Inverses and determinants for n×n matrices
Master inverses and determinants for \(n\times n\) matrices for Year 12 Specialist Mathematics in Queensland (QCAA). The determinant is one number that decides whether a square matrix has an inverse, and it is found for a \(3\times3\) matrix by cofactor expansion.
You will learn to evaluate \(3\times3\) and larger determinants, use \(\det A=0\) to test whether a matrix is singular, find the multiplicative inverse \(A^{-1}\) with technology, and apply the properties \(\det(AB)=\det A\,\det B\) and \(\det(A^{-1})=\tfrac{1}{\det A}\) — the algebra behind solving matrix equations and systems later in the course.
Theory
The determinant of a square matrix is a single number that decides whether the matrix has an inverse. In Year 12 Specialist Mathematics (QCAA, Queensland) you evaluate a determinant of a \(3\times3\) matrix by cofactor expansion (and larger orders with technology), find the multiplicative inverse \(A^{-1}\), and use \(\det A=0\) to test whether a matrix is singular.
The determinant \(\det A\) (also written \(|A|\)) is a single number attached to any square matrix. For a \(2\times2\) matrix \(\det\begin{pmatrix}a&b\\c&d\end{pmatrix}=ad-bc\). A \(3\times3\) (or larger) determinant is built from smaller ones by cofactor expansion.
The minor \(M_{ij}\) of an entry is the determinant left after deleting that entry's row and column. Its cofactor is \(C_{ij}=(-1)^{i+j}M_{ij}\), which just attaches the sign from the checkerboard pattern \(\begin{smallmatrix}+&-&+\\-&+&-\\+&-&+\end{smallmatrix}\). Expanding along any row or column, \(\det A\) is the sum of each entry times its cofactor.
A square matrix is singular when \(\det A=0\) and non-singular (invertible) when \(\det A\neq0\). The multiplicative inverse \(A^{-1}\) is the matrix with \(AA^{-1}=A^{-1}A=I\), where \(I\) is the identity. It exists exactly when \(A\) is non-singular.
The inverse is \(A^{-1}=\dfrac{1}{\det A}\,\operatorname{adj}A\), where \(\operatorname{adj}A\) is the transpose of the matrix of cofactors. For \(3\times3\) and larger orders the syllabus expects you to compute \(\det A\) and \(A^{-1}\) with technology. Useful properties for any \(n\times n\) matrices are \(\det(AB)=\det A\,\det B\), \(\det(A^{-1})=\dfrac{1}{\det A}\) and \((A^{-1})^{-1}=A\).
Expanding a \(3\times3\) determinant along the first row (entries \(a,b,c\)) with the sign pattern \(+,-,+\):
The inverse of a non-singular matrix is the reciprocal of the determinant times the adjugate (matrix of cofactors, transposed):
Determinant properties that hold for square matrices of any order:
How to find a determinant and an inverse
- Choose a row or column to expand along — one with a zero entry keeps the arithmetic short, because that entry's term vanishes.
- Form each minor by deleting the entry's row and column, then evaluate the \(2\times2\) determinant \(ad-bc\), attaching the sign from the pattern \(+,-,+\).
- Add the signed terms to get \(\det A\). If \(\det A=0\) the matrix is singular and has no inverse; otherwise it is non-singular.
- Find the inverse as \(A^{-1}=\dfrac{1}{\det A}\operatorname{adj}A\) (use technology for \(3\times3\) and larger), and check by confirming \(AA^{-1}=I\).
Expand along the first row; the third entry is \(0\), so that term drops out:
| \(\det A\) | \(=\) | \(3\begin{vmatrix}4&1\\0&2\end{vmatrix}-1\begin{vmatrix}2&1\\1&2\end{vmatrix}+0\) |
| \(=\) | \(3(4\cdot2-1\cdot0)-1(2\cdot2-1\cdot1)\) | |
| \(=\) | \(3(8)-1(3)\) | |
| \(=\) | \(24-3\) | |
| \(=\) | \(21\) |
\(\det A=21\), so \(A\) is non-singular.
Singular means \(\det=0\); expand along the first row, then solve:
| \(\det\) | \(=\) | \(x\begin{vmatrix}1&0\\2&1\end{vmatrix}-1\begin{vmatrix}1&0\\0&1\end{vmatrix}+2\begin{vmatrix}1&1\\0&2\end{vmatrix}\) |
| \(=\) | \(x(1)-1(1)+2(2)\) | |
| \(=\) | \(x+3\) | |
| \(x+3\) | \(=\) | \(0\) |
| \(x\) | \(=\) | \(-3\) |
\(x=-3\).
First the determinant (expand along the first row):
| \(\det A\) | \(=\) | \(2\begin{vmatrix}1&1\\0&2\end{vmatrix}-1\begin{vmatrix}0&1\\1&2\end{vmatrix}+0\) |
| \(=\) | \(2(2)-1(-1)\) | |
| \(=\) | \(5\) |
The inverse is \(\dfrac{1}{\det A}\) times the adjugate (from technology):
| \(A^{-1}\) | \(=\) | \(\dfrac{1}{5}\begin{pmatrix}2&-2&1\\1&4&-2\\-1&1&2\end{pmatrix}\) |
Check by multiplying — the product is the identity:
| \(AA^{-1}\) | \(=\) | \(\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}=I\) |
\(A^{-1}=\dfrac{1}{5}\begin{pmatrix}2&-2&1\\1&4&-2\\-1&1&2\end{pmatrix}\).
Use the product property for \(\det(AB)\):
| \(\det(AB)\) | \(=\) | \(\det A\,\det B\) |
| \(=\) | \((4)(-3)\) | |
| \(=\) | \(-12\) |
For \(\det(A^{-1})\), take reciprocals (since \(\det A\,\det(A^{-1})=\det I=1\)):
| \(\det(A^{-1})\) | \(=\) | \(\dfrac{1}{\det A}\) |
| \(=\) | \(\dfrac{1}{4}\) |
\(\det(AB)=-12\) and \(\det(A^{-1})=\dfrac{1}{4}\).
Common pitfalls
Frequently asked questions
How do you find the determinant of a 3x3 matrix?
Expand along a row or column: multiply each entry by the \(2\times2\) minor left after deleting its row and column, attach the signs \(+,-,+\) along that row, and add the results. Choosing a row or column with a zero shortens the work.
What does it mean for a matrix to be singular?
A square matrix is singular when its determinant is zero, \(\det A=0\). A singular matrix has no inverse. When \(\det A\neq0\) the matrix is non-singular (invertible).
What is the formula for the inverse of a matrix?
For a non-singular matrix, \(A^{-1}=\dfrac{1}{\det A}\operatorname{adj}A\), where \(\operatorname{adj}A\) is the transpose of the matrix of cofactors. It satisfies \(AA^{-1}=A^{-1}A=I\).
How do you find the determinant and inverse of larger matrices?
For \(3\times3\) and larger the QCAA syllabus expects you to use technology (a calculator or CAS): enter the matrix and read off \(\det A\) and \(A^{-1}\). The cofactor method still explains why an inverse fails to exist when \(\det A=0\).
Is det(AB) equal to det(A) times det(B)?
Yes. For square matrices of the same order \(\det(AB)=\det A\,\det B\). It follows that \(\det(A^{-1})=\dfrac{1}{\det A}\). However \(\det(A+B)\) is not \(\det A+\det B\) in general.
How do you check an inverse is correct?
Multiply the matrix by its claimed inverse: if \(AA^{-1}=I\) (the identity matrix, ones on the diagonal and zeros elsewhere) then the inverse is correct.