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Linear Algebra : Linear Algebra

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Example Questions

Example Question #551 : Linear Algebra

$\begin{align*}&\text{Calculate the eigenvalues for the matrix }\\&\begin{bmatrix}12&18\\18&7\end{bmatrix}\end{align*}$

Possible Answers:

$\lambda_{1}=-2.67;\lambda_{2}=33.67$

$\lambda_{1}=-10.67;\lambda_{2}=25.67$

$\lambda_{1}=-8.67;\lambda_{2}=27.67$

$\lambda_{1}=-17.67;\lambda_{2}=18.67$

Correct answer:

$\lambda_{1}=-8.67;\lambda_{2}=27.67$

Explanation:

$\begin{align*}&\text{Eigenvalues of a matrix A, often denoted as }\lambda\text{, are values which satisfy}\\&det(A-\lambda I)\text{, where I represents the identity matrix. For the given matrix }\begin{bmatrix}12&18\\18&7\end{bmatrix}\\&\text{We can write a matrix of the form }\begin{bmatrix}12 - \lambda&18\\18&7 - \lambda\end{bmatrix}\\&\text{For a square matrix with dimensions of }2\\&det\begin{vmatrix} a&b \\ c&d \end{vmatrix}=ad-bc\\&\text{Knowing this, we can expand and solve:}\\&\lambda ^{2} - 19\lambda - 240\\&\lambda_{1}=-8.67;\lambda_{2}=27.67\\&\text{Note that since our matrix was symmetric, our eigenvalues are real.}\end{align*}$

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Example Question #2 : Eigenvalues And Eigenvectors Of Symmetric Matrices

$\begin{align*}&\text{Calculate the eigenvalues for the matrix }\\&\begin{bmatrix}8&-2&-20\\-2&-3&-9\\-20&-9&-3\end{bmatrix}\end{align*}$

Possible Answers:

$\lambda_{1}=-31.00;\lambda_{2}=-8.84;\lambda_{3}=14.84$

$\lambda_{1}=-16.00;\lambda_{2}=6.16;\lambda_{3}=29.84$

$\lambda_{1}=-19.00;\lambda_{2}=3.16;\lambda_{3}=26.84$

$\lambda_{1}=-22.00;\lambda_{2}=0.16;\lambda_{3}=23.84$

Correct answer:

$\lambda_{1}=-22.00;\lambda_{2}=0.16;\lambda_{3}=23.84$

Explanation:

$\begin{align*}&\text{Eigenvalues of a matrix A, often denoted as }\lambda\text{, are values which satisfy}\\&det(A-\lambda I)\text{, where I represents the identity matrix. For the given matrix }\begin{bmatrix}8&-2&-20\\-2&-3&-9\\-20&-9&-3\end{bmatrix}\\&\text{We can write a matrix of the form }\begin{bmatrix}8 - \lambda&-2&-20\\-2&- \lambda - 3&-9\\-20&-9&- \lambda - 3\end{bmatrix}\\&\text{For a square matrix with dimensions of }3\\&det\begin{vmatrix} a&b&c \\ d&e&f\\g&h&i \end{vmatrix}=a(ei-fh)-b(di-fg)+c(dh-eg)\\&\text{Knowing this, we can expand and solve:}\\&(-1)\cdot (\lambda ^{3} - 2\lambda ^{2} - 524\lambda + 84)\\&\lambda_{1}=-22.00;\lambda_{2}=0.16;\lambda_{3}=23.84\\&\text{Note that since our matrix was symmetric, our eigenvalues are real.}\end{align*}$

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Example Question #553 : Linear Algebra

$\begin{align*}&\text{Calculate the eigenvalues for the matrix }\\&\begin{bmatrix}-14&4\\4&-19\end{bmatrix}\end{align*}$

Possible Answers:

$\lambda_{1}=-21.22;\lambda_{2}=-11.78$

$\lambda_{1}=-18.22;\lambda_{2}=-8.78$

$\lambda_{1}=-26.22;\lambda_{2}=-16.78$

$\lambda_{1}=-29.22;\lambda_{2}=-19.78$

Correct answer:

$\lambda_{1}=-21.22;\lambda_{2}=-11.78$

Explanation:

