[amp_mcq option1=”[-1 1 1]T” option2=”[1 2 1]T” option3=”[1 -1 2]T” option4=”[2 1 -1]T” correct=”option1″]<\/p>\n
The correct answer is $\\boxed{\\text{(C)}}$.<\/p>\n
An eigenvector of a matrix $A$ is a vector $v$ such that $Av = \\lambda v$ for some scalar $\\lambda$. In other words, an eigenvector is a vector that is scaled by a constant when it is multiplied by the matrix.<\/p>\n
To find the eigenvectors of a matrix, we can use the following formula:<\/p>\n
$$v = \\left[ {\\begin{array}{*{20}{c}} x_1 \\\\ x_2 \\\\ x_3 \\end{array}} \\right]$$<\/p>\n
$$Av = \\left[ {\\begin{array}{*{20}{c}} a_{11}x_1 + a_{12}x_2 + a_{13}x_3 \\\\ a_{21}x_1 + a_{22}x_2 + a_{23}x_3 \\\\ a_{31}x_1 + a_{32}x_2 + a_{33}x_3 \\end{array}} \\right]$$<\/p>\n
$$\\lambda v = \\left[ {\\begin{array}{*{20}{c}} \\lambda x_1 \\\\ \\lambda x_2 \\\\ \\lambda x_3 \\end{array}} \\right]$$<\/p>\n
$$\\left[ {\\begin{array}{{20}{c}} a_{11} – \\lambda & a_{12} & a_{13} \\\\ a_{21} & a_{22} – \\lambda & a_{23} \\\\ a_{31} & a_{32} & a_{33} – \\lambda \\end{array}} \\right] \\left[ {\\begin{array}{<\/em>{20}{c}} x_1 \\\\ x_2 \\\\ x_3 \\end{array}} \\right] = 0$$<\/p>\n To find the eigenvalues, we can solve the equation $|A – \\lambda I| = 0$.<\/p>\n In this case, we have:<\/p>\n $$|A – \\lambda I| = \\left| \\begin{array}{ccc} 1 & 1 & 0 \\\\ 0 & 2 & 2 \\\\ 0 & 0 & 3 – \\lambda \\end{array} \\right| = \\lambda^3 – 6 \\lambda^2 + 11 \\lambda – 6$$<\/p>\n We can factor this equation as follows:<\/p>\n $$\\lambda^3 – 6 \\lambda^2 + 11 \\lambda – 6 = (\\lambda – 2)(\\lambda – 3)(\\lambda – 2)$$<\/p>\n Therefore, the eigenvalues of $A$ are $\\lambda = 2$, $\\lambda = 3$, and $\\lambda = 2$.<\/p>\n To find the eigenvectors corresponding to each eigenvalue, we can substitute each eigenvalue into the equation $|A – \\lambda I| = 0$ and solve for $x_1$, $x_2$, and $x_3$.<\/p>\n For $\\lambda = 2$, we have:<\/p>\n $$\\left| \\begin{array}{ccc} 2 – 2 & 1 & 0 \\\\ 0 & 2 – 2 & 2 \\\\ 0 & 0 & 3 – 2 \\end{array} \\right| = \\left| \\begin{array}{ccc} 0 & 1 & 0 \\\\ 0 & 0 & 2 \\\\ 0 & 0 & 1 \\end{array} \\right| = 0$$<\/p>\n Therefore, the eigenvector corresponding to $\\lambda = 2$ is $v = [0 1 0]^T$.<\/p>\n For $\\lambda = 3$, we have:<\/p>\n $$\\left| \\begin{array}{ccc} 3 – 3 & 1 & 0 \\\\ 0 & 3 – 3 & 2 \\\\ 0 & 0 & 3 – 3 \\end{array} \\right| = \\left| \\begin{array}{ccc} 0 & 1 & 0 \\\\ 0 & 0 & 0 \\\\ 0 & 0 & 0 \\end{array} \\right| = 0$$<\/p>\n Therefore, the eigenvector corresponding to $\\lambda = 3$ is $v = [0 0 1]^T$.<\/p>\n For $\\lambda = 2$, we have:<\/p>\n $$\\left| \\begin{array}{ccc} 2 – 2 & 1 & <\/p>\n","protected":false},"excerpt":{"rendered":" [amp_mcq option1=”[-1 1 1]T” option2=”[1 2 1]T” option3=”[1 -1 2]T” option4=”[2 1 -1]T” correct=”option1″]<\/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-20034","post","type-post","status-publish","format-standard","hentry","category-linear-algebra","no-featured-image-padding"],"yoast_head":"\n