The Determinant — One Number per Square Matrix

To every square matrix $A = [a_{ij}]$ of order $n$ we attach a single number (real or complex) called its determinant, written $\det A$, $\vert A \vert$ or $\Delta$. Think of it as a function $f : M \to K$ from the set of square matrices to the set of numbers — each square matrix in, exactly one number out. Chapter 3 ended with invertibility; this number is precisely what decides it.

Two warnings before any computation (both are favourite one-mark traps):

1. $\vert A \vert$ is read determinant of $A$ — it is not a modulus, and a determinant can be negative.
2. Only square matrices have determinants. A $2 \times 3$ matrix has none.

Order one

For $A = [a]$, simply $\det A = a$. (So the determinant of $[-5]$ is $-5$, not $5$ — warning 1 already at work.)

Order two

For $A = \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix}$:

$$\vert A \vert = \begin{vmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{vmatrix} = a_{11}a_{22} - a_{21}a_{12}$$

— the product of the main diagonal minus the product of the other diagonal. For example,

$$\begin{vmatrix} 2 & 4 \\ -1 & 2 \end{vmatrix} = 2(2) - 4(-1) = 8$$

Entries can be algebraic too, and the algebra often collapses:

$$\begin{vmatrix} x & x+1 \\ x-1 & x \end{vmatrix} = x^2 - (x+1)(x-1) = x^2 - (x^2 - 1) = 1$$

— a determinant that is the same number for every $x$.

Solved Examples — Order Two

Example 1 — A plain evaluation

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

Step 1 — diagonal rule: main diagonal product minus anti-diagonal product: $2(-1) - 4(-5)$.

Step 2 — arithmetic: $-2 + 20 = 18$.

Answer: $18$.

Example 2 — Trigonometric entries

Evaluate $\begin{vmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{vmatrix}$.

Step 1 — diagonal rule: $\cos\theta \cdot \cos\theta - (-\sin\theta)(\sin\theta) = \cos^2\theta + \sin^2\theta$.

Step 2 — Pythagorean identity: $\cos^2\theta + \sin^2\theta = 1$.

Answer: $1$, for every $\theta$ — the rotation matrix from Chapter 3 always has determinant $1$.

Example 3 — Solving an equation between determinants

Find the values of $x$ for which $\begin{vmatrix} 3 & x \\ x & 1 \end{vmatrix} = \begin{vmatrix} 3 & 2 \\ 4 & 1 \end{vmatrix}$.

Step 1 — expand both sides: left side $= 3 - x^2$; right side $= 3 - 8 = -5$.

Step 2 — equate and solve: $3 - x^2 = -5$, so $x^2 = 8$.

Answer: $x = \pm 2\sqrt{2}$ — keep both signs; a quadratic in $x$ almost always rewards two roots, and dropping the negative one is the standard mark-loser.

Order Three — Expansion Along a Row or Column

A $3 \times 3$ determinant is computed by breaking it into $2 \times 2$ determinants. Pick any row (or column); each element of it contributes one term:

1. Sign: multiply the element $a_{ij}$ by $(-1)^{i+j}$ — plus when $i + j$ is even, minus when odd.
2. Minor: multiply by the $2 \times 2$ determinant left after deleting row $i$ and column $j$.
3. Add the three signed terms.

Expanding along the first row:

$$\vert A \vert = a_{11}\begin{vmatrix} a_{22} & a_{23} \\ a_{32} & a_{33} \end{vmatrix} - a_{12}\begin{vmatrix} a_{21} & a_{23} \\ a_{31} & a_{33} \end{vmatrix} + a_{13}\begin{vmatrix} a_{21} & a_{22} \\ a_{31} & a_{32} \end{vmatrix}$$

There are six possible expansions — three rows and three columns — and all six give the same value. That freedom is worth marks:

The zeros shortcut. Expand along the row or column with the most zeros: every zero kills its whole term. In

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

the third column has two zeros, so expanding along $C_3$ leaves a single term: $\Delta = 4\begin{vmatrix} -1 & 3 \\ 4 & 1 \end{vmatrix} - 0 + 0 = 4(-1 - 12) = -52$.