$\begin{align*}&\text{Eigenvalues of a matrix A, often denoted as }\lambda\text{, are values which satisfy}\\&det(A-\lambda I)\text{, where I represents the identity matrix. For the given matrix }\begin{bmatrix}-14&4\\4&-19\end{bmatrix}\\&\text{We can write a matrix of the form }\begin{bmatrix}- \lambda - 14&4\\4&- \lambda - 19\end{bmatrix}\\&\text{For a square matrix with dimensions of }2\\&det\begin{vmatrix} a&b \\ c&d \end{vmatrix}=ad-bc\\&\text{Knowing this, we can expand and solve:}\\&33\lambda + \lambda ^{2} + 250\\&\lambda_{1}=-21.22;\lambda_{2}=-11.78\\&\text{Note that since our matrix was symmetric, our eigenvalues are real.}\end{align*}$

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Example Question #554 : Linear Algebra

$\begin{align*}&\text{Calculate the eigenvalues for the matrix }\\&\begin{bmatrix}-15&8\\8&1\end{bmatrix}\end{align*}$

Possible Answers:

$\lambda_{1}=-20.31;\lambda_{2}=2.31$

$\lambda_{1}=-18.31;\lambda_{2}=4.31$

$\lambda_{1}=-13.31;\lambda_{2}=9.31$

$\lambda_{1}=-27.31;\lambda_{2}=-4.69$

Correct answer:

$\lambda_{1}=-18.31;\lambda_{2}=4.31$

Explanation:

$\begin{align*}&\text{Eigenvalues of a matrix A, often denoted as }\lambda\text{, are values which satisfy}\\&det(A-\lambda I)\text{, where I represents the identity matrix. For the given matrix }\begin{bmatrix}-15&8\\8&1\end{bmatrix}\\&\text{We can write a matrix of the form }\begin{bmatrix}- \lambda - 15&8\\8&1 - \lambda\end{bmatrix}\\&\text{For a square matrix with dimensions of }2\\&det\begin{vmatrix} a&b \\ c&d \end{vmatrix}=ad-bc\\&\text{Knowing this, we can expand and solve:}\\&14\lambda + \lambda ^{2} - 79\\&\lambda_{1}=-18.31;\lambda_{2}=4.31\\&\text{Note that since our matrix was symmetric, our eigenvalues are real.}\end{align*}$

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Example Question #555 : Linear Algebra

$\begin{align*}&\text{Find any and all eigenvalues for the matrix }\\&A=\begin{bmatrix}-16&-16\\-16&-4\end{bmatrix}\end{align*}$

Possible Answers:

$\lambda_{1}=-33.09;\lambda_{2}=1.09$

$\lambda_{1}=-25.09;\lambda_{2}=9.09$

$\lambda_{1}=-36.09;\lambda_{2}=-1.91$

$\lambda_{1}=-27.09;\lambda_{2}=7.09$

Correct answer:

$\lambda_{1}=-27.09;\lambda_{2}=7.09$

Explanation:

$\begin{align*}&\text{Eigenvalues of a matrix, typically noted as }\lambda\text{, are those values which satisfy}\\&det(A-\lambda I)\text{(I being the identity matrix). For the matrix }A=\begin{bmatrix}-16&-16\\-16&-4\end{bmatrix}\\&\text{We can represent that equation as follows }\begin{bmatrix}- \lambda - 16&-16\\-16&- \lambda - 4\end{bmatrix}\\&\text{For a square matrix with dimensions of }2\\&det\begin{vmatrix} a&b \\ c&d \end{vmatrix}=ad-bc\\&\text{From that, we can find the eigenvalues:}\\&20\lambda + \lambda ^{2} - 192\\&\lambda_{1}=-27.09;\lambda_{2}=7.09\\&\text{Note that since our matrix was symmetric, our eigenvalues are real.}\end{align*}$

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Example Question #471 : Operations And Properties

$\begin{align*}&\text{Find any and all eigenvalues for the matrix }\\&A=\begin{bmatrix}3&2&15\\2&-7&-16\\15&-16&-1\end{bmatrix}\end{align*}$

Possible Answers:

$\lambda_{1}=-34.16;\lambda_{2}=-8.74;\lambda_{3}=10.90$

$\lambda_{1}=-25.16;\lambda_{2}=0.26;\lambda_{3}=19.90$

$\lambda_{1}=-27.16;\lambda_{2}=-1.74;\lambda_{3}=17.90$

$\lambda_{1}=-30.16;\lambda_{2}=-4.74;\lambda_{3}=14.90$

Correct answer:

