[amp_mcq option1=”\\[\\left\\{ {\\begin{array}{*{20}{c}} { – 1} \\\\ 1 \\end{array}} \\right\\}\\]” option2=”\\[\\left\\{ {\\begin{array}{*{20}{c}} { – 2} \\\\ 9 \\end{array}} \\right\\}\\]” option3=”\\[\\left\\{ {\\begin{array}{*{20}{c}} 2 \\\\ { – 1} \\end{array}} \\right\\}\\]” option4=”\\[\\left\\{ {\\begin{array}{*{20}{c}} 1 \\\\ 1 \\end{array}} \\right\\}\\]” correct=”option3″]<\/p>\n
The correct answer is $\\boxed{\\left\\{ {\\begin{array}{*{20}{c}} { – 1} \\ 1 \\end{array}} \\right\\}}$.<\/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 following matrix:<\/p>\n
$$A = \\left[ {\\begin{array}{*{20}{c}} { – 5}&2 \\ { – 9}&6 \\end{array}} \\right]$$<\/p>\n
We can solve the equation $\\lambda v = A v$ for $v$ to find the eigenvectors.<\/p>\n
First, we can try to solve the equation for $v_1$:<\/p>\n
$$\\lambda v_1 = \\left[ {\\begin{array}{*{20}{c}} { – 5}&2 \\ { – 9}&6 \\end{array}} \\right] v_1$$<\/p>\n
We can reduce this equation to the following form:<\/p>\n
$$\\left[ {\\begin{array}{*{20}{c}} { – \\lambda – 5}&2 \\ { – 9}&{6 – \\lambda} \\end{array}} \\right] v_1 = 0$$<\/p>\n
We can find the eigenvalues by solving the equation $|A – \\lambda I| = 0$.<\/p>\n
In this case, we have the following equation:<\/p>\n
$$\\left| \\begin{array}{*{20}{c}} { – \\lambda – 5}&2 \\ { – 9}&{6 – \\lambda} \\end{array} \\right| = 0$$<\/p>\n
We can solve this equation to find the eigenvalues $\\lambda = -2$ and $\\lambda = 9$.<\/p>\n
Now that we know the eigenvalues, we can find the eigenvectors.<\/p>\n
To find the eigenvector corresponding to the eigenvalue $\\lambda = -2$, we can solve the equation $v_1 \\left[ {\\begin{array}{*{20}{c}} { – 2 – 5}&2 \\ { – 9}&{6 – 2} \\end{array}} \\right] = 0$.<\/p>\n
We can reduce this equation to the following form:<\/p>\n
$$v_1 \\left[ {\\begin{array}{*{20}{c}} { – 7}&2 \\ { – 9}&{4} \\end{array}} \\right] = 0$$<\/p>\n
We can solve this equation to find the eigenvector $v_1 = \\left\\{ {\\begin{array}{*{20}{c}} { – 1} \\ 1 \\end{array}} \\right\\}$.<\/p>\n
Therefore, one of the eigenvectors of the matrix $\\left[ {\\begin{array}{{20}{c}} { – 5}&2 \\ { – 9}&6 \\end{array}} \\right]$ is $\\left\\{ {\\begin{array}{<\/em>{20}{c}} { – 1} \\ 1 \\end{array}} \\right\\}$.<\/p>\n","protected":false},"excerpt":{"rendered":" [amp_mcq option1=”\\[\\left\\{ {\\begin{array}{*{20}{c}} { – 1} \\\\ 1 \\end{array}} \\right\\}\\]” option2=”\\[\\left\\{ {\\begin{array}{*{20}{c}} { – 2} \\\\ 9 \\end{array}} \\right\\}\\]” option3=”\\[\\left\\{ {\\begin{array}{*{20}{c}} 2 \\\\ { – 1} \\end{array}} \\right\\}\\]” option4=”\\[\\left\\{ {\\begin{array}{*{20}{c}} 1 \\\\ 1 \\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-19988","post","type-post","status-publish","format-standard","hentry","category-linear-algebra","no-featured-image-padding"],"yoast_head":"\n