Modules › Determinants
Determinants
Cofactor expansion, row operations, properties, adjugate and Cramer's rule.
Lecture 3 · Exercise Sheet 3 · Determinants
Textbook sections: Elementary Linear Algebra ch. 3 §3.1–3.4 · Strang ch. 5 §5.1–5.3
Textbook sections: Elementary Linear Algebra ch. 3 §3.1–3.4 · Strang ch. 5 §5.1–5.3
Determinants
Linear Algebra · Module 3
After this module you can
- Compute determinants of order up to 5 by row reduction and cofactor expansion; use Sarrus for $3\times3$.
- Use $\det(AB)$, $\det A^T$, $\det(cA)$, $\det A^{-1}$ and $\det(\operatorname{adj}A)$.
- Apply Cramer's rule and the adjugate formula; find the parameter values that make a matrix singular.
Three ways to compute $\det A$
| Method | How |
|---|---|
| row reduction | triangular form; swaps flip the sign; adding a multiple of a row changes nothing; $\det=\pm$ product of the diagonal |
| cofactor expansion | $\det A=\sum_ja_{ij}C_{ij}$, $C_{ij}=(-1)^{i+j}M_{ij}$: expand along the row/column with most zeros |
| Sarrus ($3\times3$ only) | $aei+bfg+cdh-ceg-afh-bdi$ |
The lab shows all three and checks that they agree.
Properties
| Fact | Formula |
|---|---|
| transpose | $\det A^T=\det A$ |
| product | $\det(AB)=\det A\det B$, so $\det A^k=(\det A)^k$, $\det A^{-1}=\frac1{\det A}$ |
| scalar | $\det(cA)=c^n\det A$ (e.g. $\det(2A^2)=8(\det A)^2$ for $n=3$) |
| row operations | swap: $\times(-1)$; scale a row by $c$: $\times c$; add a multiple: unchanged |
| equal / zero row | $\det=0$ |
| adjugate | $\det(\operatorname{adj}A)=(\det A)^{n-1}$, $A\operatorname{adj}A=(\det A)I$ |
Cramer's rule and the adjugate
- $A\mathbf x=\mathbf b$ with $\det A\ne0$: $x_j=\frac{\det A_j}{\det A}$ where $A_j$ has column $j$ replaced by $\mathbf b$.
- $A^{-1}=\frac1{\det A}\operatorname{adj}A$, $\operatorname{adj}A=(C_{ij})^T$ (note the transpose!).
- Example: $x_1+2x_2=5$, $-x_1+x_2=1$: $\det A=3$, $\det A_1=3$, $x_1=1$.
- The system has a unique solution iff $\det A\ne0$; when $\det A=0$ Cramer's rule says nothing (use elimination).
Determinants with a parameter
- $A(k)=\begin{bmatrix}4&1&3\\1&k&1\\-1&0&-1\end{bmatrix}$: $\det A=-k$, invertible iff $k\ne0$.
- $\begin{bmatrix}1&1&1&1\\0&1&-2&0\\2&1&0&-3\\-1&-1&0&k\end{bmatrix}$: $\det=1-4k$; for $k=2$ it is $-7$; singular iff $k=\frac14$.
- $\det\begin{bmatrix}x&1&1\\1&x&1\\1&1&x\end{bmatrix}=(x-1)^2(x+2)$: add all rows to row 1 to factor $x+2$.
- Singular matrix $\iff\det=0\iff$ rows dependent.
Typical mistakes
Cofactor signs$(-1)^{i+j}$: a checkerboard of $+,-$.
AdjugateTranspose of the cofactor matrix, then divide by $\det A$.
Scalar$\det(cA)=c^n\det A$, not $c\det A$.
Practice next
- Chapter quizzes Ch. 3 (3 sets)
- Practice bank, filtered to this chapter, then written mode
- Lab: Determinant
- Then the mock exams (Exams page)
Worked examples from the course problems
Worked example 1 · Computing determinants (Exercise Sheet 3, Ex. 4)
Find $\det A$ for $A=\begin{bmatrix}-1 & 1 & -1 & 0 & 0 \\ 0 & 1 & -1 & 0 & 1 \\ 1 & 0 & 1 & -1 & 0 \\ 0 & -1 & 0 & 1 & -1 \\ 0 & 1 & 1 & -1 & 1\end{bmatrix}$.
Show the answer and the solution
Answer: $-1$
Row-reduce (or expand along the row/column with most zeros). Result: $\det A=-1$.
Worked example 2 · Adjugate (Exercise Sheet 3, Ex. 10)
For $A=\begin{bmatrix}1 & 2 & 3 \\ 0 & 1 & -1 \\ 2 & 2 & 2\end{bmatrix}$, the entry in row $1$, column $1$ of $\operatorname{adj}(A)$ is
Show the answer and the solution
Answer: $4$
$\operatorname{adj}A$ is the transpose of the cofactor matrix: $\operatorname{adj}A=\begin{bmatrix}4 & 2 & -5 \\ -2 & -4 & 1 \\ -2 & 2 & 1\end{bmatrix}$, and $\det A=-6$.
Worked example 3 · Properties of det (Linear Algebra Summary (Gen 16), III exercise 2)
A square matrix $A$ satisfies $2022A^3+A^2-A+3I=0$. Which argument shows that $A$ is invertible?
Show the answer and the solution
Answer: $A\,(2022A^2+A-I)=-3I$, so $\det A\cdot\det(2022A^2+A-I)=(-3)^n\neq0$ and therefore $\det A\neq0$
Move $3I$ to the right and factor $A$ out of the other three terms (all are polynomials in $A$, so they commute): $A(2022A^2+A-I)=-3I$. Taking determinants, $\det A\cdot\det(2022A^2+A-I)=(-3)^n\neq0$, so $\det A\neq0$. It even gives $A^{-1}=-\frac13(2022A^2+A-I)$. The other arguments are wrong: $\det$ is not additive, and e.g. $A=0$ satisfies $A^2=0$ but is not invertible.
Worked example 4 · Cramer's rule (Exercise Sheet 3, Ex. 6)
Use Cramer's rule to find $x_1$:
$$\begin{cases}x_1 + 2x_2 = 5 \\ -x_1 + x_2 = 1\end{cases}$$
Show the answer and the solution
Answer: $1$
$\det A=3$ and replacing column 1 by $b$ gives $\det A_1=3$, so $x_1=\det A_1/\det A=1$.
Practice
Quiz set A Quiz set B Quiz set C Practice bank (this chapter) Written mode Lab: Determinant