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Linear Algebra : Operations and Properties

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Example Questions

Example Question #41 : The Determinant

$A = \begin{bmatrix}6 & -3 \\ 8 & x \end{bmatrix}$ for some real such that has an inverse.

Give $A^{-1}$ in terms of .

Possible Answers:

$\frac{1}{ 6x - 24}\begin{bmatrix}x & -3 \\ 8 & 6 \end{bmatrix}$

$\frac{1}{ 6x + 24}\begin{bmatrix}x & 3 \\ -8 & 6 \end{bmatrix}$

None of the other choices gives the correct response.

$\frac{1}{ 6x + 24}\begin{bmatrix}x & -3 \\ 8 & 6 \end{bmatrix}$

$\frac{1}{ 6x - 24}\begin{bmatrix}x & 3 \\ -8 & 6 \end{bmatrix}$

Correct answer:

$\frac{1}{ 6x + 24}\begin{bmatrix}x & 3 \\ -8 & 6 \end{bmatrix}$

Explanation:

The inverse of a two-by-two matrix

$A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21}& a_{22} \end{bmatrix}$

is the matrix

$A^{-1} =\frac{1}{D} \begin{bmatrix} a_{22} & -a_{12} \\ -a_{21}& a_{11} \end{bmatrix}$ .

is the determinant of the matrix, which is the product of the main diagonal elements minus the product of the other two:

D= 6(x)-(-3)(8)

= 6x + 24

The inverse is

$A^{-1} = \frac{1}{ 6x + 24}\begin{bmatrix}x & 3 \\ -8 & 6 \end{bmatrix}$ .

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Example Question #43 : The Determinant

$A = \begin{bmatrix} \sqrt{5} &2 \\2 \sqrt{2} & \sqrt{10} \end{bmatrix}$

Find $A^{-1}$ .

Possible Answers:

$\begin{bmatrix}- \sqrt{5}& \sqrt{2}\\ 2 &- \frac{1}{2} \sqrt{10} \end{bmatrix}$

$\begin{bmatrix} \sqrt{5}& -\sqrt{2}\\ -2 & \frac{1}{2} \sqrt{10} \end{bmatrix}$

$\begin{bmatrix} -\sqrt{5} &2 \\2 \sqrt{2} & -\sqrt{10} \end{bmatrix}$

$\begin{bmatrix} \sqrt{5} &-2 \\-2 \sqrt{2} & \sqrt{10} \end{bmatrix}$

$\begin{bmatrix} \sqrt{5}& \sqrt{2}\\ -2 & -\frac{1}{2} \sqrt{10} \end{bmatrix}$

Correct answer:

$\begin{bmatrix} \sqrt{5}& -\sqrt{2}\\ -2 & \frac{1}{2} \sqrt{10} \end{bmatrix}$

Explanation:

The inverse of a two-by-two matrix

$A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21}& a_{22} \end{bmatrix}$

is the matrix

$A^{-1} =\frac{1}{D} \begin{bmatrix} a_{22} & -a_{12} \\ -a_{21}& a_{11} \end{bmatrix}$ .

is the determinant of the matrix, which is the product of the main diagonal elements minus the product of the other two:

$D = a_{11}a_{22}- a_{12}a_{21}$

$= \sqrt{5}(\sqrt{10} )-2( 2 \sqrt{2} )$

$= 5\sqrt{2} -4 \sqrt{2}$

$=\sqrt{2}$

Therefore,

$A^{-1} =\frac{1}{\sqrt{2}} \begin{bmatrix} \sqrt{10} &-2 \\-2 \sqrt{2} & \sqrt{5} \end{bmatrix}$

$= \begin{bmatrix}\frac{1}{\sqrt{2}}( \sqrt{10}) &\frac{1}{\sqrt{2}}(-2) \\ \\ \frac{1}{\sqrt{2}} (-2 \sqrt{2}) &\frac{1}{\sqrt{2}}( \sqrt{5} )\end{bmatrix}$