$\lambda_{1}=-25.16;\lambda_{2}=0.26;\lambda_{3}=19.90$

Explanation:

$\begin{align*}&\text{Eigenvalues of a matrix, typically noted as }\lambda\text{, are those values which satisfy}\\&det(A-\lambda I)\text{(I being the identity matrix). For the matrix }A=\begin{bmatrix}3&2&15\\2&-7&-16\\15&-16&-1\end{bmatrix}\\&\text{We can represent that equation as follows }\begin{bmatrix}3 - \lambda&2&15\\2&- \lambda - 7&-16\\15&-16&- \lambda - 1\end{bmatrix}\\&\text{For a square matrix with dimensions of }3\\&det\begin{vmatrix} a&b&c \\ d&e&f\\g&h&i \end{vmatrix}=a(ei-fh)-b(di-fg)+c(dh-eg)\\&\text{From that, we can find the eigenvalues:}\\&(-1)\cdot (5\lambda ^{2} - 502\lambda + \lambda ^{3} + 128)\\&\lambda_{1}=-25.16;\lambda_{2}=0.26;\lambda_{3}=19.90\\&\text{Note that since our matrix was symmetric, our eigenvalues are real.}\end{align*}$

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Example Question #472 : Operations And Properties

$\begin{align*}&\text{Find the eigenvalues for the matrix }\\&A=\begin{bmatrix}18&-12&9\\-12&-16&4\\9&4&7\end{bmatrix}\end{align*}$

Possible Answers:

$\lambda_{1}=-16.38;\lambda_{2}=10.39;\lambda_{3}=29.99$

$\lambda_{1}=-30.38;\lambda_{2}=-3.61;\lambda_{3}=15.99$

$\lambda_{1}=-21.38;\lambda_{2}=5.39;\lambda_{3}=24.99$

$\lambda_{1}=-18.38;\lambda_{2}=8.39;\lambda_{3}=27.99$

Correct answer:

$\lambda_{1}=-21.38;\lambda_{2}=5.39;\lambda_{3}=24.99$

Explanation:

$\begin{align*}&\text{Eigenvalues of a matrix A, usually denoted as }\lambda\text{, are values which satisfy}\\&det(A-\lambda I)\text{, where I is the identity matrix. For our matrix }A=\begin{bmatrix}18&-12&9\\-12&-16&4\\9&4&7\end{bmatrix}\\&\text{We can define a matrix of the form }\begin{bmatrix}18 - \lambda&-12&9\\-12&- \lambda - 16&4\\9&4&7 - \lambda\end{bmatrix}\\&\text{For a square matrix with dimensions of }3\\&det\begin{vmatrix} a&b&c \\ d&e&f\\g&h&i \end{vmatrix}=a(ei-fh)-b(di-fg)+c(dh-eg)\\&\text{Using this, we can solve for our eigenvalues:}\\&(-1)\cdot (\lambda ^{3} - 9\lambda ^{2} - 515\lambda + 2880)\\&\lambda_{1}=-21.38;\lambda_{2}=5.39;\lambda_{3}=24.99\\&\text{Note that since our matrix was symmetric, our eigenvalues are real.}\end{align*}$

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Example Question #473 : Operations And Properties

$\begin{align*}&\text{Find any and all eigenvalues for the matrix }\\&A=\begin{bmatrix}-13&10&-11\\10&-9&11\\-11&11&-17\end{bmatrix}\end{align*}$

Possible Answers:

$\lambda_{1}=-42.75;\lambda_{2}=-11.90;\lambda_{3}=-8.35$

$\lambda_{1}=-34.75;\lambda_{2}=-3.90;\lambda_{3}=-0.35$

$\lambda_{1}=-29.75;\lambda_{2}=1.10;\lambda_{3}=4.65$

$\lambda_{1}=-36.75;\lambda_{2}=-5.90;\lambda_{3}=-2.35$

Correct answer:

$\lambda_{1}=-34.75;\lambda_{2}=-3.90;\lambda_{3}=-0.35$

Explanation:

