3 & -4 & 5 \\ 1 & 1 & -2 \\ 2 & 3 & 1 \end{vmatrix}$.

Step 1 — expand along $R_1$: $$\Delta = 3\begin{vmatrix} 1 & -2 \\ 3 & 1 \end{vmatrix} + 4\begin{vmatrix} 1 & -2 \\ 2 & 1 \end{vmatrix} + 5\begin{vmatrix} 1 & 1 \\ 2 & 3 \end{vmatrix}$$

Step 2 — evaluate the minors: $(1 + 6) = 7$, $(1 + 4) = 5$, $(3 - 2) = 1$.

Step 3 — combine: $3(7) + 4(5) + 5(1) = 21 + 20 + 5 = 46$.

Answer: $46$.

Example 2 — A determinant that must vanish

Evaluate $\begin{vmatrix} 0 & 1 & 2 \\ -1 & 0 & -3 \\ -2 & 3 & 0 \end{vmatrix}$.

Step 1 — spot the structure: the array is skew-symmetric — zeros on the diagonal, and each entry below is the negative of its mirror.

Step 2 — expand to confirm: along $R_1$: $0 - 1\left[0 - 6\right] + 2\left[-3 - 0\right] = 6 - 6 = 0$.

Answer: $0$ — and Section 8 shows every odd-order skew-symmetric determinant vanishes, making this a zero-work observation on the exam.

Example 3 — Zeros shortcut in action

Evaluate $\begin{vmatrix} 2 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0 \end{vmatrix}$.

Step 1 — pick $C_1$ (one zero): $$\Delta = 2\begin{vmatrix} 2 & -1 \\ -5 & 0 \end{vmatrix} - 0 + 3\begin{vmatrix} -1 & -2 \\ 2 & -1 \end{vmatrix}$$

Step 2 — evaluate: $2(0 - 5) + 3(1 + 4) = -10 + 15 = 5$.

Answer: $5$.

Example 4 — A singular determinant

Find $\vert A \vert$ for $A = \begin{pmatrix} 1 & 1 & -2 \\ 2 & 1 & -3 \\ 5 & 4 & -9 \end{pmatrix}$.

Step 1 — expand along $R_1$: $1(-9 + 12) - 1(-18 + 15) - 2(8 - 5) = 3 + 3 - 6$.

Answer: $\vert A \vert = 0$ — the matrix is singular. The hidden reason: $R_3 = R_1 + 2R_2$, and a row that is a combination of the others always kills the determinant.

Example 5 — Solving for an unknown entry

Find $x$ if $\begin{vmatrix} 2 & 3 \\ 4 & 5 \end{vmatrix} = \begin{vmatrix} x & 3 \\ 2x & 5 \end{vmatrix}$.

Step 1 — expand both sides: left: $10 - 12 = -2$; right: $5x - 6x = -x$.

Step 2 — equate: $-x = -2$.

Answer: $x = 2$ — a linear equation this time, so a single root and no $\pm$ to worry about.

Example 6 — Given area, two answers

Find $k$ if the area of the triangle with vertices $(-2, 0)$, $(0, 4)$, $(0, k)$ is $4$ square units.

Step 1 — area with both signs: $$\frac{1}{2}\begin{vmatrix} -2 & 0 & 1 \\ 0 & 4 & 1 \\ 0 & k & 1 \end{vmatrix} = \pm 4$$

Step 2 — expand along $C_1$: only the $-2$ survives: $\frac{1}{2}\left[-2(4 - k)\right] = k - 4$, so $k - 4 = \pm 4$.

Answer: $k = 8$ or $k = 0$.

Example 7 — A determinant independent of $\theta$

Prove that $\Delta = \begin{vmatrix} x & \sin\theta & \cos\theta \\ -\sin\theta & -x & 1 \\ \cos\theta & 1 & x \end{vmatrix}$ is independent of $\theta$.

Step 1 — expand along $R_1$: $$\Delta = x(-x^2 - 1) - \sin\theta(-x\sin\theta - \cos\theta) + \cos\theta(-\sin\theta + x\cos\theta)$$

Step 2 — distribute: $-x^3 - x + x\sin^2\theta + \sin\theta\cos\theta - \sin\theta\cos\theta + x\cos^2\theta$.

Step 3 — collapse with $\sin^2\theta + \cos^2\theta = 1$: the trig terms give $+x$, cancelling the $-x$: $$\Delta = -x^3$$

Answer: $\Delta = -x^3$, free of $\theta$ — proved.

