Skip to content

Mathematics · Ch 4 — Determinants and Matrices

Determinant of Order 3

4.1.2

Determinant of Order 3

4.1.2 Determinant of Order 3

Definition. A determinant of order 3 is a square arrangement of 9 elements enclosed between two vertical bars, arranged in 3 rows and 3 columns:

D=∣a11a12a13a21a22a23a31a32a33∣D=\begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{vmatrix}

Here aija_{ij} denotes the element in the ii-th row and jj-th column. For instance a31a_{31} is the element in the 3rd row, 1st column. A determinant is usually named by a capital letter, or by Δ\Delta (delta). The rows are R1,R2,R3R_1,R_2,R_3 and the columns C1,C2,C3C_1,C_2,C_3 — e.g. the 2nd row is [a21  a22  a23][a_{21}\ \ a_{22}\ \ a_{23}] and the 3rd column is [a13a23a33]\begin{bmatrix}a_{13}\\a_{23}\\a_{33}\end{bmatrix}.

Expansion of a determinant

A 3×3 determinant can be expanded along any of its 3 rows or 3 columns — six expansions in all, and every one gives the same value (this is proved as Property 1 in §4.2). Expanding along the first row:

D=a11∣a22a23a32a33∣−a12∣a21a23a31a33∣+a13∣a21a22a31a32∣D = 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}

Notice the alternating +,−,++,-,+ signs, and that each 2×2 determinant is obtained by deleting the row and column of the multiplying element (this "leftover" 2×2 determinant is formally called a minor, defined in §4.1.3).

Worked Examples

Example (i). Evaluate ∣3−45−1−12−2−31∣\begin{vmatrix} 3 & -4 & 5 \\ -1 & -1 & 2 \\ -2 & -3 & 1 \end{vmatrix}.

Step 1: Expand along row 1: D=3∣−12−31∣−(−4)∣−12−21∣+5∣−1−1−2−3∣D = 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 each 2×2 piece: ∣−12−31∣=(−1)(1)−(2)(−3)=−1+6=5\begin{vmatrix}-1&2\\-3&1\end{vmatrix}=(-1)(1)-(2)(-3)=-1+6=5; ∣−12−21∣=(−1)(1)−(2)(−2)=−1+4=3\begin{vmatrix}-1&2\\-2&1\end{vmatrix}=(-1)(1)-(2)(-2)=-1+4=3; ∣−1−1−2−3∣=(−1)(−3)−(−1)(−2)=3−2=1\begin{vmatrix}-1&-1\\-2&-3\end{vmatrix}=(-1)(-3)-(-1)(-2)=3-2=1.

Step 3: Combine: D=3(5)+4(3)+5(1)=15+12+5D=3(5)+4(3)+5(1)=15+12+5.

D=32D = 32

Example (ii). Evaluate ∣sec⁡θtan⁡θ0tan⁡θsec⁡θ0001∣\begin{vmatrix} \sec\theta & \tan\theta & 0 \\ \tan\theta & \sec\theta & 0 \\ 0 & 0 & 1 \end{vmatrix}.

Step 1: The third column has only one non-zero entry (11 in row 3), so expand along the third column: D=1×∣sec⁡θtan⁡θtan⁡θsec⁡θ∣D = 1\times\begin{vmatrix}\sec\theta & \tan\theta\\ \tan\theta & \sec\theta\end{vmatrix} (the sign is (−1)3+3=+1(-1)^{3+3}=+1).

Step 2: Evaluate the 2×2 piece: sec⁡2θ−tan⁡2θ\sec^2\theta - \tan^2\theta.

Step 3: Use the identity sec⁡2θ−tan⁡2θ=1\sec^2\theta-\tan^2\theta=1.

D=1D = 1

Example (iii). Evaluate ∣2−i3132−i0−1−22−i∣\begin{vmatrix} 2-i & 3 & 1 \\ 3 & 2-i & 0 \\ -1 & -2 & 2-i \end{vmatrix}, where i=−1i=\sqrt{-1}. …

Misc 4.1.2Worked Examples — evaluating three order-3 determinants

Worked out. Three worked expansions along the first row: a purely numeric 3×3 determinant, a 3×3 determinant of secθ/tanθ entries with a row of zeros that collapses quickly, and a 3×3 determinant with complex entries using i²=−1. …