[amp_mcq option1=”\[\left[ {\begin{array}{*{20}{c}} 1&1 \\ { – 1}&{ – 2} \end{array}} \right]\]” option2=”\[\left[ {\begin{array}{*{20}{c}} 1&2 \\ { – 2}&{ – 4} \end{array}} \right]\]” option3=”\[\left[ {\begin{array}{*{20}{c}} { – 1}&0 \\ 0&{ – 2} \end{array}} \right]\]” option4=”\[\left[ {\begin{array}{*{20}{c}} 0&1 \\ { – 2}&{ – 3} \end{array}} \right]\]” correct=”option3″]
The correct answer is $\boxed{\left[ {\begin{array}{*{20}{c}} 1&1 \ { – 1}&{ – 2} \end{array}} \right]}$.
To find the matrix, we can use the following formula:
$$A = \frac{v_1 v_2^T + v_2 v_1^T}{v_1^T v_2}$$
where $v_1$ and $v_2$ are the eigenvectors corresponding to the eigenvalues $-1$ and $-2$, respectively.
In this case, $v_1 = \left[ {\begin{array}{{20}{c}} 1 \ { – 1} \end{array}} \right]$ and $v_2 = \left[ {\begin{array}{{20}{c}} 1 \ { – 2} \end{array}} \right]$. Therefore,
$$A = \frac{v_1 v_2^T + v_2 v_1^T}{v_1^T v_2} = \frac{\left[ {\begin{array}{{20}{c}} 1 \ { – 1} \end{array}} \right] \left[ {\begin{array}{{20}{c}} 1 & { – 2} \end{array}} \right] + \left[ {\begin{array}{{20}{c}} 1 & { – 2} \end{array}} \right] \left[ {\begin{array}{{20}{c}} 1 \ { – 1} \end{array}} \right]}{(1, -1)^T \cdot (1, -1)} = \left[ {\begin{array}{*{20}{c}} 1&1 \ { – 1}&{ – 2} \end{array}} \right]$$