Operations and Properties - Linear Algebra

Card 1 of 2140

0

Didn't Know

0

Didn't Know

Knew It

0

1 of 2019 left

Question

Find the Eigen Values for Matrix .

$A=\begin{bmatrix} 5 & -3 \ 1& -2 \end{bmatrix}$

Tap to reveal answer

Answer

The first step into solving for eigenvalues, is adding in a $-\lambda$ along the main diagonal.

$A=\begin{bmatrix} 5-\lambda & -3 \ 1& -4-\lambda \end{bmatrix}$

Now the next step to take the determinant.

$\det(A)=(5-\lambda)(-4-\lambda)-(-3)(1)$

$\det(A)=(5-\lambda)(-4-\lambda)+3$

Now lets FOIL, and solve for $\lambda$ .

$(5-\lambda)(-4-\lambda)+3=-20-5\lambda+4\lambda+\lambda^2+3$

$=\lambda^2-\lambda-17$

Now lets use the quadratic equation to solve for $\lambda$ .

$\lambda=\frac{1\pm\sqrt{(1)^2-4(1)(-17)}}{2(1)}$

$\lambda=\frac{1\pm\sqrt{1+68}}{2}$

$\lambda=\frac{1\pm\sqrt{69}}{2}$

So our eigen values are

$\lambda=\frac{1+ \sqrt{69}}{2}$

$\lambda=\frac{1- \sqrt{69}}{2}$

← Didn't Know|Knew It →

Question

Find the eigenvalues for the matrix $\begin{bmatrix} 2 & 1\ 3& 4\end{bmatrix}$

Tap to reveal answer

Answer

The eigenvalues, $\lambda$ , for the matrix are values for which the determinant of $\begin{bmatrix} 2- \lambda & 1\ 3 & 4 - \lambda \end{bmatrix}$ is equal to zero. First, find the determinant:

$(2-\lambda)(4-\lambda) - (3)(1) = 8 - 4 \lambda - 2 \lambda + \lambda ^2 - 3 = \lambda ^2 - 6 \lambda + 5$

Now set the determinant equal to zero and solve this quadratic:

$\lambda^2 - 6 \lambda + 5 = 0$ this can be factored:

$(\lambda - 5 ) (\lambda - 1 ) = 0$

The eigenvalues are 5 and 1.

← Didn't Know|Knew It →

Question

Which is an eigenvector for $A = \begin{bmatrix} 2 &1 \3 &4 \end{bmatrix}$ , $\mathbf{u} = \begin{bmatrix} 1\3 \end{bmatrix}$ or $\mathbf{v} = \begin{bmatrix} -3\1 \end{bmatrix}$

Tap to reveal answer

Answer

To determine if something is an eignevector, multiply times A:

$\begin{bmatrix} 2 &1 \3 & 4\end{bmatrix} \begin{bmatrix} 1\ 3\end{bmatrix} = \begin{bmatrix} 15\ 5\end{bmatrix}$

Since this is equivalent to $5\begin{bmatrix} 1\ 3\end{bmatrix}$ , $\begin{bmatrix} 1\ 3\end{bmatrix}$ is an eigenvector (and 5 is an eigenvalue).

$\begin{bmatrix} 2 & 1\ 3& 4\end{bmatrix} \begin{bmatrix} -3\1 \end{bmatrix} = \begin{bmatrix} -5\ -5\end{bmatrix}$

This cannot be re-written as $\mathbf{v}$ times a scalar, so this is not an eigenvector.

← Didn't Know|Knew It →

Question

Find the eigenvalues for the matrix $\begin{bmatrix} -3 &6 \ 2& 1 \end{bmatrix}$

Tap to reveal answer

Answer

The eigenvalues are scalar quantities, $\lambda$ , where the determinant of $\begin{bmatrix} -3-\lambda & 6 \ 2& 1-\lambda \end{bmatrix}$ is equal to zero.

First, find an expression for the determinant:

$(-3-\lambda)(1-\lambda) - 12$

$=-3 - \lambda + 3 \lambda + \lambda ^2 - 12 = \lambda ^2 + 2 \lambda -15$

Now set this equal to zero, and solve:

$\lambda^2 + 2 \lambda -15 = 0$ this can be factored (or solved in another way)

$(\lambda+ 5) (\lambda-3) = 0$

The eigenvalues are -5 and 3.

