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Q6E

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Found in: Page 267

### Linear Algebra and its Applications

Book edition 5th
Author(s) David C. Lay, Steven R. Lay and Judi J. McDonald
Pages 483 pages
ISBN 978-03219822384

# Question: Is $$\left( {\begin{array}{*{20}{c}}1\\{ - 2}\\1\end{array}} \right)$$ an eigenvector of$$\left){\begin{array}{*{20}{c}}3&6&7\\3&3&7\\5&6&5\end{array}} \right)$$? If so, find the eigenvalue.

Yes, $$\left( {\begin{array}{*{20}{c}}1\\{ - 2}\\1\end{array}} \right)$$ is the eigenvector of the given matrix $$\left( {\begin{array}{*{20}{c}}3&6&7\\3&3&7\\5&6&5\end{array}} \right)$$ , and the eigenvalue is $$- 2$$.

See the step by step solution

## Step 1: Definition of eigenvector

If there exists a non-zero vector $${\bf{x}}$$ that satisfies $$A{\bf{x}} = \lambda {\bf{x}}$$ for some scaler $$\lambda$$, then $${\bf{x}}$$ be the eigenvector of an $$n \times n$$ matrix $$A$$, and if $$A{\bf{x}} = \lambda {\bf{x}}$$ exists, then scaler $$\lambda$$ is the eigenvalue of the matrix.

## Step 2: Determine whether $$\left( {\begin{array}{*{20}{c}}1\\{ - 2}\\1\end{array}} \right)$$ is the eigenvector of the given matrix

Denote the given matrix by $$A$$ and the given vector by $${\bf{x}}$$.

$$A = \left( {\begin{array}{*{20}{c}}3&6&7\\3&3&7\\5&6&5\end{array}} \right)$$, $${\bf{x}} = \left( {\begin{array}{*{20}{c}}1\\{ - 2}\\1\end{array}} \right)$$

According to the definition of eigenvalue, $${\bf{x}} = \left( {\begin{array}{*{20}{c}}1\\{ - 2}\\1\end{array}} \right)$$ is the eigenvector of the matrix $$A$$, if $$A{\bf{x}} = \lambda {\bf{x}}$$.

Find the product of $$A$$, and $${\bf{x}}$$.

$$\begin{array}{c}A{\bf{x}} = \left( {\begin{array}{*{20}{c}}3&6&7\\3&3&7\\5&6&5\end{array}} \right)\left( {\begin{array}{*{20}{c}}1\\{ - 2}\\1\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{3\left( 1 \right) + 6\left( { - 2} \right) + 7\left( 1 \right)}\\{3\left( 1 \right) + 3\left( { - 2} \right) + 7\left( 1 \right)}\\{5\left( 1 \right) + 6\left( { - 2} \right) + 5\left( 1 \right)}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{3 - 12 + 7}\\{3 - 6 + 7}\\{5 - 12 + 5}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{ - 2}\\4\\{ - 2}\end{array}} \right)\end{array}$$

The obtained matrix in the form of the given vector can be written as:

$$A{\bf{x}} = \left( { - 2} \right)\left( {\begin{array}{*{20}{c}}1\\{ - 2}\\1\end{array}} \right)$$

So, the vector $$\left( {\begin{array}{*{20}{c}}1\\{ - 2}\\1\end{array}} \right)$$ is the eigenvector of the given matrix $$\left( {\begin{array}{*{20}{c}}3&6&7\\3&3&7\\5&6&5\end{array}} \right)$$.

## Step 3: Determine the eigenvalue

As the given vector satisfies the condition $$A{\bf{x}} = \lambda {\bf{x}}$$. Which implies that $$\lambda$$ is the eigenvalue of the given matrix. So, $$\lambda = - 2$$.

So, $$- 2$$ is the eigenvalue of the given matrix $$\left( {\begin{array}{*{20}{c}}3&6&7\\3&3&7\\5&6&5\end{array}} \right)$$.