3x3 determinants and inverse matricesAQA A-Level Further Maths: Revision notes

The minor of an entry in a $3\times3$ matrix is the $2\times2$ determinant left after deleting that entry's row and column. The cofactor is the minor with a sign from the pattern $\begin{pmatrix}+&-&+\\-&+&-\\+&-&+\end{pmatrix}.$ The determinant is found by expanding along any row or column: multiply each entry by its cofactor and add. Along the first row, $\det\mathbf{A}=a_{11}C_{11}+a_{12}C_{12}+a_{13}C_{13}.$ Example: $\mathbf{A}=\begin{pmatrix}1&2&0\\3&1&1\\0&2&1\end{pmatrix}$ gives $1(1-2)-2(3-0)+0=-1-6=-7$. Choosing a row or column with a zero saves work. A matrix is singular if $\det=0$ and non-singular otherwise. With an unknown present, set the determinant to zero: for $\begin{pmatrix}1&2&3\\4&5&6\\7&8&k\end{pmatrix}$ the determinant is $27-3k$, so it is singular when $k=9$.

For a transformation with matrix $\mathbf{M}$:

• a $2\times2$ matrix multiplies areas by $|\det\mathbf{M}|$
• a $3\times3$ matrix multiplies volumes by $|\det\mathbf{M}|$. Example: $\det\mathbf{A}=-7$, so a solid of volume $4\text{ cm}^3$ is mapped to a solid of volume $4\times7=28\text{ cm}^3$. The scale factor of a volume is never negative. The sign of the determinant describes orientation. A positive determinant preserves orientation (a rotation or an enlargement keeps a shape the same way round). A negative determinant reverses orientation (as a reflection does). Check: reflection in the $y$-axis, $\begin{pmatrix}-1&0\\0&1\end{pmatrix}$, has $\det=-1$: areas are unchanged and orientation is reversed. For successive transformations the scale factors multiply.

For a non-singular $3\times3$ matrix $\mathbf{A}$:

1. Find $\det\mathbf{A}$ (it must be non-zero).
2. Find the matrix of cofactors $\mathbf{C}$ (nine $2\times2$ determinants, each with its sign).
3. Transpose it to get the adjugate, $\text{adj}\,\mathbf{A}=\mathbf{C}^{T}$.
4. Divide: $\mathbf{A}^{-1}=\frac{1}{\det\mathbf{A}}\,\text{adj}\,\mathbf{A}.$ The transpose step means the entry in row $i$, column $j$ of $\mathbf{A}^{-1}$ uses the cofactor of the entry in row $j$, column $i$ of $\mathbf{A}$. Check by confirming $\mathbf{A}\mathbf{A}^{-1}=\mathbf{I}$, or at least one row times one column.

Let $\mathbf{C}=\begin{pmatrix}1&1&1\\1&2&3\\1&4&9\end{pmatrix}$. The determinant is $1(18-12)-1(9-3)+1(4-2)=6-6+2=2$. The matrix of cofactors is $\begin{pmatrix}6&-6&2\\-5&8&-3\\1&-2&1\end{pmatrix}$ (for example the entry in row 2, column 1 is $-(1\times9-1\times4)=-5$). Transposing and dividing by $2$: $\mathbf{C}^{-1}=\frac12\begin{pmatrix}6&-5&1\\-6&8&-2\\2&-3&1\end{pmatrix}=\begin{pmatrix}3&-\frac52&\frac12\\-3&4&-1\\1&-\frac32&\frac12\end{pmatrix}.$ Check the first row of $\mathbf{C}$ against the first column of $\mathbf{C}^{-1}$: $1(3)+1(-3)+1(1)=1$, as required.

Typical questions combine these ideas.

• Determinant with a constant $k$. For $\begin{pmatrix}k&1&0\\0&2&1\\1&0&k\end{pmatrix}$, $\det=2k^2+1$, which is at least $1$ for every real $k$, so the matrix is always non-singular and always preserves orientation.
• Scale factor to find $k$. If the volume scale factor is $19$, then $2k^2+1=19$, so $k=\pm3$.
• Show that questions: show every line of working in the expansion, because the answer is given. Write the sign pattern next to the matrix before you start.