[amp_mcq option1=”\\[\\left[ {\\begin{array}{*{20}{c}} 1 \\\\ 2 \\end{array}} \\right]\\left[ {\\begin{array}{*{20}{c}} 1 \\\\ { – 2} \\end{array}} \\right]\\]” option2=”\\[\\left[ {\\begin{array}{*{20}{c}} 2 \\\\ 1 \\end{array}} \\right]\\left[ {\\begin{array}{*{20}{c}} 1 \\\\ { – 2} \\end{array}} \\right]\\]” option3=”\\[\\left[ {\\begin{array}{*{20}{c}} 2 \\\\ { – 1} \\end{array}} \\right]\\left[ {\\begin{array}{*{20}{c}} 1 \\\\ { – 2} \\end{array}} \\right]\\]” option4=”\\[\\left[ {\\begin{array}{*{20}{c}} { – 2} \\\\ 1 \\end{array}} \\right]\\left[ {\\begin{array}{*{20}{c}} 1 \\\\ 2 \\end{array}} \\right]\\]” correct=”option3″]<\/p>\n
The correct answer is $\\boxed{\\left[ {\\begin{array}{{20}{c}} 2 \\ 1 \\end{array}} \\right]\\left[ {\\begin{array}{<\/em>{20}{c}} 1 \\ { – 2} \\end{array}} \\right]}$.<\/p>\n To find the eigenvector pair of a matrix, we can use the following formula:<\/p>\n $$\\mathbf{v} = \\left[ {\\begin{array}{*{20}{c}} x_1 \\ x_2 \\end{array}} \\right]$$<\/p>\n where $\\mathbf{v}$ is the eigenvector, $x_1$ and $x_2$ are the eigenvalues, and $A$ is the matrix.<\/p>\n In this case, we have the following matrix:<\/p>\n $$A = \\left[ {\\begin{array}{*{20}{c}} 3&4 \\ 4&{ – 3} \\end{array}} \\right]$$<\/p>\n The eigenvalues of $A$ are found by solving the equation $|A – \\lambda I| = 0$. In this case, we get the following eigenvalues:<\/p>\n $$\\lambda_1 = 2 \\text{ and } \\lambda_2 = -3$$<\/p>\n Now that we have the eigenvalues, we can find the eigenvectors. To do this, we can use the following formula:<\/p>\n $$\\mathbf{v} = \\frac{1}{\\lambda – A}\\mathbf{u}$$<\/p>\n where $\\mathbf{u}$ is a vector that is not a multiple of the null vector.<\/p>\n In this case, we can choose $\\mathbf{u} = \\left[ {\\begin{array}{*{20}{c}} 1 \\ 0 \\end{array}} \\right]$. Substituting this into the formula, we get the following eigenvector:<\/p>\n