← Didn't Know|Knew It →

Question

Which is an eigenvector for $A = \begin{bmatrix} -3 & 6\ 2& 1\end{bmatrix}$ , $\mathbf{u} = \begin{bmatrix} -3\1 \end{bmatrix}$ or $\mathbf{v} = \begin{bmatrix} 1\ 1\end{bmatrix}$ ?

Tap to reveal answer

Answer

To determine if something is an eigenvector, multiply by the matrix A:

$\begin{bmatrix} -3 & 6\ 2& 1\end{bmatrix} \begin{bmatrix} -3\ 1\end{bmatrix} = \begin{bmatrix} 15\ -5 \end{bmatrix}$

This is equivalent to $-5\begin{bmatrix} -3\ 1\end{bmatrix}$ so this is an eigenvector.

$\begin{bmatrix} -3 & 6\ 2& 1\end{bmatrix} \begin{bmatrix} 1\ 1\end{bmatrix} = \begin{bmatrix} 3\ 3 \end{bmatrix}$

This is equivalent to $3\begin{bmatrix} 1\ 1\end{bmatrix}$ so this is also an eigenvector.

← Didn't Know|Knew It →

Question

Determine the eigenvalues for the matrix $\begin{bmatrix} -5 &2 \ 3 &-4 \end{bmatrix}$

Tap to reveal answer

Answer

The eigenvalues are scalar quantities $\lambda$ where the determinant of $\begin{bmatrix} -5-\lambda & 2\ 3& -4-\lambda\end{bmatrix}$ is equal to zero. First, write an expression for the determinant:

$(-5-\lambda) (-4-\lambda) -6 = 0$

$20 + 4\lambda+5 \lambda + \lambda^2 - 6 = 0$

$\lambda^2 + 9 \lambda +14 = 0$ this can be solved by factoring:

$(\lambda+2) (\lambda+7) = 0$

The solutions are -2 and -7

← Didn't Know|Knew It →

Question

Which is an eigenvector for the matrix $A = \begin{bmatrix} -5 &2 \ 3& -4\end{bmatrix}$ , $\mathbf{u} = \begin{bmatrix} -2\ 3\end{bmatrix}$ or $\mathbf{v} = \begin{bmatrix} 2\ 3\end{bmatrix}$

Tap to reveal answer

Answer

To determine if a vector is an eigenvector, multiply with A:

$\begin{bmatrix} -5 & 2\ 3& -4 \end{bmatrix} \begin{bmatrix} -2\ 3\end{bmatrix} = \begin{bmatrix} 16\-18 \end{bmatrix}$ . This cannot be expressed as an integer times $\mathbf{u}$ , so $\mathbf{u}$ is not an eigenvector

$\begin{bmatrix} -5 & 2\ 3& -4 \end{bmatrix} \begin{bmatrix} 2\ 3\end{bmatrix} = \begin{bmatrix} -4\-6 \end{bmatrix}$ This can be expressed as $-2\begin{bmatrix} 2\ 3\end{bmatrix}$ , so $\mathbf{v}$ is an eigenvector.

← Didn't Know|Knew It →

Question

$\begin{align}&\text{Find any and all eigenvalues for the matrix }\&A=\begin{bmatrix}6&-13\8&-19\end{bmatrix}\end{align}$

Tap to reveal answer

Answer

$\begin{align}&\text{Eigenvalues of a matrix, typically noted as }\lambda\text{, are those values which satisfy}\&det(A-\lambda I)\text{(I being the identity matrix). For the matrix }A=\begin{bmatrix}6&-13\8&-19\end{bmatrix}\&\text{We can represent that equation as follows }\begin{bmatrix}6 - \lambda &-13\8&- \lambda - 19\end{bmatrix} \&\text{For a square matrix with dimensions of }2\&det\begin{vmatrix} a&b \ c&d \end{vmatrix}=ad-bc\&\text{From that, we can find the eigenvalues:}\&13\lambda + \lambda ^{2} - 10\&\lambda_{1}=0.73;\lambda_{2}=-13.73\end{align}$

← Didn't Know|Knew It →

Question

$\begin{align}&\text{Calculate the eigenvalues for the matrix }\&\begin{bmatrix}0&-2\6&9\end{bmatrix}\end{align}$

