Let the Eigen vector of the matrix \[\left[ {\begin{array}{*{20}{c}} 1&2 \\ 0&2 \end{array}} \right]\] be written in the form \[\left[ {\begin{array}{*{20}{c}} 1 \\ {\text{a}} \end{array}} \right]\] and \[\left[ {\begin{array}{*{20}{c}} 1 \\ {\text{b}} \end{array}} \right]\]. What is the value of (a + b) = ? A. 0 B. \[\frac{1}{2}\] C. 1 D. 2

” option3=”1″ option4=”2″ correct=”option1″]

The correct answer is $\boxed{1}$.

The eigenvalues of the matrix $\left[ {\begin{array}{{20}{c}} 1&2 \ 0&2 \end{array}} \right]$ are $1$ and $2$. The eigenvectors corresponding to the eigenvalues $1$ and $2$ are $\left[ {\begin{array}{{20}{c}} 1 \ 0 \end{array}} \right]$ and $\left[ {\begin{array}{*{20}{c}} 0 \ 1 \end{array}} \right]$, respectively. Therefore, the value of $(a+b)$ is $1+1=\boxed{2}$.

Here is a more detailed explanation of each option:

• Option A: $0$. This is not possible because the eigenvectors of a matrix are always nonzero vectors.
• Option B: $\frac{1}{2}$. This is also not possible because the eigenvectors of a matrix are always orthogonal to each other, and the vectors $\left[ {\begin{array}{{20}{c}} 1 \ 0 \end{array}} \right]$ and $\left[ {\begin{array}{{20}{c}} 0 \ 1 \end{array}} \right]$ are not orthogonal.
• Option C: $1$. This is the correct answer. As explained above, the eigenvectors of the matrix $\left[ {\begin{array}{{20}{c}} 1&2 \ 0&2 \end{array}} \right]$ are $\left[ {\begin{array}{{20}{c}} 1 \ 0 \end{array}} \right]$ and $\left[ {\begin{array}{*{20}{c}} 0 \ 1 \end{array}} \right]$. Therefore, the value of $(a+b)$ is $1+1=\boxed{2}$.
• Option D: $2$. This is also not possible because the eigenvectors of a matrix are always linearly independent, and the vectors $\left[ {\begin{array}{{20}{c}} 1 \ 0 \end{array}} \right]$ and $\left[ {\begin{array}{{20}{c}} 0 \ 1 \end{array}} \right]$ are linearly dependent.