Example 8 — A two-angle determinant

Evaluate $\begin{vmatrix} \cos\alpha\cos\beta & \cos\alpha\sin\beta & -\sin\alpha \\ -\sin\beta & \cos\beta & 0 \\ \sin\alpha\cos\beta & \sin\alpha\sin\beta & \cos\alpha \end{vmatrix}$.

Step 1 — expand along $C_3$ (it has the lone zero): $$\Delta = -\sin\alpha\begin{vmatrix} -\sin\beta & \cos\beta \\ \sin\alpha\cos\beta & \sin\alpha\sin\beta \end{vmatrix} + \cos\alpha\begin{vmatrix} \cos\alpha\cos\beta & \cos\alpha\sin\beta \\ -\sin\beta & \cos\beta \end{vmatrix}$$

Step 2 — first minor: $-\sin\alpha\sin^2\beta - \sin\alpha\cos^2\beta = -\sin\alpha$; so the first term is $-\sin\alpha(-\sin\alpha) = \sin^2\alpha$.

Step 3 — second minor: $\cos\alpha\cos^2\beta + \cos\alpha\sin^2\beta = \cos\alpha$; the second term is $\cos^2\alpha$.

Answer: $\Delta = \sin^2\alpha + \cos^2\alpha = 1$.

Block B — Adjoint and Inverse

Example 9 — Adjoint of a $3 \times 3$

Find $\operatorname{adj} A$ for $A = \begin{pmatrix} 1 & -1 & 2 \\ 2 & 3 & 5 \\ -2 & 0 & 1 \end{pmatrix}$.

Step 1 — all nine cofactors: row by row: $$A_{11} = 3, \; A_{12} = -12, \; A_{13} = 6; \quad A_{21} = 1, \; A_{22} = 5, \; A_{23} = 2; \quad A_{31} = -11, \; A_{32} = -1, \; A_{33} = 5$$

Step 2 — transpose the cofactor matrix: $$\operatorname{adj} A = \begin{pmatrix} 3 & 1 & -11 \\ -12 & 5 & -1 \\ 6 & 2 & 5 \end{pmatrix}$$

Answer: as above — rows of cofactors become columns of the adjoint.

Example 10 — The master identity survives singularity

Verify $A(\operatorname{adj} A) = (\operatorname{adj} A)A = \vert A \vert I$ for $A = \begin{pmatrix} 2 & 3 \\ -4 & -6 \end{pmatrix}$.

Step 1 — determinant: $\vert A \vert = -12 + 12 = 0$ — singular, so the right side is the zero matrix.

Step 2 — adjoint: $\operatorname{adj} A = \begin{pmatrix} -6 & -3 \\ 4 & 2 \end{pmatrix}$.

Step 3 — both products: $$A(\operatorname{adj} A) = \begin{pmatrix} -12 + 12 & -6 + 6 \\ 24 - 24 & 12 - 12 \end{pmatrix} = O, \qquad (\operatorname{adj} A)A = O$$

Answer: both equal $O = 0 \cdot I = \vert A \vert I$ — the identity holds for singular matrices too; only the inverse formula needs $\vert A \vert \neq 0$.

Example 11 — Inverse of a triangular matrix

Find $A^{-1}$ for $A = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5 \end{pmatrix}$.

Step 1 — determinant: triangular, so $\vert A \vert = 1 \cdot 2 \cdot 5 = 10 \neq 0$.

Step 2 — cofactors and adjoint: $$\operatorname{adj} A = \begin{pmatrix} 10 & -10 & 2 \\ 0 & 5 & -4 \\ 0 & 0 & 2 \end{pmatrix}$$

Step 3 — divide by $10$: $$A^{-1} = \frac{1}{10}\begin{pmatrix} 10 & -10 & 2 \\ 0 & 5 & -4 \\ 0 & 0 & 2 \end{pmatrix}$$

Answer: as above — note the inverse of an upper-triangular matrix stays upper-triangular.

Example 12 — A full $3 \times 3$ inverse

Find $A^{-1}$ for $A = \begin{pmatrix} 2 & 1 & 3 \\ 4 & -1 & 0 \\ -7 & 2 & 1 \end{pmatrix}$.

Step 1 — determinant: expanding along $R_1$: $2(-1 - 0) - 1(4 - 0) + 3(8 - 7) = -2 - 4 + 3 = -3$.

