If
\[ A = \left[ \begin{array}{r r r} 6 & - 4 & 0 \\ 4 & - 2 & 0 \\ - 1 & 0 & 3 \end{array} \right] \]
find all solutions of AX = 2X and all solutions of AX = 3X. (The symbol cX denotes the matrix each entry of which is c times the corresponding entry of X.)
Solution
The equation \(AX=cX\) is equivalent to \[ (A-cI)X=0. \]
1. The solutions of \(AX=2X\) are\[ X=t\begin{bmatrix}1\\1\\1\end{bmatrix}, \qquad t\in F. \]
For \(c=2\), \[ A-2I= \begin{bmatrix} 4&-4&0\\ 4&-4&0\\ -1&0&1 \end{bmatrix} \sim \begin{bmatrix} 1&0&-1\\ 0&1&-1\\ 0&0&0 \end{bmatrix}. \] Hence \(x_1=x_3\) and \(x_2=x_3\), giving the stated one-dimensional solution space.
2. The solutions of \(AX=3X\) are\[ X=t\begin{bmatrix}0\\0\\1\end{bmatrix}, \qquad t\in F. \]
For \(c=3\), \[ A-3I= \begin{bmatrix} 3&-4&0\\ 4&-5&0\\ -1&0&0 \end{bmatrix} \sim \begin{bmatrix} 1&0&0\\ 0&1&0\\ 0&0&0 \end{bmatrix}. \] Thus \(x_1=x_2=0\) and \(x_3\) is arbitrary.