Square and inverse of a 2x2 matrix

Problem

Calculate $A^2$ and $A^{-1}$ for $A = \begin{pmatrix} 2 & 1 \\ 1 & 1\end{pmatrix}$.

Solution

Square (row-by-column product of $A$ with itself): $A^2 = \begin{pmatrix} 2 & 1 \\ 1 & 1\end{pmatrix}\begin{pmatrix} 2 & 1 \\ 1 & 1\end{pmatrix} = \begin{pmatrix} 4+1 & 2+1 \\ 2+1 & 1+1\end{pmatrix} = \boxed{\begin{pmatrix} 5 & 3 \\ 3 & 2\end{pmatrix}}.$ Inverse with the $2\times 2$ formula. The determinant is $\det A = 2\cdot 1 - 1\cdot 1 = 1\ne 0$, therefore: $A^{-1} = \frac{1}{1}\begin{pmatrix} 1 & -1 \\ -1 & 2\end{pmatrix} = \boxed{\begin{pmatrix} 1 & -1 \\ -1 & 2\end{pmatrix}}.$ Check: $A\,A^{-1} = \begin{pmatrix} 2 & 1 \\ 1 & 1\end{pmatrix}\begin{pmatrix} 1 & -1 \\ -1 & 2\end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1\end{pmatrix} = I_2$ ✓.

Links

Topics: Linear systems
Concepts: Determinant · Inverse matrix · Matrix product
Methods: 2x2 inverse matrix
Skills: Calculate · Use formulae
Exercise type: Expression calculation