Algebra

Matrices and determinants

Rectangular arrays of numbers and the scalar that measures how they scale volume.

IntuitionIntuition: a matrix as a machine that reshapes space

A matrix is a rectangular table of numbers, but it is most useful as a rule for turning one vector into another by a fixed linear recipe: stretch here, rotate there, maybe flip. The determinant is a single number attached to a square matrix that answers one very concrete question: by what factor does this matrix scale areas (in the plane) or volumes (in space), and does it flip orientation (left-handed becomes right-handed) along the way? A determinant of $0$ means the matrix squashes space flat, losing a dimension entirely — that single fact turns out to control almost everything else about the matrix, including whether it can be undone.

How the

2\times 2

matrix

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

deforms the plane: the unit square becomes a parallelogram of signed area

\det A = a_{11}a_{22} - a_{12}a_{21}

SchoolMatrix multiplication and the determinant formula

Definition: Matrix multiplication

If $A$ is $m\times n$ and $B$ is $n\times p$ , their product $C=AB$ is the $m\times p$ matrix whose entries are $C_{ij} = \sum_{k=1}^{n} A_{ik}B_{kj}$ : entry $(i,j)$ of $C$ is the dot product of row $i$ of $A$ with column $j$ of $B$ .

C_{ij} = \sum_{k=1}^{n} A_{ik}B_{kj}

Here $A_{ik}$ is the entry in row $i$ , column $k$ of $A$ , and $B_{kj}$ is the entry in row $k$ , column $j$ of $B$ ; the sum runs over the shared inner index $k$ , which is why $A$ 's column count must equal $B$ 's row count for the product to be defined at all.

Definition: Determinant (Leibniz formula)

For a square $n\times n$ matrix $A$ , the determinant is defined by $\det(A)=\sum_{\sigma\in S_n}\operatorname{sgn}(\sigma)\prod_{i=1}^n a_{i,\sigma(i)}$ , summing over every permutation $\sigma$ of $\{1,\dots,n\}$ , with $\operatorname{sgn}(\sigma)=+1$ or $-1$ depending on whether $\sigma$ is even or odd. For $n=2,3$ this expands into the familiar cofactor formulas below.

\det\begin{pmatrix}a&b\\ c&d\end{pmatrix}=ad-bc

For a $2\times 2$ matrix, $\det\begin{pmatrix}a&b\\ c&d\end{pmatrix}=ad-bc$ is exactly the signed area of the parallelogram spanned by the row vectors $(a,b)$ and $(c,d)$ : the two permutations of $\{1,2\}$ are the identity (contributing $+ad$ ) and the swap (contributing $-bc$ ).

\det\begin{pmatrix}a_{11}&a_{12}&a_{13}\\ a_{21}&a_{22}&a_{23}\\ a_{31}&a_{32}&a_{33}\end{pmatrix}=a_{11}(a_{22}a_{33}-a_{23}a_{32})-a_{12}(a_{21}a_{33}-a_{23}a_{31})+a_{13}(a_{21}a_{32}-a_{22}a_{31})

For a $3\times 3$ matrix, $\det\begin{pmatrix}a_{11}&a_{12}&a_{13}\\ a_{21}&a_{22}&a_{23}\\ a_{31}&a_{32}&a_{33}\end{pmatrix}=a_{11}(a_{22}a_{33}-a_{23}a_{32})-a_{12}(a_{21}a_{33}-a_{23}a_{31})+a_{13}(a_{21}a_{32}-a_{22}a_{31})$ is the cofactor expansion along the first row: expand along row $1$ , multiply each entry $a_{1j}$ by the $2\times 2$ determinant of the matrix left after deleting row $1$ and column $j$ , and alternate the sign $+,-,+$ across $j=1,2,3$ .

How elementary row operations affect the determinant
Row operation	Effect on $\det$
Swap two rows	multiplies $\det$ by $-1$
Scale a row by $k$	multiplies $\det$ by $k$
Add $k$ times a row to another	$\det$ unchanged
Two rows equal or proportional	$\det=0$

UndergraduateTheorems: multiplicativity and the invertibility criterion

Theorem: Multiplicativity of the determinant

For any two $n\times n$ matrices $A,B$ , $\det(AB)=\det(A)\det(B)$ .

Why is it true?

Since the determinant measures how much a linear map scales volume, applying map $B$ first and then map $A$ should scale volume by $B$ 's factor and then by $A$ 's factor, i.e. by the product of the two factors — multiplicativity is exactly the algebraic statement that composing linear maps composes their volume-scaling factors.

Proof

First recall the effect of the three elementary row operations on the determinant (established directly from the Leibniz sum $\det(A)=\sum_{\sigma\in S_n}\operatorname{sgn}(\sigma)\prod_{i=1}^n a_{i,\sigma(i)}$ , since each term of that sum is linear in each row separately): swapping two rows multiplies $\det$ by $-1$ , scaling one row by $k$ multiplies $\det$ by $k$ , and adding a multiple of one row to another leaves $\det$ unchanged.

