[amp_mcq option1=”2, -5″ option2=”3, -5″ option3=”2, 5″ option4=”3, 5″ correct=”option3″]<\/p>\n
The correct answer is $\\boxed{\\text{(C)}}$.<\/p>\n
To find the eigenvalues of a matrix, we can use the following formula:<\/p>\n
$$\\lambda = \\det(A – \\lambda I)$$<\/p>\n
where $A$ is the matrix and $\\lambda$ is the eigenvalue.<\/p>\n
In this case, we have:<\/p>\n
$$\\begin{align} Since $\\lambda = 3$ is a root of the quadratic polynomial $\\lambda^2 – 4\\lambda + 40$, then the other two eigenvalues must be the roots of the quadratic polynomial $(\\lambda – 3)(\\lambda^2 – 4\\lambda + 40) = 0$.<\/p>\n Solving for the roots of this quadratic polynomial, we find that the other two eigenvalues are $\\boxed{2, 5}$.<\/p>\n","protected":false},"excerpt":{"rendered":" [amp_mcq option1=”2, -5″ option2=”3, -5″ option3=”2, 5″ option4=”3, 5″ correct=”option3″]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[489],"tags":[],"class_list":["post-20023","post","type-post","status-publish","format-standard","hentry","category-linear-algebra","no-featured-image-padding"],"yoast_head":"\n
\n\\det(A – \\lambda I) &= \\det \\left[ {\\begin{array}{<\/em>{20}{c}} 2&{ – 2}&3 \\ { – 2}&{ – 1}&6 \\ 1&2&0 \\end{array}} – \\lambda \\begin{array}{{20}{c}} 1&0&0 \\ 0&1&0 \\ 0&0&1 \\end{array} \\right] \\
\n&= \\det \\left[ {\\begin{array}{<\/em>{20}{c}} 2-\\lambda &{ – 2}&3 \\ { – 2}&{ – 1}-\\lambda &6 \\ 1&2&-\\lambda \\end{array}} \\right] \\
\n&= -\\lambda^3 + 4\\lambda^2 – 23\\lambda + 120 \\
\n&= (\\lambda – 3)(\\lambda^2 – 4\\lambda + 40)
\n\\end{align*}$$<\/p>\n