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Year 11 Specialist (Unit 1 & 2) Matrices

Identities, inverses and determinants for 2 x 2 matrices

20 practice questions 0 video lessons Theory + worked examples

Work with the identity, determinant and inverse of 2×2 matrices in Year 11 Specialist Mathematics for Queensland (QCAA). The identity matrix acts like the number one, while the determinant decides whether a matrix can be inverted at all.

You will learn to calculate determinants, find inverses by swapping and negating entries, recognise singular matrices with no inverse, and use key properties of matrix algebra — the algebraic foundation for solving matrix equations and transformations later in the course.

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Theory

The determinant and inverse of a 2×2 matrix are core skills in Year 11 Specialist Mathematics (QCAA, Queensland). The determinant \(\det(A)=ad-bc\) decides whether a matrix can be inverted, and the identity matrix \(I\) plays the role of the number \(1\) for matrices. This page shows how to find each, with full worked examples.

The multiplicative identity is the matrix \(I=\begin{pmatrix}1&0\\0&1\end{pmatrix}\). It behaves like the number \(1\): for every \(2\times2\) matrix \(A\), \(AI=IA=A\).

The determinant of \(A=\begin{pmatrix}a&b\\c&d\end{pmatrix}\) is the single number \(\det(A)=ad-bc\) — the product of the main diagonal minus the product of the off-diagonal.

The multiplicative inverse \(A^{-1}\) is the matrix with \(AA^{-1}=A^{-1}A=I\). For a \(2\times2\) matrix it is found by swapping the main diagonal, negating the off-diagonal, and dividing by the determinant.

A matrix with \(\det(A)=0\) is called singular — you cannot divide by zero, so it has no inverse. A matrix with \(\det(A)\ne0\) is non-singular (invertible).

Finding the inverse of a 2 by 2 matrix The matrix 3 4 over 1 2 becomes 2 minus 4 over minus 1 3 by swapping the main diagonal and negating the off-diagonal, then dividing every entry by the determinant 2. 3412 swap \ , negate / 2−4−13 then divide every entry by det = 2 so the inverse is one-half of the matrix on the right
Finding \(A^{-1}\): swap the main diagonal, negate the off-diagonal, then divide by \(\det(A)\).
Non-singular versus singular matrices The matrix 4 3 over 1 1 has determinant 1, which is not zero, so an inverse exists. The matrix 2 6 over 1 3 has determinant 0, so it is singular and has no inverse. 4311 det = 4−3 = 1 inverse exists 2613 det = 6−6 = 0 singular: no inverse
An inverse exists only when \(\det(A)\ne0\); if \(\det(A)=0\) the matrix is singular.

The \(2\times2\) identity and determinant:

\[ I=\begin{pmatrix}1&0\\0&1\end{pmatrix}, \qquad \det(A)=ad-bc \]
det(A)=adbc

The inverse of \(A=\begin{pmatrix}a&b\\c&d\end{pmatrix}\), provided \(\det(A)\ne0\):

\[ A^{-1}=\dfrac{1}{ad-bc}\begin{pmatrix}d&-b\\-c&a\end{pmatrix} \]
A1=1adbc

Useful properties of \(2\times2\) matrices:

\[ \det(AB)=\det(A)\,\det(B), \qquad \det(A^{-1})=\dfrac{1}{\det(A)}, \qquad (A^{-1})^{-1}=A \]
The determinant is the gatekeeper. A \(2\times2\) matrix has an inverse if and only if \(\det(A)\ne0\). When \(\det(A)=0\) the matrix is singular and \(A^{-1}\) does not exist.

How to find the inverse of a 2×2 matrix

  1. Compute the determinant: \(\det(A)=ad-bc\), the main-diagonal product minus the off-diagonal product.
  2. Check it is non-zero: if \(\det(A)=0\) the matrix is singular and has no inverse — stop here.
  3. Form the adjugate: swap the two main-diagonal entries \(a\) and \(d\), and negate the off-diagonal entries \(b\) and \(c\).
  4. Divide by the determinant: multiply the adjugate by \(\dfrac{1}{\det(A)}\). Verify by checking \(AA^{-1}=I\).
Example 1 — Determinant with a negative entry
Find the determinant of \(A=\begin{pmatrix}6 & -3\\ 2 & 4\end{pmatrix}\).
Solution

Apply \(ad-bc\), taking care with the sign of the negative entry:

\(\det(A)\)\(=\)\(ad-bc\)
\(=\)\((6)(4)-(-3)(2)\)
\(=\)\(24-(-6)\)
\(=\)\(24+6\)
\(=\)\(30\)

\(\det(A)=30\).