$= \begin{bmatrix} \sqrt{5}& -\sqrt{2}\\ -2 & \frac{1}{2} \sqrt{10} \end{bmatrix}$

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Example Question #1 : Quadratic Forms And Positive Semidefinite Matrices

$\begin{align*}&\text{See with eigenvalues if the following matrix is positive semi-definite:}\begin{bmatrix}-19&-18\\-18&3\end{bmatrix}\end{align*}$

Possible Answers:

$\text{The matrix is not positive semi-definite.}$

$\text{The matrix is positive semi-definite.}$

Correct answer:

$\text{The matrix is not positive semi-definite.}$

Explanation:

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Example Question #1 : Quadratic Forms And Positive Semidefinite Matrices

$\begin{align*}&\text{Decide whether or not the matrix }\begin{bmatrix}4&17&-13\\17&-9&19\\-13&19&-2\end{bmatrix}\\&\text{is positive semi-definite using eigenvalues.}\end{align*}$

Possible Answers:

$\text{The matrix is not positive semi-definite.}$

$\text{The matrix is positive semi-definite.}$

Correct answer:

$\text{The matrix is not positive semi-definite.}$

Explanation:

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Example Question #3 : Quadratic Forms And Positive Semidefinite Matrices

$\begin{align*}&\text{Decide whether or not the matrix }\begin{bmatrix}13&-4\\-4&3\end{bmatrix}\\&\text{is positive semi-definite using eigenvalues.}\end{align*}$

Possible Answers:

$\text{The matrix is positive semi-definite.}$

$\text{The matrix is not positive semi-definite.}$

Correct answer:

$\text{The matrix is positive semi-definite.}$

Explanation:

$\begin{align*}&\text{The matrix given in the problem is symmetric; that is to say,}\\&\text{it is identical to its transpose. If a matrix is symmetric,}\\&\text{then it is also positive semi-definite if all of its eigenvalues}\\&\text{are non-negative. Computing these eigenvalues:}\\&\text{Eigenvalues of a matrix A, usually denoted as }\lambda\text{, are values which satisfy}\\&det(A-\lambda I)\text{, where I is the identity matrix. For our matrix }A=\begin{bmatrix}13&-4\\-4&3\end{bmatrix}\\&\text{We can define a matrix of the form }\begin{bmatrix}13 - \lambda&-4\\-4&3 - \lambda\end{bmatrix}\\&\text{For a square matrix with dimensions of }2\\&det\begin{vmatrix} a&b \\ c&d \end{vmatrix}=ad-bc\\&\lambda ^{2} - 16\lambda + 23\\&\lambda_{1}=1.60;\lambda_{2}=14.40\\&\text{The matrix is positive semi-definite.}\end{align*}$

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Example Question #4 : Quadratic Forms And Positive Semidefinite Matrices

$\begin{align*}&\text{See with eigenvalues if the following matrix is positive semi-definite:}\begin{bmatrix}-9&-15&8\\-15&12&6\\8&6&6\end{bmatrix}\end{align*}$

Possible Answers:

$\text{The matrix is positive semi-definite.}$

$\text{The matrix is not positive semi-definite.}$

Correct answer:

$\text{The matrix is not positive semi-definite.}$

Explanation:

$\begin{align*}&\text{Observation of the matrix given will reveal that it is symmetric.}\\&\text{In other words, it is equal to its transpose. If a matrix is}\\&\text{symmetric, then it is also positive semi-definite if all of}\\&\text{its eigenvalues are non-negative. So then, let us compute these}\\&\text{eigenvalues:}\\&\text{Eigenvalues of a matrix A, usually denoted as }\lambda\text{, are values which satisfy}\\&det(A-\lambda I)\text{, where I is the identity matrix. For our matrix }A=\begin{bmatrix}-9&-15&8\\-15&12&6\\8&6&6\end{bmatrix}\\&\text{We can define a matrix of the form }\begin{bmatrix}- \lambda - 9&-15&8\\-15&12 - \lambda&6\\8&6&6 - \lambda\end{bmatrix}\\&\text{For a square matrix with dimensions of }3\\&det\begin{vmatrix} a&b&c \\ d&e&f\\g&h&i \end{vmatrix}=a(ei-fh)-b(di-fg)+c(dh-eg)\\&(-1)\cdot (\lambda ^{3} - 9\lambda ^{2} - 415\lambda + 3882)\\&\lambda_{1}=-20.49;\lambda_{2}=9.45;\lambda_{3}=20.04\\&\text{The matrix is not positive semi-definite.}\end{align*}$

