Determinants
Determinant and Its Properties
Every square matrix $A$ has a single number attached to it, its determinant $\det A$ (written $|A|$). It measures whether the matrix is invertible and, geometrically, how it scales area or volume.
Evaluating small determinants
For $2\times2$: $\begin{vmatrix}a&b\\c&d\end{vmatrix}=ad-bc.$ For $3\times3$, expand along any row or column using signed cofactors:
$$\begin{vmatrix}a_1&b_1&c_1\\a_2&b_2&c_2\\a_3&b_3&c_3\end{vmatrix}=a_1\begin{vmatrix}b_2&c_2\\b_3&c_3\end{vmatrix}-b_1\begin{vmatrix}a_2&c_2\\a_3&c_3\end{vmatrix}+c_1\begin{vmatrix}a_2&b_2\\a_3&b_3\end{vmatrix}.$$
Properties that save work
- $|A^{T}|=|A|$ — rows and columns play symmetric roles.
- Swapping two rows (or columns) multiplies the determinant by $-1$.
- If two rows (or columns) are identical or proportional, $|A|=0$.
- Multiplying one row by $k$ multiplies $|A|$ by $k$; hence $|kA|=k^{n}|A|$ for $n\times n$.
- Adding a multiple of one row to another leaves $|A|$ unchanged (the key simplification tool).
- $|AB|=|A||B|$.
$=3\cdot4-1\cdot2=12-2=10.$
The matrix is upper-triangular, so the determinant is the product of the diagonal: $1\cdot1\cdot5=5$. (You can confirm by expanding along the first column.)
Column $2$ is $2\times$ column $1$ (proportional columns), so by the proportionality property the determinant is $0$. (Check: $2\cdot6-4\cdot3=0$.)
$|kA|=k^{n}|A|$ with $n=3$, $k=2$: $|2A|=2^{3}\cdot5=8\cdot5=40.$
- $2\times2$: $ad-bc$; expand larger determinants by signed cofactors along any row/column.
- $|A^{T}|=|A|$; swapping two rows/columns changes the sign.
- Identical or proportional rows/columns $\Rightarrow |A|=0$.
- $|kA|=k^{n}|A|$ for $n\times n$; $|AB|=|A||B|$.
- Row operations of the type $R_i\to R_i+\lambda R_j$ do not change the determinant.
Area, Minors and Cofactors
To build the inverse of a matrix we need two related quantities for each entry: its minor and its cofactor.
Minors and cofactors
The minor $M_{ij}$ of entry $a_{ij}$ is the determinant left after deleting row $i$ and column $j$. The cofactor attaches a sign:
$$A_{ij}=(-1)^{i+j}M_{ij}.$$
The sign pattern alternates like a checkerboard: $+\ -\ +\ \dots$ A determinant can be written as the sum of entries times their cofactors along any line: $|A|=\sum_j a_{ij}A_{ij}$.
Adjoint of a matrix
The adjoint $\operatorname{adj}A$ is the transpose of the matrix of cofactors. Its defining property is
$$A\,(\operatorname{adj}A)=(\operatorname{adj}A)\,A=|A|\,I.$$
For an $n\times n$ matrix, $|\operatorname{adj}A|=|A|^{\,n-1}$.
Area of a triangle
The area of a triangle with vertices $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ is
$$\text{Area}=\frac12\left|\begin{matrix}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{matrix}\right|.$$
If this determinant is $0$, the three points are collinear — a quick test you will use often.
Delete row $1$, column $2$: $M_{12}=\begin{vmatrix}4&6\\7&9\end{vmatrix}=36-42=-6$. Cofactor $A_{12}=(-1)^{1+2}M_{12}=-(-6)=6$.
For a $2\times2$ matrix the adjoint swaps the diagonal and negates the off-diagonal: $\operatorname{adj}A=\begin{bmatrix}4&-3\\-1&2\end{bmatrix}$. Check $A(\operatorname{adj}A)=|A|I=(5)I$.
$\text{Area}=\tfrac12\left|\begin{matrix}0&0&1\\4&0&1\\0&3&1\end{matrix}\right|=\tfrac12\,|\,0-0+1\cdot(12-0)\,|=\tfrac12\cdot12=6$ square units.
$\begin{vmatrix}1&2&1\\2&4&1\\3&6&1\end{vmatrix}$: columns $1$ and $2$ are proportional ($y=2x$), so the determinant is $0$, giving zero area — the points are collinear.
- Minor $M_{ij}$ = determinant after deleting row $i$, column $j$; cofactor $A_{ij}=(-1)^{i+j}M_{ij}$.
- $\operatorname{adj}A$ = transpose of the cofactor matrix; $A(\operatorname{adj}A)=|A|I$.
- $|\operatorname{adj}A|=|A|^{n-1}$ for $n\times n$.
- Triangle area $=\tfrac12|\det|$ of the coordinate matrix; determinant $0 \Rightarrow$ collinear.
Adjoint, Inverse and Systems
The adjoint gives a direct formula for the inverse, and the inverse in turn solves systems of linear equations.
The inverse formula
If $|A|\ne 0$ (we call $A$ non-singular), then
$$A^{-1}=\frac{1}{|A|}\,\operatorname{adj}A.$$
If $|A|=0$ the matrix is singular and has no inverse. Always compute $|A|$ first.
Matrix method for $AX=B$
A system of linear equations can be written $AX=B$, where $A$ is the coefficient matrix, $X$ the unknowns and $B$ the constants. If $|A|\ne0$ the unique solution is
$$X=A^{-1}B.$$
Consistency — what $|A|$ tells you
| Case | Meaning |
|---|---|
| $|A|\ne 0$ | Unique solution (consistent, independent). |
| $|A|=0$ and $(\operatorname{adj}A)B=O$ | Either no solution or infinitely many. |
| $|A|=0$ and $(\operatorname{adj}A)B\ne O$ | No solution (inconsistent). |
$|A|=2\cdot4-3\cdot1=5\ne0$. $\operatorname{adj}A=\begin{bmatrix}4&-3\\-1&2\end{bmatrix}$, so $A^{-1}=\dfrac{1}{5}\begin{bmatrix}4&-3\\-1&2\end{bmatrix}=\begin{bmatrix}\tfrac45&-\tfrac35\\[2pt]-\tfrac15&\tfrac25\end{bmatrix}.$
Here $A=\begin{bmatrix}2&3\\1&4\end{bmatrix}$, $B=\begin{bmatrix}8\\9\end{bmatrix}$. Using $A^{-1}$ from Example 1, $X=A^{-1}B=\dfrac15\begin{bmatrix}4&-3\\-1&2\end{bmatrix}\begin{bmatrix}8\\9\end{bmatrix}=\dfrac15\begin{bmatrix}32-27\\-8+18\end{bmatrix}=\dfrac15\begin{bmatrix}5\\10\end{bmatrix}=\begin{bmatrix}1\\2\end{bmatrix}.$ So $x=1,\ y=2$.
$|A|=1\cdot4-2\cdot2=0$, so $A$ is singular and has no inverse.
If $|A|\ne0$ the system $AX=B$ has the unique solution $X=A^{-1}B$ — it is consistent and independent.
- $A^{-1}=\dfrac{1}{|A|}\operatorname{adj}A$, valid only when $|A|\ne0$ (non-singular).
- Write a linear system as $AX=B$; if $|A|\ne0$ then $X=A^{-1}B$ (unique solution).
- $|A|=0$ means singular: no inverse, and the system is either inconsistent or has infinitely many solutions.
- Always evaluate $|A|$ before attempting an inverse.