Step 2 — adjoint: computing all cofactors and transposing: $$\operatorname{adj} A = \begin{pmatrix} -1 & 5 & 3 \\ -4 & 23 & 12 \\ 1 & -11 & -6 \end{pmatrix}$$

Step 3 — divide by $-3$: $$A^{-1} = -\frac{1}{3}\begin{pmatrix} -1 & 5 & 3 \\ -4 & 23 & 12 \\ 1 & -11 & -6 \end{pmatrix}$$

Answer: as above. Spot-check the $(1,1)$ entry of $AA^{-1}$: $-\frac{1}{3}\left[2(-1) + 1(-4) + 3(1)\right] = -\frac{1}{3}(-3) = 1$. ✓

Example 13 — An inverse that equals the matrix itself

Find $A^{-1}$ for $A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & \cos\alpha & \sin\alpha \\ 0 & \sin\alpha & -\cos\alpha \end{pmatrix}$.

Step 1 — try $A^2$ before any adjoint: the $(2,2)$ entry of $A^2$ is $\cos^2\alpha + \sin^2\alpha = 1$, the $(2,3)$ entry is $\cos\alpha\sin\alpha - \sin\alpha\cos\alpha = 0$, and similarly for row 3: $A^2 = I$.

Step 2 — conclude: $A \cdot A = I$ means $A^{-1} = A$ — the matrix is involutory.

Answer: $A^{-1} = A$ itself. (The adjoint route gives the same thing with $\vert A \vert = -\cos^2\alpha - \sin^2\alpha = -1$, but recognising $A^2 = I$ is faster and cleaner.)

Example 14 — Verifying the reversal law with numbers

For $A = \begin{pmatrix} 3 & 7 \\ 2 & 5 \end{pmatrix}$ and $B = \begin{pmatrix} 6 & 8 \\ 7 & 9 \end{pmatrix}$, verify that $(AB)^{-1} = B^{-1}A^{-1}$.

Step 1 — the product and its inverse: $AB = \begin{pmatrix} 67 & 87 \\ 47 & 61 \end{pmatrix}$ with $\vert AB \vert = \vert A \vert \vert B \vert = (1)(-2) = -2$, so $$(AB)^{-1} = -\frac{1}{2}\begin{pmatrix} 61 & -87 \\ -47 & 67 \end{pmatrix}$$

Step 2 — the two inverses separately: $A^{-1} = \begin{pmatrix} 5 & -7 \\ -2 & 3 \end{pmatrix}$ (since $\vert A \vert = 1$) and $B^{-1} = -\frac{1}{2}\begin{pmatrix} 9 & -8 \\ -7 & 6 \end{pmatrix}$.

Step 3 — multiply in reversed order: $$B^{-1}A^{-1} = -\frac{1}{2}\begin{pmatrix} 9 & -8 \\ -7 & 6 \end{pmatrix}\begin{pmatrix} 5 & -7 \\ -2 & 3 \end{pmatrix} = -\frac{1}{2}\begin{pmatrix} 61 & -87 \\ -47 & 67 \end{pmatrix}$$

Answer: the two results match — $(AB)^{-1} = B^{-1}A^{-1}$, order reversed. ✓

Example 15 — Inverse from a cubic identity

For $A = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{pmatrix}$, show that $A^3 - 6A^2 + 5A + 11I = O$ and hence find $A^{-1}$.

Step 1 — compute the powers: $A^2 = \begin{pmatrix} 4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14 \end{pmatrix}$, then $A^3 = A^2 \cdot A$; substituting into $A^3 - 6A^2 + 5A + 11I$ gives the zero matrix entry by entry. ✓

Step 2 — rearrange for the inverse: $A^3 - 6A^2 + 5A = -11I$, so $A(A^2 - 6A + 5I) = -11I$ and $$A^{-1} = -\frac{1}{11}\left(A^2 - 6A + 5I\right)$$

Step 3 — evaluate: $$A^2 - 6A + 5I = \begin{pmatrix} 3 & -4 & -5 \\ -9 & 1 & 4 \\ -5 & 3 & 1 \end{pmatrix} \quad\Rightarrow\quad A^{-1} = \frac{1}{11}\begin{pmatrix} -3 & 4 & 5 \\ 9 & -1 & -4 \\ 5 & -3 & -1 \end{pmatrix}$$

Answer: as above — a cubic identity converts a nine-cofactor slog into two matrix multiplications.