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Example Question #5 : Quadratic Forms And Positive Semidefinite Matrices

$\begin{align*}&\text{Use eigenvalues to determine if the following matrix is positive semi-definite:}\begin{bmatrix}12&-5\\-5&9\end{bmatrix}\end{align*}$

Possible Answers:

$\text{The matrix is positive semi-definite.}$

$\text{The matrix is not positive semi-definite.}$

Correct answer:

$\text{The matrix is positive semi-definite.}$

Explanation:

$\begin{align*}&\text{The matrix given in the problem is symmetric; that is to say,}\\&\text{it is identical to its transpose. If a matrix is symmetric,}\\&\text{then it is also positive semi-definite if all of its eigenvalues}\\&\text{are non-negative. Computing these eigenvalues:}\\&\text{Eigenvalues of a matrix A, usually denoted as }\lambda\text{, are values which satisfy}\\&det(A-\lambda I)\text{, where I is the identity matrix. For our matrix }A=\begin{bmatrix}12&-5\\-5&9\end{bmatrix}\\&\text{We can define a matrix of the form }\begin{bmatrix}12 - \lambda&-5\\-5&9 - \lambda\end{bmatrix}\\&\text{For a square matrix with dimensions of }2\\&det\begin{vmatrix} a&b \\ c&d \end{vmatrix}=ad-bc\\&\lambda ^{2} - 21\lambda + 83\\&\lambda_{1}=5.28;\lambda_{2}=15.72\\&\text{The matrix is positive semi-definite.}\end{align*}$

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Example Question #1 : Quadratic Forms And Positive Semidefinite Matrices

$\begin{align*}&\text{Decide whether or not the matrix }\begin{bmatrix}4&14\\14&9\end{bmatrix}\\&\text{is positive semi-definite using eigenvalues.}\end{align*}$

Possible Answers:

$\text{The matrix is positive semi-definite.}$

$\text{The matrix is not positive semi-definite.}$

Correct answer:

$\text{The matrix is not positive semi-definite.}$

Explanation:

$\begin{align*}&\text{The matrix given in the problem is symmetric; that is to say,}\\&\text{it is identical to its transpose. If a matrix is symmetric,}\\&\text{then it is also positive semi-definite if all of its eigenvalues}\\&\text{are non-negative. Computing these eigenvalues:}\\&\text{Eigenvalues of a matrix A, usually denoted as }\lambda\text{, are values which satisfy}\\&det(A-\lambda I)=0\text{, where I is the identity matrix. For our matrix }A=\begin{bmatrix}4&14\\14&9\end{bmatrix}\\&\text{We can define a matrix of the form }\begin{bmatrix}4 - \lambda&14\\14&9 - \lambda\end{bmatrix}\\&\text{For a square matrix with dimensions of }2\\&det\begin{vmatrix} a&b \\ c&d \end{vmatrix}=ad-bc\\&\lambda ^{2} - 13\lambda - 160=0\\&\lambda_{1}=-7.72;\lambda_{2}=20.72\\&\text{The matrix is not positive semi-definite.}\end{align*}$

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Example Question #6 : Quadratic Forms And Positive Semidefinite Matrices

$\begin{align*}&\text{Use eigenvalues to determine if the following matrix is positive semi-definite:}\begin{bmatrix}6&5&12\\5&19&3\\12&3&7\end{bmatrix}\end{align*}$

