[amp_mcq option1=”\\[\\left[ {\\begin{array}{*{20}{c}} 3 \\\\ 2 \\end{array}} \\right]\\]” option2=”\\[\\left[ {\\begin{array}{*{20}{c}} 4 \\\\ 3 \\end{array}} \\right]\\]” option3=”\\[\\left[ {\\begin{array}{*{20}{c}} 2 \\\\ { – 1} \\end{array}} \\right]\\]” option4=”\\[\\left[ {\\begin{array}{*{20}{c}} { – 1} \\\\ 2 \\end{array}} \\right]\\]” correct=”option3″]<\/p>\n
The correct answer is $\\boxed{\\left[ {\\begin{array}{*{20}{c}} 2 \\ { – 1} \\end{array}} \\right]}$.<\/p>\n
To find the eigenvalues and eigenvectors of a matrix, we can use the following formula:<\/p>\n
$$\\lambda v = A v$$<\/p>\n
where $\\lambda$ is the eigenvalue, $v$ is the eigenvector, and $A$ is the matrix.<\/p>\n
In this case, we have the matrix $A = \\left[ {\\begin{array}{*{20}{c}} { – 4}&2 \\ 4&3 \\end{array}} \\right]$.<\/p>\n
To find the eigenvalues, we can solve the equation $Av = \\lambda v$.<\/p>\n
We can do this by first finding the characteristic polynomial of $A$, which is given by:<\/p>\n
$$| A – \\lambda I |$$<\/p>\n
In this case, the characteristic polynomial is:<\/p>\n
$$| \\left[ {\\begin{array}{{20}{c}} { – 4}&2 \\ 4&3 \\end{array}} \\right] – \\lambda \\left[ {\\begin{array}{<\/em>{20}{c}} 1&0 \\ 0&1 \\end{array}} \\right] | = \\left[ {\\begin{array}{*{20}{c}} { – 4}-\\lambda &2 \\ 4&3-\\lambda \\end{array}} \\right]$$<\/p>\n We can then solve the equation $| A – \\lambda I | = 0$ to find the eigenvalues.<\/p>\n In this case, the eigenvalues are $\\lambda = 2$ and $\\lambda = 5$.<\/p>\n Once we have found the eigenvalues, we can find the eigenvectors by solving the equation $Av = \\lambda v$ for each eigenvalue.<\/p>\n In this case, we have the following equations:<\/p>\n $$\\left[ {\\begin{array}{{20}{c}} { – 4}&2 \\ 4&3 \\end{array}} \\right] \\left[ {\\begin{array}{<\/em>{20}{c}} x_1 \\ x_2 \\end{array}} \\right] = \\left[ {\\begin{array}{*{20}{c}} 2 \\ 5 \\end{array}} \\right]$$<\/p>\n and<\/p>\n $$\\left[ {\\begin{array}{{20}{c}} { – 4}&2 \\ 4&3 \\end{array}} \\right] \\left[ {\\begin{array}{<\/em>{20}{c}} y_1 \\ y_2 \\end{array}} \\right] = \\left[ {\\begin{array}{*{20}{c}} 5 \\ 2 \\end{array}} \\right]$$<\/p>\n We can solve these equations for $x_1$ and $x_2$, and $y_1$ and $y_2$, respectively.<\/p>\n In this case, we find that the eigenvectors are $\\left[ {\\begin{array}{{20}{c}} 2 \\ { – 1} \\end{array}} \\right]$ and $\\left[ {\\begin{array}{<\/em>{20}{c}} 1 \\ 2 \\end{array}} \\right]$.<\/p>\n Therefore, the correct answer is $\\boxed{\\left[ {\\begin{array}{*{20}{c}} 2 \\ { – 1} \\end{array}} \\right]}$.<\/p>\n","protected":false},"excerpt":{"rendered":" [amp_mcq option1=”\\[\\left[ {\\begin{array}{*{20}{c}} 3 \\\\ 2 \\end{array}} \\right]\\]” option2=”\\[\\left[ {\\begin{array}{*{20}{c}} 4 \\\\ 3 \\end{array}} \\right]\\]” option3=”\\[\\left[ {\\begin{array}{*{20}{c}} 2 \\\\ { – 1} \\end{array}} \\right]\\]” option4=”\\[\\left[ {\\begin{array}{*{20}{c}} { – 1} \\\\ 2 \\end{array}} \\right]\\]” correct=”option3″]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[489],"tags":[],"class_list":["post-20035","post","type-post","status-publish","format-standard","hentry","category-linear-algebra","no-featured-image-padding"],"yoast_head":"\n