The Determinant and Its Expansion

Every square matrix $A$ carries one number, $\det A = \vert A \vert$. Two warnings first: $\vert A \vert$ is not a modulus (determinants can be negative), and only square matrices have determinants. For order $1$, $\vert [a] \vert = a$; for order $2$, $$\begin{vmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{vmatrix} = a_{11}a_{22} - a_{12}a_{21}$$

For order $3$, expand along any row or column: each element is multiplied by its sign $(-1)^{i+j}$ and by the $2 \times 2$ determinant left after deleting its row and column, and the three signed terms are added. All six expansions agree, so pick the line with the most zeros — every zero kills a whole term.

The scaling law: $\vert kA \vert = k^n\vert A \vert$ for order $n$ — one factor of $k$ per row, so writing $\vert kA \vert = k\vert A \vert$ is the chapter's classic error. A triangular matrix's determinant is just the product of its diagonal entries.

Area of a Triangle and Collinearity

$$\Delta = \frac{1}{2}\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}$$

Three remarks carry the marks: take the absolute value (area is positive); when the area is given, equate the half-determinant to $\pm$ the area and solve both signs; and collinear points give determinant zero — which doubles as a way to find the equation of a line through two points (set the flat-triangle determinant with $(x, y)$ to zero).

Minors, Cofactors, Adjoint and Inverse

Minor $M_{ij}$: delete row $i$ and column $j$. Cofactor $A_{ij} = (-1)^{i+j}M_{ij}$ — the chessboard sign attached. Then elements times their own cofactors sum to $\Delta$ (along any line), while elements times another row's cofactors sum to $0$ — the own/alien dichotomy.

Adjoint $\operatorname{adj} A$ = transpose of the cofactor matrix. For $2 \times 2$: swap the diagonal, negate the off-diagonal. The master identity and its consequences: $$A(\operatorname{adj} A) = (\operatorname{adj} A)A = \vert A \vert I, \qquad A^{-1} = \frac{1}{\vert A \vert}\operatorname{adj} A \;\; (\vert A \vert \neq 0)$$

Singular means $\vert A \vert = 0$ (no inverse); nonsingular means $\vert A \vert \neq 0$ (invertible). The supporting formulas: $\vert AB \vert = \vert A \vert\vert B \vert$, $\det(A^{-1}) = \frac{1}{\det A}$, and $\vert \operatorname{adj} A \vert = \vert A \vert^{n-1}$ (so $\vert A \vert^2$ at order $3$).

Solving Systems — the Matrix Method

Write the system as $AX = B$ (coefficients, unknowns, constants). If $\vert A \vert \neq 0$: unique solution $$X = A^{-1}B$$ (premultiply — never $BA^{-1}$). If $\vert A \vert = 0$: compute $(\operatorname{adj} A)B$ — nonzero means inconsistent (no solution); zero means infinitely many solutions or none, to be examined directly. Board presentation: state $\vert A \vert$ before inverting, show $\operatorname{adj} A$, and end by substituting the solution back into one equation.

Worked Examples — Evaluation and Area

Example 1 — A $2 \times 2$ with signs

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

Step 1 — diagonal rule: $2(-1) - 4(-5) = -2 + 20$.

Answer: $18$ — the second product is subtracted, so its minus becomes a plus.

Example 2 — The zeros shortcut

Evaluate $\Delta = \begin{vmatrix} 1 & 2 & 4 \\ -1 & 3 & 0 \\ 4 & 1 & 0 \end{vmatrix}$.

Step 1 — pick the third column (two zeros): only one term survives, with chessboard sign $+$ at position $(1,3)$: $$\Delta = 4\begin{vmatrix} -1 & 3 \\ 4 & 1 \end{vmatrix} - 0 + 0$$

Step 2 — finish: $4(-1 - 12) = -52$.

Answer: $-52$ — one $2 \times 2$ instead of three, because the column was chosen well.

Example 3 — Solving for an unknown entry

Find $x$ if $\begin{vmatrix} 2x & 4 \\ 6 & x \end{vmatrix} = \begin{vmatrix} 2 & 4 \\ 5 & 1 \end{vmatrix}$.

Step 1 — expand both sides: $2x^2 - 24 = 2 - 20 = -18$.

Step 2 — solve: $2x^2 = 6$, so $x^2 = 3$.

Answer: $x = \pm\sqrt{3}$ — a quadratic gives both roots; dropping the negative one is the standard mark-loser.

Example 4 — Area of a triangle

Find the area of the triangle with vertices $(3, 8)$, $(-4, 2)$, $(5, 1)$.

Step 1 — set up and expand: $$\Delta = \frac{1}{2}\left[3(2 - 1) - 8(-4 - 5) + 1(-4 - 10)\right] = \frac{1}{2}(3 + 72 - 14)$$

Answer: $\dfrac{61}{2}$ square units — and had the bracket come out negative, the absolute value would fix the sign, not a recalculation.

Example 5 — Given area, two answers

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

Step 1 — half-determinant with both signs: the determinant simplifies (base on the $x$-axis, height $2$) to give $4 - k = \pm 4$.

Step 2 — solve both: $k = 0$ or $k = 8$.

Answer: $k = 0$ or $8$ — one triangle on each side; a single answer is half the marks.

Worked Examples — Cofactors, Adjoint, Inverse

Example 6 — A cofactor table with the alien check

For $\Delta = \begin{vmatrix} 2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7 \end{vmatrix}$, find the first-row cofactors and the value of $\Delta$.

