The inverse of a 2 × 2 matrix
A 2 × 2 inverse is defined only when the determinant is non-zero. After finding the determinant, swap the leading-diagonal entries, negate the other two entries and divide every entry by the determinant
- Check the determinant If the determinant is zero, no inverse exists \(\displaystyle \det(A)=2\times3-1\times4=2\ne0\)
- Swap the leading diagonal Exchange the top-left and bottom-right entries \(\displaystyle \begin{bmatrix}2&1\\4&3\end{bmatrix}\mapsto\begin{bmatrix}3&1\\4&2\end{bmatrix}\)
- Negate the other entries Change the signs of the top-right and bottom-left entries in their existing positions \(\displaystyle \begin{bmatrix}3&1\\4&2\end{bmatrix}\mapsto\begin{bmatrix}3&-1\\-4&2\end{bmatrix}\)
- Divide every entry Multiply the rearranged matrix by the reciprocal of the determinant \(\displaystyle A^{-1}=\frac{1}{2}\begin{bmatrix}3&-1\\-4&2\end{bmatrix}=\begin{bmatrix}\frac{3}{2}&-\frac{1}{2}\\-2&1\end{bmatrix}\)
Worked example
Find the inverse of the matrix B and give the entries as exact fractions
- \(\displaystyle \det(B)=4\times2-1\times3=8-3=5\)The determinant is 5, so the inverse exists
- \(\displaystyle \begin{bmatrix}4&1\\3&2\end{bmatrix}\mapsto\begin{bmatrix}2&-1\\-3&4\end{bmatrix}\)Swap the leading-diagonal entries and negate the other two entries
- \(\displaystyle B^{-1}=\frac{1}{5}\begin{bmatrix}2&-1\\-3&4\end{bmatrix}=\begin{bmatrix}\frac{2}{5}&-\frac{1}{5}\\-\frac{3}{5}&\frac{4}{5}\end{bmatrix}\)Divide every rearranged entry by 5 and leave the fractions in lowest terms
Common mistakes
Difficulty levels
Questions about this topic
Why must the determinant be non-zero?+
The inverse rule divides by the determinant. Division by zero is undefined, so a zero determinant means that no inverse exists
Which entries are swapped in the inverse rule?+
The top-left and bottom-right entries are swapped. The top-right and bottom-left entries stay in their positions and have their signs changed
How can an inverse be checked?+
Multiply the matrix by its proposed inverse. A correct inverse gives the identity matrix, with 1s on the leading diagonal and 0s elsewhere
Can I use a calculator?+
No. This topic is practised without a calculator
Which year level and chapter is this?+
Year 11, VCE Units 1 & 2, General Mathematics, chapter 4: Matrices