2x2 determinants and inverse matricesAQA A-Level Further Maths: Revision notes

For $\mathbf{A}=\begin{pmatrix}a&b\\ c&d\end{pmatrix}$ the determinant is $\det\mathbf{A}=|\mathbf{A}|=ad-bc.$ Multiply the leading diagonal and subtract the product of the other diagonal. Example: $\begin{pmatrix}3&-2\\5&4\end{pmatrix}$ has determinant $(3)(4)-(-2)(5)=12+10=22$. Take care with negative entries: $-(-2)(5)=+10$. Determinants often contain an unknown. For $\begin{pmatrix}k&6\\2&k+1\end{pmatrix}$, $\det=k(k+1)-12=k^2+k-12$.

A matrix is singular if $\det\mathbf{A}=0$ and non-singular if $\det\mathbf{A}\neq0$. A singular matrix has no inverse; a non-singular matrix has exactly one. To find when a matrix with an unknown is singular, set the determinant to zero and solve. For $\mathbf{B}=\begin{pmatrix}k&6\\2&k+1\end{pmatrix}$: $k^2+k-12=0$, so $(k+4)(k-3)=0$ and $\mathbf{B}$ is singular for $k=-4$ and $k=3$. For every other value of $k$ it is non-singular. In a singular matrix the two rows (and the two columns) are multiples of each other, for example $\begin{pmatrix}2&4\\3&6\end{pmatrix}$.

The inverse of $\mathbf{A}$ is the matrix $\mathbf{A}^{-1}$ with $\mathbf{A}\mathbf{A}^{-1}=\mathbf{A}^{-1}\mathbf{A}=\mathbf{I}$, where $\mathbf{I}=\begin{pmatrix}1&0\\0&1\end{pmatrix}$. For a non-singular $2\times2$ matrix $\mathbf{A}^{-1}=\frac{1}{ad-bc}\begin{pmatrix}d&-b\\-c&a\end{pmatrix}.$ In words: swap the leading diagonal, negate the other two entries, and divide by the determinant. Example: for $\begin{pmatrix}1&2\\3&5\end{pmatrix}$, $\det=5-6=-1$, so the inverse is $\frac{1}{-1}\begin{pmatrix}5&-2\\-3&1\end{pmatrix}=\begin{pmatrix}-5&2\\3&-1\end{pmatrix}$. Check by multiplying: $\begin{pmatrix}1&2\\3&5\end{pmatrix}\begin{pmatrix}-5&2\\3&-1\end{pmatrix}=\begin{pmatrix}1&0\\0&1\end{pmatrix}$.

For non-singular matrices $\mathbf{A}$ and $\mathbf{B}$ of the same size:

• $\mathbf{A}\mathbf{A}^{-1}=\mathbf{A}^{-1}\mathbf{A}=\mathbf{I}$
• $(\mathbf{A}^{-1})^{-1}=\mathbf{A}$
• $(\mathbf{A}\mathbf{B})^{-1}=\mathbf{B}^{-1}\mathbf{A}^{-1}$, the order reverses. The reversal is like taking off shoes and socks: the last thing done is the first thing undone. Example: $\mathbf{P}=\begin{pmatrix}2&1\\3&2\end{pmatrix}$ and $\mathbf{Q}=\begin{pmatrix}1&-1\\0&2\end{pmatrix}$ give $\mathbf{P}^{-1}=\begin{pmatrix}2&-1\\-3&2\end{pmatrix}$ and $\mathbf{Q}^{-1}=\begin{pmatrix}1&\frac12\\0&\frac12\end{pmatrix}$, so $(\mathbf{P}\mathbf{Q})^{-1}=\mathbf{Q}^{-1}\mathbf{P}^{-1}=\begin{pmatrix}\frac12&0\\-\frac32&1\end{pmatrix}$.

Finding an original point. If $\mathbf{A}\mathbf{x}=\mathbf{b}$, then $\mathbf{x}=\mathbf{A}^{-1}\mathbf{b}$. Under $\mathbf{A}=\begin{pmatrix}3&-2\\5&4\end{pmatrix}$ the point $(2,1)$ maps to $(4,14)$; conversely $\mathbf{A}^{-1}\begin{pmatrix}4\\14\end{pmatrix}=\begin{pmatrix}2\\1\end{pmatrix}$. Matrix equations. To solve $\mathbf{A}\mathbf{X}=\mathbf{C}$, multiply on the left by $\mathbf{A}^{-1}$: $\mathbf{X}=\mathbf{A}^{-1}\mathbf{C}$. To solve $\mathbf{X}\mathbf{A}=\mathbf{C}$, multiply on the right: $\mathbf{X}=\mathbf{C}\mathbf{A}^{-1}$. The two answers are usually different because matrix multiplication is not commutative. Two simultaneous equations. Write $3x-2y=4$, $5x+4y=14$ as $\begin{pmatrix}3&-2\\5&4\end{pmatrix}\begin{pmatrix}x\\ y\end{pmatrix}=\begin{pmatrix}4\\14\end{pmatrix}$ and multiply by the inverse to get $x=2$, $y=1$. This works only if the determinant is non-zero.