The 2x2 Determinant

Every square matrix has a single number attached to it called the determinant. That number answers two questions at once: is the matrix invertible? and how much does the matrix stretch area? The formula for the $2 \times 2$ case is short enough to memorise; the picture behind it is short enough to draw on a napkin. Together they unlock the rest of the chapter.

The formula

For a $2 \times 2$ matrix,

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

Cross-multiply along the main diagonal (top-left to bottom-right), subtract the cross-multiply along the off-diagonal. Either notation works: $\det(A)$ or the vertical-bar shorthand \begin{vmatrix} a & b \\ c & d \end{vmatrix}.

The geometric meaning

Take the two columns of $A$ as vectors in the plane: $\mathbf{u} = (a, c)$ and $\mathbf{v} = (b, d)$. They span a parallelogram. The signed area of that parallelogram equals $\det(A)$. Absolute value gives the area; the sign tells you orientation, positive means $\mathbf{u}$ rotates counterclockwise to reach $\mathbf{v}$, negative means clockwise.

• Computer Graphics: The signed area of a parallelogram is \det\begin{pmatrix} a & c \\ b & d \end{pmatrix}; this is how a polygon-fill algorithm tests whether a triangle is wound clockwise or counter-clockwise.
• Physics: The Jacobian determinant $\partial(x,y)/\partial(u,v)$ measures how a change-of-coordinates scales areas; this is what makes polar-coordinate integration $r\,dr\,d\theta$ give the right answer.
• Computer Vision: Detecting features (Harris corner detector) uses the determinant of the structure tensor; a vanishing determinant signals a flat patch with no detectable corner.

(Drag the green $\mathbf{u}$ or orange $\mathbf{v}$. The shaded parallelogram is what the unit square [0,1] \times [0,1] becomes after the linear map $A$. When $\mathbf{u}$ and $\mathbf{v}$ are nearly parallel the area collapses toward zero; when they cross over each other the centroid label changes to (flipped), that is the sign of the determinant changing.)

The invertibility test

The determinant is zero exactly when the two columns are parallel, when $\mathbf{u}$ and $\mathbf{v}$ lie along the same line. In that case the matrix squashes the whole plane onto a single line (or a single point), and there is no way to undo the squashing: the matrix is singular. Otherwise the matrix is invertible, and the inverse has a tidy closed form:

A^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}.

Swap the diagonal entries, negate the off-diagonal entries, divide through by the determinant. Verify by multiplying: $A A^{-1} = I$ falls out.

Two useful identities

Without proof at this stage, two identities you will use again and again:

• Product rule: $\det(AB) = \det(A) \det(B)$.
• Transpose rule: $\det(A^T) = \det(A)$.

The product rule is striking. Matrices generally do not commute, yet their determinants do, because the determinant is a single number, and numbers always commute. Geometrically: if $A$ scales area by $\det(A)$ and $B$ scales by $\det(B)$, then applying them in either order scales by the product.

Try it

• Predict first: for $\mathbf{u} = (3, 1)$ and $\mathbf{v} = (1, 4)$, what is \det\begin{pmatrix} 3 & 1 \\ 1 & 4 \end{pmatrix}? Drag the columns to set those values and verify the readout shows $\det = 11$ with parallelogram area $= 11$.
• Predict first: when $\mathbf{v}$ is a scalar multiple of $\mathbf{u}$, what should $\det$ equal, and what does that say about the matrix? Slide $\mathbf{v}$ until it lines up with $\mathbf{u}$ and verify the parallelogram collapses and $\det = 0$ (the matrix is singular).
• Drag the columns to a configuration where $\det = 6$ exactly (e.g., $\mathbf{u} = (3, 0)$, $\mathbf{v} = (0, 2)$).
• By hand, find the inverse of \begin{pmatrix} 2 & 1 \\ 5 & 3 \end{pmatrix}. Check $A^{-1} A = I$.

Pause: if $\det(A) = 0$, the matrix sends the unit square to something with zero area. Geometrically, what does that mean for the equation $A\mathbf{x} = \mathbf{b}$, can you always recover $\mathbf{x}$ uniquely?

Try it in code

A trap to watch for

The product rule $\det(AB) = \det(A)\det(B)$ tempts beginners to write $\det(A + B) = \det(A) + \det(B)$. Wrong. Test with $A = B = I_2$: $\det(A + B) = \det(2I_2) = 4$, but $\det(A) + \det(B) = 1 + 1 = 2$. The determinant is multiplicative under matrix multiplication, not under addition. There is no useful linear identity for sums of matrices.