Example 16 — The same trick, cleaner numbers

For $A = \begin{pmatrix} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{pmatrix}$, verify $A^3 - 6A^2 + 9A - 4I = O$ and find $A^{-1}$.

Step 1 — powers: $A^2 = \begin{pmatrix} 6 & -5 & 5 \\ -5 & 6 & -5 \\ 5 & -5 & 6 \end{pmatrix}$; computing $A^3 = A^2 A$ and substituting confirms the identity. ✓

Step 2 — rearrange: $A(A^2 - 6A + 9I) = 4I$, so $A^{-1} = \frac{1}{4}(A^2 - 6A + 9I)$.

Step 3 — evaluate: $$A^{-1} = \frac{1}{4}\begin{pmatrix} 3 & 1 & -1 \\ 1 & 3 & 1 \\ -1 & 1 & 3 \end{pmatrix}$$

Answer: as above. Check one entry of $AA^{-1}$: the $(1,1)$ entry is $\frac{1}{4}\left[2(3) + (-1)(1) + 1(-1)\right] = 1$. ✓

Example 17 — Quick-fire determinant facts

A is a nonsingular matrix of order $3$ with $\vert A \vert = 5$. Find $\vert \operatorname{adj} A \vert$, $\det(A^{-1})$ and $\vert 3A \vert$.

Step 1 — adjoint: $\vert \operatorname{adj} A \vert = \vert A \vert^{n-1} = 5^2 = 25$.

Step 2 — inverse: $\det(A^{-1}) = \frac{1}{\det A} = \frac{1}{5}$.

Step 3 — scaling: $\vert 3A \vert = 3^3 \vert A \vert = 135$.

Answer: $25$, $\frac{1}{5}$, $135$ — three one-liners that appear verbatim in JEE Main year after year.

Block C — Systems of Equations and Miscellaneous Classics

Example 18 — A gifted inverse

Use the product $\begin{pmatrix} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{pmatrix}\begin{pmatrix} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{pmatrix}$ to solve $x - y + 2z = 1$, $2y - 3z = 1$, $3x - 2y + 4z = 2$.

Step 1 — multiply the given matrices: every entry works out to the identity pattern: $$AB = I \quad\Rightarrow\quad A^{-1} = B = \begin{pmatrix} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{pmatrix}$$

Step 2 — the system is $AX = (1, 1, 2)^{\prime}$: so $$X = A^{-1}\begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} = \begin{pmatrix} -2 + 0 + 2 \\ 9 + 2 - 6 \\ 6 + 1 - 4 \end{pmatrix} = \begin{pmatrix} 0 \\ 5 \\ 3 \end{pmatrix}$$

Answer: $x = 0$, $y = 5$, $z = 3$. When an exam question hands you a product, it is handing you the inverse — never recompute it.

Example 19 — Consistency check that fails

Examine the consistency of $3x - y - 2z = 2$, $2y - z = -1$, $3x - 5y = 3$.

Step 1 — determinant: $A = \begin{pmatrix} 3 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0 \end{pmatrix}$; expanding along $C_1$: $3(0 - 5) + 3(1 + 4) = -15 + 15 = 0$.

Step 2 — adjoint test: computing $(\operatorname{adj} A)B$ with $B = (2, -1, 3)^{\prime}$ gives $(-5, -3, -6)^{\prime} \neq O$.

Answer: inconsistent — the system has no solution.

Example 20 — Fractional solutions are normal

Solve $2x - y = -2$, $3x + 4y = 3$ by the matrix method.

Step 1 — determinant: $\vert A \vert = 8 + 3 = 11$.

Step 2 — solve: $$X = \frac{1}{11}\begin{pmatrix} 4 & 1 \\ -3 & 2 \end{pmatrix}\begin{pmatrix} -2 \\ 3 \end{pmatrix} = \frac{1}{11}\begin{pmatrix} -8 + 3 \\ 6 + 6 \end{pmatrix} = \begin{pmatrix} -\frac{5}{11} \\ \frac{12}{11} \end{pmatrix}$$

Answer: $x = -\frac{5}{11}$, $y = \frac{12}{11}$ — do not panic at fractions; check instead: $2\left(-\frac{5}{11}\right) - \frac{12}{11} = -\frac{22}{11} = -2$. ✓

Example 21 — Another $2 \times 2$ with fractions

Solve $4x - 3y = 3$, $3x - 5y = 7$ by the matrix method.

Step 1 — determinant: $\vert A \vert = -20 + 9 = -11$.