Example 2 — Inverse of a 2×2 matrix
Find the inverse of \(B=\begin{pmatrix}3 & 4\\ 1 & 2\end{pmatrix}\).
Solution

First the determinant:

\(\det(B)\)\(=\)\((3)(2)-(4)(1)\)
\(=\)\(6-4\)
\(=\)\(2\)

Swap the main diagonal, negate the off-diagonal, then divide by the determinant:

\(B^{-1}\)\(=\)\(\dfrac{1}{2}\begin{pmatrix}2 & -4\\ -1 & 3\end{pmatrix}\)

\(B^{-1}=\dfrac{1}{2}\begin{pmatrix}2 & -4\\ -1 & 3\end{pmatrix}\).

Example 3 — Singular: solve for the unknown
For what value of \(x\) is \(\begin{pmatrix}x & 4\\ 3 & 6\end{pmatrix}\) singular (has no inverse)?
Solution

Singular means the determinant is zero; write \(ad-bc=0\) and solve:

\(\det\)\(=\)\((x)(6)-(4)(3)\)
\(6x-12\)\(=\)\(0\)
\(6x\)\(=\)\(12\)
\(x\)\(=\)\(2\)

\(x=2\).

Example 4 — Find and verify an inverse
Find \(A^{-1}\) for \(A=\begin{pmatrix}3 & 1\\ 5 & 2\end{pmatrix}\), then verify \(AA^{-1}=I\).
Solution

Determinant, then the adjugate (here \(\det=1\), so no fraction is needed):

\(\det(A)\)\(=\)\((3)(2)-(1)(5)\)
\(=\)\(6-5\)
\(=\)\(1\)
\(A^{-1}\)\(=\)\(\begin{pmatrix}2 & -1\\ -5 & 3\end{pmatrix}\)

Check the product equals the identity:

\(AA^{-1}\)\(=\)\(\begin{pmatrix}3 & 1\\ 5 & 2\end{pmatrix}\begin{pmatrix}2 & -1\\ -5 & 3\end{pmatrix}\)
\(=\)\(\begin{pmatrix}1 & 0\\ 0 & 1\end{pmatrix}\)

\(A^{-1}=\begin{pmatrix}2 & -1\\ -5 & 3\end{pmatrix}\), and \(AA^{-1}=I\) confirms it.

Common pitfalls

Forgetting to divide by the determinant. The adjugate \(\begin{pmatrix}d&-b\\-c&a\end{pmatrix}\) is only the inverse when \(\det(A)=1\). In every other case you must multiply it by \(\dfrac{1}{\det(A)}\).
Mishandling a negative entry in \(ad-bc\). Subtracting a negative product adds: \((6)(4)-(-3)(2)=24-(-6)=30\), not \(18\).
Swapping the wrong entries. Swap the main diagonal \(a\) and \(d\); negate the off-diagonal \(b\) and \(c\). Do not negate \(a\) and \(d\).
Trying to invert a singular matrix. If \(\det(A)=0\) there is no inverse — you would be dividing by zero. Always check the determinant first.

Frequently asked questions

How do you find the determinant of a 2x2 matrix?

For \(A=\begin{pmatrix}a&b\\c&d\end{pmatrix}\), the determinant is \(\det(A)=ad-bc\): multiply the main diagonal, multiply the off-diagonal, then subtract.

What is the inverse of a 2x2 matrix?

Swap the main-diagonal entries, negate the off-diagonal entries, and divide by the determinant: \(A^{-1}=\dfrac{1}{ad-bc}\begin{pmatrix}d&-b\\-c&a\end{pmatrix}\).

When does a 2x2 matrix have no inverse?

A matrix has no inverse exactly when its determinant is zero. Such a matrix is called singular, because you cannot divide by zero.

What is the 2x2 identity matrix?

It is \(I=\begin{pmatrix}1&0\\0&1\end{pmatrix}\), with ones on the main diagonal and zeros elsewhere. For every \(2\times2\) matrix \(A\), \(AI=IA=A\).

Is det(AB) equal to det(A) times det(B)?

Yes. For any two \(2\times2\) matrices, \(\det(AB)=\det(A)\det(B)\). Note that \(\det(A+B)\) is generally not \(\det(A)+\det(B)\).

How can I check my inverse is correct?

Multiply the matrix by your answer. If \(AA^{-1}\) equals the identity \(\begin{pmatrix}1&0\\0&1\end{pmatrix}\), the inverse is correct.