Each elementary row operation on $A$ is the same as left-multiplying $A$ by a corresponding elementary matrix $E$ (obtained by applying that same operation to the identity matrix $I$ ). Comparing with the previous paragraph, $\det(E)$ equals exactly the scaling factor of that operation ( $-1$ , $k$ , or $1$ respectively), so for every elementary matrix $E$ , $\det(EA)=\det(E)\det(A)$ .

Case $A$ invertible: Gaussian elimination reduces any invertible matrix to the identity using a finite sequence of elementary row operations, i.e. $E_m\cdots E_1A=I$ for elementary matrices $E_1,\dots,E_m$ , so $A=E_1^{-1}\cdots E_m^{-1}$ , itself a product of elementary matrices (the inverse of an elementary matrix is again elementary, of the same type). Applying the previous paragraph's identity repeatedly, $\det(A)=\det(E_1^{-1})\cdots\det(E_m^{-1})$ , and applying it again to $AB=E_1^{-1}\cdots E_m^{-1}B$ gives $\det(AB)=\det(E_1^{-1})\cdots\det(E_m^{-1})\det(B)=\det(A)\det(B)$ .

Case $A$ singular: then $\det(A)=0$ (a singular matrix cannot be reduced all the way to the identity, and the row-operation rules above show every reachable row-echelon form still has a zero row, forcing $\det=0$ by cofactor expansion along that row). Singularity also means $\operatorname{rank}(A)<n$ , and since $\operatorname{rank}(AB)\leq\operatorname{rank}(A)<n$ for any $B$ , the product $AB$ is singular too, so $\det(AB)=0$ as well. Hence $\det(AB)=0=0\cdot\det(B)=\det(A)\det(B)$ , and the identity holds in this case too, completing the proof for every $A$ .

Theorem: Invertibility criterion via the determinant

A square matrix $A$ is invertible if and only if $\det(A)\neq 0$ ; when it is invertible, $A^{-1}=\dfrac{1}{\det(A)}\operatorname{adj}(A)$ , where $\operatorname{adj}(A)$ is the adjugate (transpose of the cofactor matrix).

Why is it true?

A linear map can be undone exactly when it does not collapse space to a lower dimension — since the determinant already measures exactly this "does it collapse volume to zero" question, invertibility and $\det\neq 0$ are really the same statement seen from two angles: one algebraic (does an inverse matrix exist), one geometric (is the volume-scaling factor nonzero).

Proof

( $\Rightarrow$ ) Suppose $A$ is invertible, so there is a matrix $A^{-1}$ with $AA^{-1}=I$ . Applying the multiplicativity theorem proved above, $\det(A)\det(A^{-1})=\det(AA^{-1})=\det(I)=1$ . Since the product of the two numbers $\det(A)$ and $\det(A^{-1})$ equals $1$ , neither of them can be $0$ — so $\det(A)\neq 0$ .

( $\Leftarrow$ ), worked concretely for $2\times 2$ : suppose $A=\begin{pmatrix}a&b\\ c&d\end{pmatrix}$ has $\det(A)=ad-bc\neq 0$ . Define $\operatorname{adj}(A)=\begin{pmatrix}d&-b\\ -c&a\end{pmatrix}$ . Multiplying directly, $A\cdot\operatorname{adj}(A)=\begin{pmatrix}a&b\\ c&d\end{pmatrix}\begin{pmatrix}d&-b\\ -c&a\end{pmatrix}=\begin{pmatrix}ad-bc&-ab+ab\\ cd-cd&-cb+ad\end{pmatrix}=\begin{pmatrix}ad-bc&0\\ 0&ad-bc\end{pmatrix}=\det(A)\,I$ , exactly the identity $A\cdot\operatorname{adj}(A)=\det(A)\,I$ . The same computation in the other order gives $\operatorname{adj}(A)\cdot A=\det(A)\,I$ as well.

Since $\det(A)\neq 0$ , divide both sides of $A\cdot\operatorname{adj}(A)=\det(A)I$ by the scalar $\det(A)$ to get $A\cdot\dfrac{\operatorname{adj}(A)}{\det(A)}=I$ , and likewise from the other order $\dfrac{\operatorname{adj}(A)}{\det(A)}\cdot A=I$ . This exhibits an explicit two-sided inverse $A^{-1}=\dfrac{1}{\det(A)}\operatorname{adj}(A)$ , so $A$ is invertible.

