Singular and non-singular matrix

What it is. Every square matrix has a single number attached to it called the determinant, written $\det\mathbf{A}$ or $|\mathbf{A}|$. It is the gateway to the inverse, to solving systems, and to the geometry of the transformation.

The $2\times2$ rule (this one IS in MF19, but learn it cold): $\det\begin{pmatrix} a & b \\ c & d \end{pmatrix} = ad - bc$ Multiply down the main diagonal ($a\times d$), then subtract the product up the anti-diagonal ($b\times c$).

The $3\times3$ rule — cofactor expansion along the top row. Each top-row entry multiplies the $2\times2$ determinant left when you delete that entry's row and column, with signs alternating $+\,-\,+$: $\det\begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} = a\begin{vmatrix} e & f \\ h & i \end{vmatrix} - b\begin{vmatrix} d & f \\ g & i \end{vmatrix} + c\begin{vmatrix} d & e \\ g & h \end{vmatrix}$

You may expand along any row or column (each has its own sign pattern from the checkerboard $\begin{smallmatrix} + & - & + \\ - & + & - \\ + & - & + \end{smallmatrix}$). Choose the row or column with the most zeros to cut the arithmetic.

Shortcut — triangular matrices. If every entry below (or above) the diagonal is zero, the determinant is simply the product of the diagonal entries: $\det\begin{pmatrix} a & * & * \\ 0 & e & * \\ 0 & 0 & i \end{pmatrix} = aei$.

Cambridge tip. The $3\times3$ cofactor formula is not printed in MF19, so the $+\,-\,+$ sign pattern must come from memory. Write those three signs above the top row before you expand — it is the single best guard against a sign slip.

Once you have the determinant, the classification is immediate.

Non-singular ($\det\mathbf{A} \neq 0$):

• the inverse $\mathbf{A}^{-1}$ exists;
• the system $\mathbf{A}\mathbf{x} = \mathbf{b}$ has exactly one solution, $\mathbf{x} = \mathbf{A}^{-1}\mathbf{b}$.

Singular ($\det\mathbf{A} = 0$):

• the matrix has no inverse (you would be dividing by zero in the inverse formula);
• the system $\mathbf{A}\mathbf{x} = \mathbf{b}$ has no unique solution — depending on $\mathbf{b}$, either no solution at all or infinitely many.

A fast singularity check. If one row is a scalar multiple of another (or one is a combination of the others), the determinant is automatically zero. For example, $\begin{pmatrix} 2 & 6 \\ 1 & 3 \end{pmatrix}$ has its top row equal to twice the bottom row, so $\det = 0$ without any arithmetic.

Find-the-unknown questions. A classic Paper 1 task gives a matrix with a letter in it and asks for the value(s) making it singular. The instruction "singular" is a command to set $\det = 0$ and solve the resulting equation — often a quadratic or cubic, so give all the roots.

Cambridge tip. When a question says "show that $\mathbf{A}$ is singular", the expected working is "$\det\mathbf{A} = \ldots = 0$, hence singular" — the word singular on its own earns nothing without the determinant shown.

The determinant is not just an algebraic gadget — it measures how the transformation $\mathbf{x} \mapsto \mathbf{A}\mathbf{x}$ stretches space.

Two dimensions. A $2\times2$ matrix scales every area by the factor $|\det\mathbf{A}|$. The unit square (area $1$) maps to a parallelogram of area $|\det\mathbf{A}|$, so any shape of area $S$ becomes one of area $|\det\mathbf{A}|\times S$.

Three dimensions. A $3\times3$ matrix scales every volume by $|\det\mathbf{A}|$: the unit cube becomes a parallelepiped of volume $|\det\mathbf{A}|$.

Why the modulus? Area and volume are never negative. A negative determinant means the transformation also reverses orientation — it includes a reflection — but the magnitude of the scaling is still $|\det|$.

Why $\det = 0$ kills the inverse. If $\det\mathbf{A} = 0$, the area (or volume) scale factor is $0$: the transformation squashes the plane onto a line, or space onto a plane. A whole dimension is lost, so many different points pile up onto the same image. The mapping is no longer one-to-one, and you cannot reverse it — which is exactly why a singular matrix has no inverse.

Multiplicative property. Determinants multiply: $\det(\mathbf{AB}) = \det\mathbf{A}\,\det\mathbf{B}$. Geometrically this just says "scale by $\det\mathbf{B}$, then by $\det\mathbf{A}$". Two useful corollaries follow: $\det(\mathbf{A}^{-1}) = \dfrac{1}{\det\mathbf{A}}$, and scaling a whole $n\times n$ matrix gives $\det(k\mathbf{A}) = k^n\det\mathbf{A}$ (every one of the $n$ rows is stretched).

Cambridge tip. In a worded transformation part, always answer with $|\det\mathbf{A}|$ for the scale factor and add "the transformation reverses orientation" if the determinant is negative — that extra sentence often carries a mark.

This is the most heavily examined application of the topic, so make the routine automatic.

The method, every time:

1. Compute the determinant with the letter left in (a $2\times2$ gives a quadratic; a $3\times3$ usually gives a cubic).
2. Set it equal to zero — "singular" means $\det = 0$.
3. Solve the equation, and give all the values (factorise the quadratic/cubic; do not stop at one root).

A $2\times2$ example sketch. For $\begin{pmatrix} k & 3 \\ 2 & k-1 \end{pmatrix}$: $\det = k(k-1) - 6 = k^2 - k - 6 = 0 \;\Rightarrow\; (k-3)(k+2) = 0 \;\Rightarrow\; k = 3 \text{ or } k = -2.$

Why two answers? A $2\times2$ determinant is quadratic in the unknown, so there are usually two values; a $3\times3$ is cubic, so up to three. Reporting only one loses the remaining accuracy marks. If a root repeats, say so.

Sanity check. Substitute a found value back: the matrix should now have proportional rows/columns (or, for a $3\times3$, identical or dependent rows). That confirms singularity without re-expanding the whole determinant.

Cambridge tip. Lay the work out as an equation — "$\det = 0$" — not as a stray evaluation. Examiners look for the line where the determinant is explicitly equated to zero before solving; it secures the method mark even if your arithmetic later slips.