Matrices — practice questions

1. First, we set up the matrix $(\mathbf{A} - \lambda\mathbf{I})$. $(\mathbf{A} - \lambda\mathbf{I}) = \begin{pmatrix} 4 & -2 \\ 1 & 1 \end{pmatrix} - \lambda\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 4-\lambda & -2 \\ 1 & 1-\lambda \end{pmatrix}$.
2. Now, we find the determinant and set it to zero to form the characteristic equation. $\det(\mathbf{A} - \lambda\mathbf{I}) = (4-\lambda)(1-\lambda) - (-2)(1) = 0$ $4 - 4\lambda - \lambda + \lambda^2 + 2 = 0$ $\lambda^2 - 5\lambda + 6 = 0$.
3. We solve this quadratic equation for $\lambda$. $(\lambda - 2)(\lambda - 3) = 0$. The eigenvalues are $\lambda_1 = 2$ and $\lambda_2 = 3$.

We solve $(\mathbf{A} - \lambda\mathbf{I})\mathbf{x} = \mathbf{0}$ for each eigenvalue. Let $\mathbf{x} = \begin{pmatrix} x \\ y \end{pmatrix}$.

Case 1: $\lambda_1 = 2$ $(\mathbf{A} - 2\mathbf{I})\mathbf{x} = \mathbf{0}$ $\begin{pmatrix} 4-2 & -2 \\ 1 & 1-2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$ $\begin{pmatrix} 2 & -2 \\ 1 & -1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$ This gives two identical equations: $2x - 2y = 0$ (which simplifies to $x - y = 0$) and $x - y = 0$. So, $x = y$. We can choose any non-zero value. Let $y=1$, then $x=1$. An eigenvector corresponding to $\lambda_1 = 2$ is $\mathbf{e}_1 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$.

Case 2: $\lambda_2 = 3$ $(\mathbf{A} - 3\mathbf{I})\mathbf{x} = \mathbf{0}$ $\begin{pmatrix} 4-3 & -2 \\ 1 & 1-3 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$ $\begin{pmatrix} 1 & -2 \\ 1 & -2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$ This gives the equation $x - 2y = 0$, so $x = 2y$. Let $y=1$, then $x=2$. An eigenvector corresponding to $\lambda_2 = 3$ is $\mathbf{e}_2 = \begin{pmatrix} 2 \\ 1 \end{pmatrix}$.

By the Cayley-Hamilton theorem, the matrix $\mathbf{M}$ satisfies its own characteristic equation: $\mathbf{M}^3 - 2\mathbf{M}^2 - 5\mathbf{M} + 6\mathbf{I} = \mathbf{0}$.

(a) To find $\mathbf{M}^{-1}$, we multiply the entire equation by $\mathbf{M}^{-1}$: $\mathbf{M}^{-1}(\mathbf{M}^3 - 2\mathbf{M}^2 - 5\mathbf{M} + 6\mathbf{I}) = \mathbf{M}^{-1}\mathbf{0}$ $\mathbf{M}^2 - 2\mathbf{M} - 5\mathbf{I} + 6\mathbf{M}^{-1} = \mathbf{0}$ Now, we rearrange to make $\mathbf{M}^{-1}$ the subject: $6\mathbf{M}^{-1} = -\mathbf{M}^2 + 2\mathbf{M} + 5\mathbf{I}$ $\mathbf{M}^{-1} = \frac{1}{6}(-\mathbf{M}^2 + 2\mathbf{M} + 5\mathbf{I})$.

(b) To find $\mathbf{M}^4$, we first rearrange the characteristic equation to express $\mathbf{M}^3$: $\mathbf{M}^3 = 2\mathbf{M}^2 + 5\mathbf{M} - 6\mathbf{I}$. Now, multiply by $\mathbf{M}$ to get $\mathbf{M}^4$: $\mathbf{M}^4 = \mathbf{M}(\mathbf{M}^3) = \mathbf{M}(2\mathbf{M}^2 + 5\mathbf{M} - 6\mathbf{I})$ $\mathbf{M}^4 = 2\mathbf{M}^3 + 5\mathbf{M}^2 - 6\mathbf{M}$. We can substitute the expression for $\mathbf{M}^3$ back into this equation: $\mathbf{M}^4 = 2(2\mathbf{M}^2 + 5\mathbf{M} - 6\mathbf{I}) + 5\mathbf{M}^2 - 6\mathbf{M}$ $\mathbf{M}^4 = 4\mathbf{M}^2 + 10\mathbf{M} - 12\mathbf{I} + 5\mathbf{M}^2 - 6\mathbf{M}$ $\mathbf{M}^4 = 9\mathbf{M}^2 + 4\mathbf{M} - 12\mathbf{I}$.