Step 2 — solve: $$X = -\frac{1}{11}\begin{pmatrix} -5 & 3 \\ -3 & 4 \end{pmatrix}\begin{pmatrix} 3 \\ 7 \end{pmatrix} = -\frac{1}{11}\begin{pmatrix} -15 + 21 \\ -9 + 28 \end{pmatrix} = \begin{pmatrix} -\frac{6}{11} \\ -\frac{19}{11} \end{pmatrix}$$

Answer: $x = -\frac{6}{11}$, $y = -\frac{19}{11}$.

Example 22 — A $3 \times 3$ system, start to finish

Solve $2x + 3y + 3z = 5$, $x - 2y + z = -4$, $3x - y - 2z = 3$.

Step 1 — pack and test: $\vert A \vert = 2(4 + 1) - 3(-2 - 3) + 3(-1 + 6) = 10 + 15 + 15 = 40 \neq 0$.

Step 2 — solve $X = A^{-1}B$: carrying out the adjoint computation and multiplication gives $$X = \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}$$

Step 3 — check every equation: $2 + 6 - 3 = 5$ ✓, $1 - 4 - 1 = -4$ ✓, $3 - 2 + 2 = 3$ ✓.

Answer: $x = 1$, $y = 2$, $z = -1$.

Example 23 — Find the inverse, then reuse it

For $A = \begin{pmatrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{pmatrix}$, find $A^{-1}$ and use it to solve $2x - 3y + 5z = 11$, $3x + 2y - 4z = -5$, $x + y - 2z = -3$.

Step 1 — determinant: $2(-4 + 4) + 3(-6 + 4) + 5(3 - 2) = 0 - 6 + 5 = -1$.

Step 2 — adjoint and inverse: $$\operatorname{adj} A = \begin{pmatrix} 0 & -1 & 2 \\ 2 & -9 & 23 \\ 1 & -5 & 13 \end{pmatrix} \quad\Rightarrow\quad A^{-1} = \begin{pmatrix} 0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13 \end{pmatrix}$$

Step 3 — multiply into $B = (11, -5, -3)^{\prime}$: $$X = \begin{pmatrix} 0 - 5 + 6 \\ -22 - 45 + 69 \\ -11 - 25 + 39 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}$$

Answer: $x = 1$, $y = 2$, $z = 3$.

Example 24 — Reciprocal substitution

Solve $\dfrac{2}{x} + \dfrac{3}{y} + \dfrac{10}{z} = 4$, $\;\dfrac{4}{x} - \dfrac{6}{y} + \dfrac{5}{z} = 1$, $\;\dfrac{6}{x} + \dfrac{9}{y} - \dfrac{20}{z} = 2$.

Step 1 — substitute $u = \frac{1}{x}$, $v = \frac{1}{y}$, $w = \frac{1}{z}$: the system becomes linear in $u, v, w$ with coefficient matrix $\begin{pmatrix} 2 & 3 & 10 \\ 4 & -6 & 5 \\ 6 & 9 & -20 \end{pmatrix}$, whose determinant is $1200 \neq 0$.

Step 2 — solve the linear system: the matrix method gives $u = \frac{1}{2}$, $v = \frac{1}{3}$, $w = \frac{1}{5}$.

Step 3 — undo the substitution: $x = \frac{1}{u} = 2$, $y = 3$, $z = 5$.

Answer: $x = 2$, $y = 3$, $z = 5$ — the substitution is the whole trick; the matrix method does the rest.

Example 25 — Price-per-kg by matrices

$4$ kg onion, $3$ kg wheat and $2$ kg rice cost ₹60. $2$ kg onion, $4$ kg wheat and $6$ kg rice cost ₹90. $6$ kg onion, $2$ kg wheat and $3$ kg rice cost ₹70. Find the cost of each per kg.

Step 1 — translate: with prices $x, y, z$ per kg: $4x + 3y + 2z = 60$, $2x + 4y + 6z = 90$, $6x + 2y + 3z = 70$.

Step 2 — test: $\vert A \vert = 4(12 - 12) - 3(6 - 36) + 2(4 - 24) = 0 + 90 - 40 = 50 \neq 0$.

Step 3 — solve: $X = A^{-1}B = (5, 8, 8)^{\prime}$.

Answer: onion ₹5 per kg, wheat ₹8 per kg, rice ₹8 per kg. Check the first purchase: $4(5) + 3(8) + 2(8) = 60$. ✓