Determinants
Cofactor expansion, the property toolkit, triangular and block shortcuts — the single number that decides invertibility
§1Why Determinants? A Bit of Motivation
Imagine you have two arrows (vectors) drawn from the origin. Together they make a parallelogram. A natural question is: how big is the area of that parallelogram? It turns out there is one single number, built from the coordinates of the two arrows, that gives you this area. That number is the determinant.
The same idea scales up:
• In 2D, the determinant of a $2\times2$ matrix is the (signed) area of the parallelogram formed by its rows (or columns).
• In 3D, the determinant of a $3\times3$ matrix is the (signed) volume of the parallelepiped (a slanted box) formed by its three rows.
• If the determinant is zero, the arrows are squashed flat — they lie on the same line or plane — and the box has no area/volume. This is exactly the situation where a matrix has no inverse.
Why should you care? Two quick applications
1. Does a system have a unique solution? For a square system $A\mathbf{x} = \mathbf{b}$, there is exactly one solution if and only if $\det A \neq 0$. So a single number tells you whether your equations are "well behaved" or not.
2. Change of variables in calculus (the Jacobian). When you switch coordinate systems in a multiple integral (for example from rectangular to polar), the factor that corrects the stretching of area or volume is a determinant called the Jacobian. Without it, your integral would give the wrong answer.
§2Notation — How We Write a Determinant
For a square matrix $A$, its determinant is a single number. We write it in two common ways: $\det A$ or $|A|$. When we use the bar notation, we replace the square brackets of the matrix by straight vertical bars. For example,
$$A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \;\Longrightarrow\; \det A = \begin{vmatrix} a & b \\ c & d \end{vmatrix}.$$
The brackets $[\,\;]$ mean a matrix (a table of numbers). The bars $|\,\;|$ mean the determinant (a single number). They are not the same object, so keep the notation straight.
§3The Cofactor Expansion
The determinant is only defined for square matrices. We build it up step by step: first for tiny matrices, then for bigger ones.
The 1×1 case
If $A = [a]$, we simply define $\det A = a$. (And $A$ is invertible exactly when $a \neq 0$.)
The 2×2 case
For a 2×2 matrix the rule is short and worth memorising:
$$\det \begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc.$$
In words: multiply the main diagonal ($a$ and $d$), then subtract the product of the other diagonal ($b$ and $c$).
$\begin{vmatrix} 3 & 5 \\ 2 & 4 \end{vmatrix} = (3)(4) - (5)(2) = 12 - 10 = 2.$
$\begin{vmatrix} 1 & -2 \\ 4 & 0 \end{vmatrix} = (1)(0) - (-2)(4) = 0 + 8 = 8.$
Building the 3×3 case: the main idea
The clever idea is this: we compute a big determinant by reducing it to smaller ones. For a 3×3 matrix, we break it into 2×2 determinants, which we already know how to do. Take
$$A = \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix}.$$
The recipe is: go along the first row. For each entry, multiply that entry by the 2×2 determinant you get by deleting the row and column that the entry sits in, and attach a $+$ or $-$ sign. Then add the results.
$$\det A = a\begin{vmatrix} e & f \\ h & i \end{vmatrix} - b\begin{vmatrix} d & f \\ g & i \end{vmatrix} + c\begin{vmatrix} d & e \\ g & h \end{vmatrix}.$$
Carrying out the three small determinants gives the full formula:
$$\det A = a(ei - fh) - b(di - fg) + c(dh - eg) = aei + bfg + cdh - ceg - afh - bdi.$$
§4Minors, Cofactors, and the Sign Rule
Let $A$ be a square matrix.
• The minor of the entry in row $i$, column $j$ is the determinant of the smaller matrix you get by deleting row $i$ and column $j$. We write it $M_{ij}$.
• The cofactor of that entry is the minor together with a sign: $c_{ij} = (-1)^{i+j} M_{ij}$.
Understanding the sign. The factor $(-1)^{i+j}$ is just a tidy way to write "plus or minus." If $i+j$ is even, the sign is $+$. If $i+j$ is odd, the sign is $-$.
Keeping track of the signs — more practice. For a $3\times3$ matrix the sign pattern is $\begin{bmatrix} + & - & + \\ - & + & - \\ + & - & + \end{bmatrix}$. A few quick checks:
• Position $(2,3)$: $i+j = 5$ (odd) $\Rightarrow$ sign is $-$.
• Position $(3,1)$: $i+j = 4$ (even) $\Rightarrow$ sign is $+$.
• Position $(2,2)$: $i+j = 4$ (even) $\Rightarrow$ sign is $+$.
For a $4\times4$ matrix:
• Position $(4,3)$: $i+j = 7$ (odd) $\Rightarrow -$.