Possible Answers:

$\text{The matrix is not positive semi-definite.}$

$\text{The matrix is positive semi-definite.}$

Correct answer:

$\text{The matrix is not positive semi-definite.}$

Explanation:

$\begin{align*}&\text{The matrix given in the problem is symmetric; that is to say,}\\&\text{it is identical to its transpose. If a matrix is symmetric,}\\&\text{then it is also positive semi-definite if all of its eigenvalues}\\&\text{are non-negative. Computing these eigenvalues:}\\&\text{Eigenvalues of a matrix A, usually denoted as }\lambda\text{, are values which satisfy}\\&det(A-\lambda I)=0\text{, where I is the identity matrix. For our matrix }A=\begin{bmatrix}6&5&12\\5&19&3\\12&3&7\end{bmatrix}\\&\text{We can define a matrix of the form }\begin{bmatrix}6 - \lambda&5&12\\5&19 - \lambda&3\\12&3&7 - \lambda\end{bmatrix}\\&\text{For a square matrix with dimensions of }3\\&det\begin{vmatrix} a&b&c \\ d&e&f\\g&h&i \end{vmatrix}=a(ei-fh)-b(di-fg)+c(dh-eg)\\&(-1)\cdot (111\lambda - 32\lambda ^{2} + \lambda ^{3} + 1807)=0\\&\lambda_{1}=-5.61;\lambda_{2}=13.18;\lambda_{3}=24.43\\&\text{The matrix is not positive semi-definite.}\end{align*}$

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Example Question #7 : Quadratic Forms And Positive Semidefinite Matrices

$\begin{align*}&\text{See with eigenvalues if the following matrix is positive semi-definite:}\begin{bmatrix}-14&2\\2&-15\end{bmatrix}\end{align*}$

Possible Answers:

$\text{The matrix is positive semi-definite.}$

$\text{The matrix is not positive semi-definite.}$

Correct answer:

$\text{The matrix is not positive semi-definite.}$

Explanation:

$\begin{align*}&\text{Observation of the matrix given will reveal that it is symmetric.}\\&\text{In other words, it is equal to its transpose. If a matrix is}\\&\text{symmetric, then it is also positive semi-definite if all of}\\&\text{its eigenvalues are non-negative. So then, let us compute these}\\&\text{eigenvalues:}\\&\text{Eigenvalues of a matrix A, usually denoted as }\lambda\text{, are values which satisfy}\\&det(A-\lambda I)=0\text{, where I is the identity matrix. For our matrix }A=\begin{bmatrix}-14&2\\2&-15\end{bmatrix}\\&\text{We can define a matrix of the form }\begin{bmatrix}- \lambda - 14&2\\2&- \lambda - 15\end{bmatrix}\\&\text{For a square matrix with dimensions of }2\\&det\begin{vmatrix} a&b \\ c&d \end{vmatrix}=ad-bc\\&29\lambda + \lambda ^{2} + 206=0\\&\lambda_{1}=-16.56;\lambda_{2}=-12.44\\&\text{The matrix is not positive semi-definite.}\end{align*}$

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Linear Algebra : Operations and Properties

Study concepts, example questions & explanations for Linear Algebra

All Linear Algebra Resources

Example Questions

Example Question #41 : The Determinant

Example Question #43 : The Determinant

Example Question #1 : Quadratic Forms And Positive Semidefinite Matrices

Example Question #1 : Quadratic Forms And Positive Semidefinite Matrices

Example Question #3 : Quadratic Forms And Positive Semidefinite Matrices

Example Question #4 : Quadratic Forms And Positive Semidefinite Matrices

Example Question #5 : Quadratic Forms And Positive Semidefinite Matrices

Example Question #1 : Quadratic Forms And Positive Semidefinite Matrices

Example Question #6 : Quadratic Forms And Positive Semidefinite Matrices

Example Question #7 : Quadratic Forms And Positive Semidefinite Matrices

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