Step 1 — the three cofactors: $$A_{11} = +\begin{vmatrix} 0 & 4 \\ 5 & -7 \end{vmatrix} = -20, \quad A_{12} = -\begin{vmatrix} 6 & 4 \\ 1 & -7 \end{vmatrix} = 46, \quad A_{13} = +\begin{vmatrix} 6 & 0 \\ 1 & 5 \end{vmatrix} = 30$$

Step 2 — own-cofactor sum: $\Delta = 2(-20) + (-3)(46) + 5(30) = -40 - 138 + 150 = -28$.

Answer: $\Delta = -28$ — and pairing these elements with another row's cofactors would sum to $0$, the alien-cofactor rule that powers the adjoint identity.

Example 7 — Evaluating along a chosen row

Using the cofactors of the second row, evaluate $\Delta = \begin{vmatrix} 5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3 \end{vmatrix}$.

Step 1 — second-row cofactors: $A_{21} = -\begin{vmatrix} 3 & 8 \\ 2 & 3 \end{vmatrix} = 7$, $A_{22} = +\begin{vmatrix} 5 & 8 \\ 1 & 3 \end{vmatrix} = 7$, $A_{23} = -\begin{vmatrix} 5 & 3 \\ 1 & 2 \end{vmatrix} = -7$.

Step 2 — sum with the row's elements: $2(7) + 0(7) + 1(-7) = 7$.

Answer: $\Delta = 7$ — the zero in the row was exactly why this row was worth choosing.

Example 8 — Adjoint by the shortcut

Find $\operatorname{adj} A$ for $A = \begin{pmatrix} 2 & 3 \\ 1 & 4 \end{pmatrix}$.

Step 1 — swap and negate: diagonal $2 \leftrightarrow 4$; off-diagonal $3, 1$ become $-3, -1$ in place.

Answer: $\operatorname{adj} A = \begin{pmatrix} 4 & -3 \\ -1 & 2 \end{pmatrix}$ — five seconds, no cofactor table.

Example 9 — Verifying the master identity

For $A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$, verify $A(\operatorname{adj} A) = \vert A \vert I$.

Step 1 — the pieces: $\vert A \vert = 4 - 6 = -2$ and $\operatorname{adj} A = \begin{pmatrix} 4 & -2 \\ -3 & 1 \end{pmatrix}$.

Step 2 — multiply: $A(\operatorname{adj} A) = \begin{pmatrix} 4 - 6 & -2 + 2 \\ 12 - 12 & -6 + 4 \end{pmatrix} = \begin{pmatrix} -2 & 0 \\ 0 & -2 \end{pmatrix} = -2I$.

Answer: verified — the diagonal carries $\vert A \vert$, the off-diagonal vanishes by the alien-cofactor rule.

Example 10 — Inverse by the formula

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

Step 1 — determinant first: $\vert A \vert = 6 + 8 = 14 \neq 0$ — the inverse exists.

Step 2 — adjoint and divide: $$A^{-1} = \frac{1}{14}\begin{pmatrix} 3 & 2 \\ -4 & 2 \end{pmatrix}$$

Answer: as displayed — quick check: the $(1,1)$ entry of $AA^{-1}$ is $\frac{1}{14}(6 + 8) = 1$. ✓

Worked Examples — Systems of Equations

Example 11 — Matrix method, two variables

Solve $2x + 5y = 1$, $3x + 2y = 7$.

Step 1 — pack and test: $A = \begin{pmatrix} 2 & 5 \\ 3 & 2 \end{pmatrix}$, $\vert A \vert = 4 - 15 = -11 \neq 0$.

Step 2 — solve: $$X = A^{-1}B = -\frac{1}{11}\begin{pmatrix} 2 & -5 \\ -3 & 2 \end{pmatrix}\begin{pmatrix} 1 \\ 7 \end{pmatrix} = -\frac{1}{11}\begin{pmatrix} -33 \\ 11 \end{pmatrix} = \begin{pmatrix} 3 \\ -1 \end{pmatrix}$$

Answer: $x = 3$, $y = -1$; check in the second equation: $9 - 2 = 7$. ✓

Example 12 — Matrix method, three variables

Solve $3x - 2y + 3z = 8$, $2x + y - z = 1$, $4x - 3y + 2z = 4$.

Step 1 — determinant: expanding along $R_1$: $3(2 - 3) + 2(4 + 4) + 3(-6 - 4) = -3 + 16 - 30 = -17 \neq 0$.

Step 2 — adjoint, inverse, multiply: computing the cofactor matrix, transposing, dividing by $-17$ and multiplying into $B = (8, 1, 4)'$ gives $$X = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}$$

Answer: $x = 1$, $y = 2$, $z = 3$; check the first equation: $3 - 4 + 9 = 8$. ✓ This full pipeline — determinant, adjoint, inverse, product, verification — is the standard 5-mark question of the chapter.

Example 13 — Detecting inconsistency

Examine the consistency of $x + 3y = 5$, $2x + 6y = 8$.

Step 1 — determinant: $\vert A \vert = 6 - 6 = 0$ — the matrix method stops; move to the adjoint test.

Step 2 — the test product: $$(\operatorname{adj} A)B = \begin{pmatrix} 6 & -3 \\ -2 & 1 \end{pmatrix}\begin{pmatrix} 5 \\ 8 \end{pmatrix} = \begin{pmatrix} 6 \\ -2 \end{pmatrix} \neq O$$

Answer: inconsistent — no solution; geometrically the two lines are parallel (doubling the first equation demands $2x + 6y = 10 \neq 8$).