9 The matrix \(\mathbf { M }\), where $$\mathbf { M } = \left( \begin{array} { r r r } - 2 & 2 & 2 \\ 2 & 1 & 2 \\ - 3 & - 6 & - 7 \end{array} \right)$$ has an eigenvector \(\left( \begin{array} { r } 0 \\ 1 \\ - 1 \end{array} \right)\). Find the corresponding eigenvalue. It is given that if the eigenvalues of a general \(3 \times 3\) matrix \(\mathbf { A }\), where $$\mathbf { A } = \left( \begin{array} { l l l } a & b & c \\ d & e & f \\ g & h & i \end{array} \right)$$ are \(\lambda _ { 1 } , \lambda _ { 2 }\) and \(\lambda _ { 3 }\) then $$\lambda _ { 1 } + \lambda _ { 2 } + \lambda _ { 3 } = a + e + i$$ and the determinant of \(\mathbf { A }\) has the value \(\lambda _ { 1 } \lambda _ { 2 } \lambda _ { 3 }\). Use these results to find the other two eigenvalues of the matrix \(\mathbf { M }\), and find corresponding eigenvectors.

# Question 9:

| Answer/Working | Mark | Notes |
|---|---|---|
| $\mathbf{M}\begin{pmatrix}0\\1\\-1\end{pmatrix} = \begin{pmatrix}0\\-1\\1\end{pmatrix} \Rightarrow$ eigenvalue $= -1$ $(=\lambda_1)$ | M1A1 [2] | |
| $\lambda_1 + \lambda_2 + \lambda_3 = -8 \Rightarrow \lambda_2 + \lambda_3 = -7$ | M1 | |
| $-\lambda_2\lambda_3 = \begin{vmatrix}-2 & 2 & 2\\2 & 1 & 2\\-3 & -6 & -7\end{vmatrix} = -10+16-18 = -12 \Rightarrow \lambda_2\lambda_3 = 12$ | M1A1 A1A1 | |
| $\lambda_2 = -3$ and $\lambda_3 = -4$ or vice versa | | S.C. Award M1A1 for $\lambda^3 + 8\lambda^2 + 19\lambda + 12 = 0$ and A1 for both values |
| $\lambda_2 = -3 \Rightarrow \mathbf{e}_2 = \begin{pmatrix}2\\-1\\0\end{pmatrix}$ (OE) | M1A1 | |
| $\lambda_3 = -4 \Rightarrow \mathbf{e}_3 = \begin{pmatrix}1\\0\\-1\end{pmatrix}$ (OE) | A1 [8] | |

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9 The matrix $\mathbf { M }$, where

$$\mathbf { M } = \left( \begin{array} { r r r } 
- 2 & 2 & 2 \\
2 & 1 & 2 \\
- 3 & - 6 & - 7
\end{array} \right)$$

has an eigenvector $\left( \begin{array} { r } 0 \\ 1 \\ - 1 \end{array} \right)$. Find the corresponding eigenvalue.

It is given that if the eigenvalues of a general $3 \times 3$ matrix $\mathbf { A }$, where

$$\mathbf { A } = \left( \begin{array} { l l l } 
a & b & c \\
d & e & f \\
g & h & i
\end{array} \right)$$

are $\lambda _ { 1 } , \lambda _ { 2 }$ and $\lambda _ { 3 }$ then

$$\lambda _ { 1 } + \lambda _ { 2 } + \lambda _ { 3 } = a + e + i$$

and the determinant of $\mathbf { A }$ has the value $\lambda _ { 1 } \lambda _ { 2 } \lambda _ { 3 }$.

Use these results to find the other two eigenvalues of the matrix $\mathbf { M }$, and find corresponding eigenvectors.

\hfill \mbox{\textit{CAIE FP1 2014 Q9 [10]}}