• Position $(1,4)$: $i+j = 5$ (odd) $\Rightarrow -$.
To compute $\det A$, pick any single row or any single column. Multiply each entry in that row/column by its cofactor, and add the results.
Expanding along row $i$: $\;\det A = a_{i1}c_{i1} + a_{i2}c_{i2} + \cdots + a_{in}c_{in}.$
Expanding along column $j$: $\;\det A = a_{1j}c_{1j} + a_{2j}c_{2j} + \cdots + a_{nj}c_{nj}.$
§5A Worked 3×3 Example (Same Answer, Different Rows/Columns)
$$A = \begin{bmatrix} 2 & -1 & 3 \\ 0 & 4 & 1 \\ 5 & 0 & -2 \end{bmatrix}.$$
$$\det A = 2\begin{vmatrix} 4 & 1 \\ 0 & -2 \end{vmatrix} - (-1)\begin{vmatrix} 0 & 1 \\ 5 & -2 \end{vmatrix} + 3\begin{vmatrix} 0 & 4 \\ 5 & 0 \end{vmatrix}$$
$$= 2(-8 - 0) + 1(0 - 5) + 3(0 - 20) = -16 - 5 - 60 = -81.$$
The sign pattern down column 2 is $-, +, -$.
$$\det A = -(-1)\begin{vmatrix} 0 & 1 \\ 5 & -2 \end{vmatrix} + 4\begin{vmatrix} 2 & 3 \\ 5 & -2 \end{vmatrix} - 0\begin{vmatrix} 2 & 3 \\ 0 & 1 \end{vmatrix}$$
$$= 1(0 - 5) + 4(-4 - 15) - 0 = -5 + 4(-19) = -5 - 76 = -81.$$
$$\det A = 2\begin{vmatrix} 4 & 1 \\ 0 & -2 \end{vmatrix} - 0\begin{vmatrix} -1 & 3 \\ 0 & -2 \end{vmatrix} + 5\begin{vmatrix} -1 & 3 \\ 4 & 1 \end{vmatrix}$$
$$= 2(-8) - 0 + 5(-1 - 12) = -16 + 5(-13) = -16 - 65 = -81.$$
All three give $\det A = -81$. The choice of row or column does not change the answer — it only changes how much arithmetic you do.
§6Properties of Determinants
These rules let you compute determinants faster and predict how a determinant changes when you modify the matrix.
Let $A$ be an $n\times n$ matrix.
$\textbf{1.}$ If $A$ has a row or column of zeros, then $\det A = 0$.
$\textbf{2.}$ If two distinct rows (or columns) are interchanged, the determinant becomes $-\det A$.
$\textbf{3.}$ If a row (or column) is multiplied by a constant $u$, the new determinant is $u(\det A)$.
$\textbf{4.}$ If two distinct rows (or columns) are identical, then $\det A = 0$.
$\textbf{5.}$ If a multiple of one row is added to a different row (or column to column), the determinant is unchanged; it stays $\det A$.
A few more properties worth knowing. The following are also true and very commonly used; they fit naturally with the list above.
$\textbf{6. Transpose.}$ $\det(A^{\mathsf{T}}) = \det A$. (This is why every statement above works equally for rows and columns.)
$\textbf{7. Product rule.}$ $\det(AB) = (\det A)(\det B)$ for square $A, B$ of the same size.
$\textbf{8. Scalar multiple of whole matrix.}$ $\det(uA) = u^n \det A$ for an $n\times n$ matrix. Be careful: multiplying every entry by $u$ multiplies the determinant by $u^n$, not by $u$.
$\textbf{9. Invertibility.}$ $A$ is invertible if and only if $\det A \neq 0$, and then $\det(A^{-1}) = \dfrac{1}{\det A}$.
$\textbf{10. Triangular matrices.}$ The determinant of a triangular matrix is the product of its diagonal entries.
If $A$ is an $n\times n$ matrix, then $\det(uA) = u^n \det A$ for any number $u$.
§7Worked Examples and Practice (Property-Style)
Suppose $\det \begin{bmatrix} a & b & c \\ p & q & r \\ x & y & z \end{bmatrix} = 6$. Evaluate $\det A$, where $A = \begin{bmatrix} a+x & b+y & c+z \\ 3x & 3y & 3z \\ -p & -q & -r \end{bmatrix}$.
Throughout, suppose $\det \begin{bmatrix} a & b & c \\ p & q & r \\ x & y & z \end{bmatrix} = 4$. Evaluate each using the properties.
Find all $x$ with $\det A = 0$, where $A = \begin{bmatrix} 1 & x & x \\ x & 1 & x \\ x & x & 1 \end{bmatrix}$.
Show that $\det \begin{bmatrix} 1 & a_1 & a_1^2 \\ 1 & a_2 & a_2^2 \\ 1 & a_3 & a_3^2 \end{bmatrix} = (a_3 - a_1)(a_3 - a_2)(a_2 - a_1)$.