$\begin{align*}&\text{Eigenvalues of a matrix, typically noted as }\lambda\text{, are those values which satisfy}\\&det(A-\lambda I)\text{(I being the identity matrix). For the matrix }A=\begin{bmatrix}-13&10&-11\\10&-9&11\\-11&11&-17\end{bmatrix}\\&\text{We can represent that equation as follows }\begin{bmatrix}- \lambda - 13&10&-11\\10&- \lambda - 9&11\\-11&11&- \lambda - 17\end{bmatrix}\\&\text{For a square matrix with dimensions of }3\\&det\begin{vmatrix} a&b&c \\ d&e&f\\g&h&i \end{vmatrix}=a(ei-fh)-b(di-fg)+c(dh-eg)\\&\text{From that, we can find the eigenvalues:}\\&(-1)\cdot (149\lambda + 39\lambda ^{2} + \lambda ^{3} + 47)\\&\lambda_{1}=-34.75;\lambda_{2}=-3.90;\lambda_{3}=-0.35\\&\text{Note that since our matrix was symmetric, our eigenvalues are real.}\end{align*}$

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Example Question #11 : Eigenvalues And Eigenvectors Of Symmetric Matrices

$\begin{align*}&\text{Calculate the eigenvalues for the matrix }\\&\begin{bmatrix}-5&17&-20\\17&-3&-2\\-20&-2&12\end{bmatrix}\end{align*}$

Possible Answers:

$\lambda_{1}=-18.37;\lambda_{2}=8.91;\lambda_{3}=37.46$

$\lambda_{1}=-26.37;\lambda_{2}=0.91;\lambda_{3}=29.46$

$\lambda_{1}=-24.37;\lambda_{2}=2.91;\lambda_{3}=31.46$

$\lambda_{1}=-30.37;\lambda_{2}=-3.09;\lambda_{3}=25.46$

Correct answer:

$\lambda_{1}=-26.37;\lambda_{2}=0.91;\lambda_{3}=29.46$

Explanation:

$\begin{align*}&\text{Eigenvalues of a matrix A, often denoted as }\lambda\text{, are values which satisfy}\\&det(A-\lambda I)\text{, where I represents the identity matrix. For the given matrix }\begin{bmatrix}-5&17&-20\\17&-3&-2\\-20&-2&12\end{bmatrix}\\&\text{We can write a matrix of the form }\begin{bmatrix}- \lambda - 5&17&-20\\17&- \lambda - 3&-2\\-20&-2&12 - \lambda\end{bmatrix}\\&\text{For a square matrix with dimensions of }3\\&det\begin{vmatrix} a&b&c \\ d&e&f\\g&h&i \end{vmatrix}=a(ei-fh)-b(di-fg)+c(dh-eg)\\&\text{Knowing this, we can expand and solve:}\\&(-1)\cdot (\lambda ^{3} - 4\lambda ^{2} - 774\lambda + 708)\\&\lambda_{1}=-26.37;\lambda_{2}=0.91;\lambda_{3}=29.46\\&\text{Note that since our matrix was symmetric, our eigenvalues are real.}\end{align*}$

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Example Question #12 : Eigenvalues And Eigenvectors Of Symmetric Matrices

$C= \begin{bmatrix} 4 & a \\ a & 8 \end{bmatrix}$ ,

where is a real number.

For to have two real eigenvalues, what must be true for ?

Possible Answers:

$-4 \sqrt{2} < a < 4 \sqrt{2}$

can be any real number.

a < -4 or a > 4

-4 < a < 4

$a < - 4 \sqrt{2}$ or $a > 4 \sqrt{2}$

Correct answer:

can be any real number.

Explanation:

Any real value of makes a symmetric matrix with real entries. It holds that any eigenvalues of must be real regardless of the value of .

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Linear Algebra : Linear Algebra

Study concepts, example questions & explanations for Linear Algebra

All Linear Algebra Resources

Example Questions

Example Question #551 : Linear Algebra

Example Question #2 : Eigenvalues And Eigenvectors Of Symmetric Matrices

Example Question #553 : Linear Algebra

Example Question #554 : Linear Algebra

Example Question #555 : Linear Algebra

Example Question #471 : Operations And Properties

Example Question #472 : Operations And Properties

Example Question #473 : Operations And Properties

Example Question #11 : Eigenvalues And Eigenvectors Of Symmetric Matrices

Example Question #12 : Eigenvalues And Eigenvectors Of Symmetric Matrices

All Linear Algebra Resources

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