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For the matrix $\mathbf{A}=\left[\begin{array}{ll}5 & 3 \\ 1 & 3\end{array}\right],$ ONE of the normalized eigen vectors is given as
(A) $\left(\begin{array}{c}\frac{1}{2} \\ \frac{\sqrt{3}}{2}\end{array}\right)$
(B) $\left(\begin{array}{l}\frac{1}{\sqrt{2}} \\ \frac{-1}{\sqrt{2}}\end{array}\right)$
(C) $\left(\begin{array}{c}\frac{3}{\sqrt{10}} \\ \frac{-1}{\sqrt{10}}\end{array}\right)$
(D) $\left(\begin{array}{l}\frac{1}{\sqrt{5}} \\ \frac{2}{\sqrt{5}}\end{array}\right)$



Question ID - 157475 | SaraNextGen Top Answer

For the matrix $\mathbf{A}=\left[\begin{array}{ll}5 & 3 \\ 1 & 3\end{array}\right],$ ONE of the normalized eigen vectors is given as
(A) $\left(\begin{array}{c}\frac{1}{2} \\ \frac{\sqrt{3}}{2}\end{array}\right)$
(B) $\left(\begin{array}{l}\frac{1}{\sqrt{2}} \\ \frac{-1}{\sqrt{2}}\end{array}\right)$
(C) $\left(\begin{array}{c}\frac{3}{\sqrt{10}} \\ \frac{-1}{\sqrt{10}}\end{array}\right)$
(D) $\left(\begin{array}{l}\frac{1}{\sqrt{5}} \\ \frac{2}{\sqrt{5}}\end{array}\right)$

1 Answer
127 votes
Answer Key / Explanation : (B) -

$\left(\begin{array}{l}\frac{1}{\sqrt{2}} \\ \frac{-1}{\sqrt{2}}\end{array}\right)$

127 votes


127