The sign chessboard. Instead of computing $(-1)^{i+j}$ each time, remember the pattern $\begin{smallmatrix} + & - & + \\ - & + & - \\ + & - & + \end{smallmatrix}$ — corners and centre positive, edges negative.

The scaling law

Multiplying the whole matrix by $k$ multiplies every row by $k$, and each of the $n$ rows contributes one factor:

$$\boxed{\;\vert kA \vert = k^n \vert A \vert \quad (A \text{ of order } n)\;}$$

So for a $2 \times 2$ matrix $\vert 2A \vert = 4\vert A \vert$, and for a $3 \times 3$ matrix $\vert 3A \vert = 27 \vert A \vert$. Writing $\vert kA \vert = k\vert A \vert$ is the classic error of this section — determinants are not linear in the matrix.

One more freebie: a triangular matrix (all zeros below, or above, the diagonal) has determinant equal to the product of its diagonal entries — expand repeatedly along the zero-heavy column and watch every other term vanish.

Solved Examples — Order Three

Example 4 — Using the zeros shortcut

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

Step 1 — choose the row with most zeros: $R_2$ has two zeros, so expand along it. Chessboard signs for row 2 are $-, +, -$.

Step 2 — only one term survives: $$\Delta = -0\begin{vmatrix} -1 & -2 \\ -5 & 0 \end{vmatrix} + 0\begin{vmatrix} 3 & -2 \\ 3 & 0 \end{vmatrix} - (-1)\begin{vmatrix} 3 & -1 \\ 3 & -5 \end{vmatrix}$$

Step 3 — finish: $\Delta = 1 \cdot \left(3(-5) - (-1)(3)\right) = -15 + 3 = -12$.

Answer: $-12$.

Example 5 — A determinant that vanishes identically

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

Step 1 — expand along $R_1$: $$\Delta = 0 - \sin\alpha \begin{vmatrix} -\sin\alpha & \sin\beta \\ \cos\alpha & 0 \end{vmatrix} + (-\cos\alpha)\begin{vmatrix} -\sin\alpha & 0 \\ \cos\alpha & -\sin\beta \end{vmatrix}$$

Step 2 — evaluate the two minors: first bracket: $0 - \sin\beta\cos\alpha = -\sin\beta\cos\alpha$; second bracket: $\sin\alpha\sin\beta - 0 = \sin\alpha\sin\beta$.

Step 3 — combine: $\Delta = \sin\alpha\sin\beta\cos\alpha - \cos\alpha\sin\alpha\sin\beta = 0$.

Answer: $0$. (Look at the matrix again: it is skew-symmetric of odd order — Section 8 turns this observation into a one-line rule.)

Example 6 — Verifying the scaling law

For $A = \begin{pmatrix} 1 & 2 \\ 4 & 2 \end{pmatrix}$, show that $\vert 2A \vert = 4\vert A \vert$.

Step 1 — left side: $2A = \begin{pmatrix} 2 & 4 \\ 8 & 4 \end{pmatrix}$, so $\vert 2A \vert = 8 - 32 = -24$.

Step 2 — right side: $\vert A \vert = 2 - 8 = -6$, so $4\vert A \vert = -24$.

Answer: both sides equal $-24$; in general $\vert kA \vert = k^2\vert A \vert$ for order $2$.

Example 7 — A triangular determinant

Evaluate $\vert A \vert$ for $A = \begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{pmatrix}$, and hence find $\vert 3A \vert$.

Step 1 — expand along $C_1$ (two zeros): $\vert A \vert = 1\begin{vmatrix} 1 & 2 \\ 0 & 4 \end{vmatrix} = 4$ — exactly the product of the diagonal entries $1 \cdot 1 \cdot 4$.

Step 2 — scaling law with $n = 3$: $\vert 3A \vert = 3^3 \vert A \vert = 27 \times 4 = 108$.

Answer: $\vert A \vert = 4$ and $\vert 3A \vert = 108$.