For general $n$ (illustrated at $n=3$ ), the same argument works because the Laplace cofactor expansion gives $\sum_j a_{ij}C_{ij}=\det(A)$ (expanding along row $i$ ) while $\sum_j a_{ij}C_{kj}=0$ for $i\neq k$ (this sum is the cofactor expansion of a matrix with row $i$ copied into row $k$ , which has two equal rows and hence determinant $0$ by the table above). Assembling these row-by-row identities into matrix form gives exactly $A\cdot\operatorname{adj}(A)=\det(A)I$ for every size $n$ , so whenever $\det(A)\neq 0$ the same division argument as in the $2\times 2$ case produces $A^{-1}=\operatorname{adj}(A)/\det(A)$ ; conversely, if $\det(A)=0$ , the first paragraph's contrapositive rules out invertibility entirely, since $\det(A)\det(A^{-1})=1$ would force $\det(A)\neq 0$ . This proves the criterion in both directions.

UndergraduateReal-World Applications and Worked Examples

Determinants let engineers solve small linear systems by an explicit formula (Cramer's rule) instead of elimination, and let computer graphics programs check in one number whether a transformation preserves or flips orientation and by how much it scales area — both are direct uses of $\det\neq 0$ meaning "invertible" and $|\det|$ meaning "scale factor."

Example: Engineering: solving a two-loop circuit with Cramer's rule

Kirchhoff's laws applied to a two-loop circuit give the linear system $3x+y=11$ and $x+2y=8$ for the two loop currents $x,y$ (in amperes). Solve for $x$ and $y$ using determinants.

Solution

Write the system as a matrix equation $\begin{pmatrix}3&1\\1&2\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}11\\8\end{pmatrix}$ . The coefficient matrix has $\det=3\times2-1\times1=6-1=5\neq 0$ , so by the invertibility criterion the system has a unique solution, and Cramer's rule applies.

Cramer's rule says $x=\dfrac{\det\begin{pmatrix}11&1\\8&2\end{pmatrix}}{\det\begin{pmatrix}3&1\\1&2\end{pmatrix}}$ : replace the first column of the coefficient matrix by the right-hand side, take its determinant, and divide by the original determinant. The numerator is $11\times2-1\times8=22-8=14$ , so $x=14/5=2.8$ A.

Similarly $y=\dfrac{\det\begin{pmatrix}3&11\\1&8\end{pmatrix}}{5}$ : replace the second column instead. The numerator is $3\times8-11\times1=24-11=13$ , so $y=13/5=2.6$ A.

Checking: $3(2.8)+2.6=8.4+2.6=11$ ✓ and $2.8+2(2.6)=2.8+5.2=8$ ✓, confirming $x=2.8$ A and $y=2.6$ A solve the circuit equations exactly.

Example: Computer graphics: does a transformation flip a shape, and by how much does it scale area?

A 2D graphics engine applies the transformation matrix $M=\begin{pmatrix}2&1\\4&3\end{pmatrix}$ to every point of a shape of area $5$ square units. Determine whether the transformed shape is mirror-flipped compared to the original, and compute the area of the transformed shape.

Solution

Compute the determinant of $M$ : $\det(M)=2\times3-1\times4=6-4=2$ .

The sign of $\det(M)$ tells us about orientation: since $\det(M)=2>0$ , the transformation preserves orientation, so the shape is not mirror-flipped (a negative determinant would indicate a flip, since it means an odd number of reflections are baked into the matrix).

The magnitude $|\det(M)|$ tells us the area-scaling factor: every region's area gets multiplied by $|\det(M)|=2$ after the transformation, regardless of the region's shape, because this is exactly the geometric meaning of the determinant established earlier.

So the transformed shape has area $5\times|\det(M)|=5\times2=10$ square units, and it keeps the same orientation (not flipped) as the original shape.

Common mistake. Common mistakes: (1) assuming

\det(A+B)=\det(A)+\det(B)

— this is false in general; the determinant is multiplicative, not additive. (2) Forgetting the alternating checkerboard sign pattern when doing cofactor expansion, e.g. writing

+a_{12}(\cdots)

instead of

-a_{12}(\cdots)

in the

3\times 3

formula

\det\begin{pmatrix}a_{11}&a_{12}&a_{13}\\ a_{21}&a_{22}&a_{23}\\ a_{31}&a_{32}&a_{33}\end{pmatrix}=a_{11}(a_{22}a_{33}-a_{23}a_{32})-a_{12}(a_{21}a_{33}-a_{23}a_{31})+a_{13}(a_{21}a_{32}-a_{22}a_{31})

. (3) Concluding a matrix is "probably invertible" because none of its entries are

0

— invertibility depends only on

\det(A)\neq 0

, and a matrix full of nonzero entries can easily have rows that are multiples of each other, giving

\det(A)=0

What is $\det\begin{pmatrix}2&5\\1&3\end{pmatrix}$ ?

If $\det(A)=3$ and $\det(B)=4$ for two $n\times n$ matrices, what is $\det(AB)$ ?

For which matrix does an inverse fail to exist?

A graphics transformation matrix has $\det=-3$ . What does this tell us?

References

Eric W. Weisstein (MathWorld) (2024). Determinant
Gilbert Strang (2016). Introduction to Linear Algebra