§8Triangular Matrices and Their Determinants
A square matrix is lower triangular if every entry above the main diagonal is zero, and upper triangular if every entry below the main diagonal is zero. A triangular matrix is either upper or lower triangular. For example, $\begin{bmatrix} a & 0 & 0 \\ u & b & 0 \\ v & w & c \end{bmatrix}$ is lower triangular and $\begin{bmatrix} a & u & v \\ 0 & b & w \\ 0 & 0 & c \end{bmatrix}$ is upper triangular.
If $A$ is a square triangular matrix, then $\det A$ is the product of the entries on the main diagonal.
Evaluate $\det A$ for the lower triangular matrix $A = \begin{bmatrix} a & 0 & 0 & 0 \\ u & b & 0 & 0 \\ v & w & c & 0 \\ x & y & z & d \end{bmatrix}$.
Expand along the first row. Only the entry $a$ survives:
$$\det A = a\begin{vmatrix} b & 0 & 0 \\ w & c & 0 \\ y & z & d \end{vmatrix} = a\,b\begin{vmatrix} c & 0 \\ z & d \end{vmatrix} = a\,b\,(cd) = abcd.$$
The determinant is just the product of the diagonal entries $a, b, c, d$.
$\begin{vmatrix} 2 & 7 & -1 \\ 0 & 3 & 5 \\ 0 & 0 & 4 \end{vmatrix} = 2\cdot 3\cdot 4 = 24, \qquad \begin{vmatrix} 5 & 0 & 0 \\ 9 & -2 & 0 \\ 1 & 6 & 3 \end{vmatrix} = 5\cdot(-2)\cdot 3 = -30.$
Note: if any diagonal entry is $0$, the whole product is $0$, so a triangular matrix with a zero on its diagonal has determinant $0$.
§9Block (Partitioned) Matrices
When a matrix is built out of smaller square blocks arranged triangularly, there is a shortcut.
Let $A$ and $B$ be square matrices, and let $X, Y$ be blocks of the right sizes. Then $$\det\begin{bmatrix} A & X \\ 0 & B \end{bmatrix} = (\det A)(\det B), \qquad \det\begin{bmatrix} A & 0 \\ Y & B \end{bmatrix} = (\det A)(\det B).$$
Compute $\det \begin{bmatrix} 2 & 3 & 1 & 3 \\ 1 & -2 & -1 & 1 \\ 0 & 1 & 0 & 1 \\ 0 & 4 & 0 & 1 \end{bmatrix}$.
- Determinant of $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ is $ad - bc$.
- For larger matrices, use cofactor expansion: entry × cofactor, summed along any row or column.
- Cofactor $= (-1)^{i+j} \times$ (minor); remember the checkerboard of signs.
- You may expand along any row or column — choose the one with the most zeros.
- Know the property list (Theorem 3.1.2) cold; it turns hard determinants into easy ones.
- Triangular matrix $\Rightarrow$ determinant = product of the diagonal.
- $\det A \neq 0 \iff A$ is invertible.
§10Solutions to Section 3.1 Exercises
$\textbf{Method.}$ Use row operations. Adding a multiple of one row to another does not change the determinant (Property 5). Once upper triangular, the determinant is the product of the diagonal (Theorem 3.1.4). Track any row swaps (each contributes a factor $-1$).
$\textbf{Strategy.}$ Pull common factors out of each row (Property 3), then restore the rows $a,b,c / p,q,r / x,y,z$ using "add a multiple of a row" (Property 5, no change) and row swaps (Property 2, factor $-1$). Let $D = -1$ be the given determinant.
$\textbf{Method.}$ Use only Property 5 (add a multiple of one row to another, no change), Property 3 (factor a common row factor), and Property 2 (swap rows, factor $-1$). Write the target rows as $R_1, R_2, R_3$.
(a) Find $\det A$ if $A$ is $3\times3$ and $\det(2A) = 6$.
(b) Under what conditions is $\det(-A) = \det A$?
(a) Find $b$ if $\det\begin{bmatrix} 5 & -1 & x \\ 2 & 6 & y \\ -5 & 4 & z \end{bmatrix} = ax + by + cz$.
(b) Find $c$ if $\det\begin{bmatrix} 2 & x & -1 \\ 1 & y & 3 \\ -3 & z & 4 \end{bmatrix} = ax + by + cz$.
You can now compute any determinant and bend it with the property toolkit. Next we put determinants to work.
The determinant unlocks the adjugate formula for the inverse, Cramer's rule for solving systems, and the geometric meaning of area and volume scaling — and later, it is the key to finding eigenvalues. Everything you mastered today feeds directly into what comes next.