Tap to reveal answer

Answer

$\begin{align}&\text{Eigenvalues of a matrix A, often denoted as }\lambda\text{, are values which satisfy}\&det(A-\lambda I)\text{, where I represents the identity matrix. For the given matrix }\begin{bmatrix}0&-2\6&9\end{bmatrix}\&\text{We can write a matrix of the form }\begin{bmatrix}-\lambda&-2\6&9 - \lambda\end{bmatrix}\&\text{For a square matrix with dimensions of }2\&det\begin{vmatrix} a&b \ c&d \end{vmatrix}=ad-bc\&\text{Knowing this, we can expand and solve:}\&\lambda ^{2} - 9\lambda + 12\&\lambda_{1}=1.63;\lambda_{2}=7.37\end{align}$

← Didn't Know|Knew It →

Question

$\begin{align}&\text{Find any and all eigenvalues for the matrix }\&A=\begin{bmatrix}-7&3&-11\10&-10&0\8&16&19\end{bmatrix}\end{align}$

Tap to reveal answer

Answer

$\begin{align}&\text{Eigenvalues of a matrix, typically noted as }\lambda\text{, are those values which satisfy}\&det(A-\lambda I)\text{(I being the identity matrix). For the matrix }A=\begin{bmatrix}-7&3&-11\10&-10&0\8&16&19\end{bmatrix}\&\text{We can represent that equation as follows }\begin{bmatrix}- \lambda - 7&3&-11\10&- \lambda - 10&0\8&16&19 - \lambda\end{bmatrix}\&\text{For a square matrix with dimensions of }3\&det\begin{vmatrix} a&b&c \ d&e&f\g&h&i \end{vmatrix}=a(ei-fh)-b(di-fg)+c(dh-eg)\&\text{From that, we can find the eigenvalues:}\&(-1)\cdot (\lambda ^{3} - 2\lambda ^{2} - 195\lambda + 1880)\&\lambda_{1}=9.29+5.19i;\lambda_{2}=9.29-5.19i;\lambda_{3}=-16.59\end{align}$

← Didn't Know|Knew It →

Question

$\begin{align}&\text{Find any and all eigenvalues for the matrix }\&A=\begin{bmatrix}18&-6\-12&-10\end{bmatrix}\end{align}$

Tap to reveal answer

Answer

$\begin{align}&\text{Eigenvalues of a matrix, typically noted as }\lambda\text{, are those values which satisfy}\&det(A-\lambda I)\text{(I being the identity matrix). For the matrix }A=\begin{bmatrix}18&-6\-12&-10\end{bmatrix}\&\text{We can represent that equation as follows }\begin{bmatrix}18 - \lambda&-6\-12&- \lambda - 10\end{bmatrix}\&\text{For a square matrix with dimensions of }2\&det\begin{vmatrix} a&b \ c&d \end{vmatrix}=ad-bc\&\text{From that, we can find the eigenvalues:}\&\lambda ^{2} - 8\lambda - 252\&\lambda_{1}=20.37;\lambda_{2}=-12.37\end{align}$

← Didn't Know|Knew It →

Question

$\begin{align}&\text{Calculate the eigenvalues for the matrix }\&\begin{bmatrix}11&10\-5&3\end{bmatrix}\end{align}$

Tap to reveal answer

Answer

$\begin{align}&\text{Eigenvalues of a matrix A, often denoted as }\lambda\text{, are values which satisfy}\&det(A-\lambda I)\text{, where I represents the identity matrix. For the given matrix }\begin{bmatrix}11&10\-5&3\end{bmatrix}\&\text{We can write a matrix of the form }\begin{bmatrix}11 - \lambda&10\-5&3 - \lambda\end{bmatrix}\&\text{For a square matrix with dimensions of }2\&det\begin{vmatrix} a&b \ c&d \end{vmatrix}=ad-bc\&\text{Knowing this, we can expand and solve:}\&\lambda ^{2} - 14\lambda + 83\&\lambda_{1}=7.00+5.83i;\lambda_{2}=7.00-5.83i\end{align}$

← Didn't Know|Knew It →

Question

$\begin{align}&\text{Find any and all eigenvalues for the matrix }\&A=\begin{bmatrix}-20&-7\-14&12\end{bmatrix}\end{align}$

Tap to reveal answer

Answer

$\begin{align}&\text{Eigenvalues of a matrix, typically noted as }\lambda\text{, are those values which satisfy}\&det(A-\lambda I)\text{(I being the identity matrix). For the matrix }A=\begin{bmatrix}-20&-7\-14&12\end{bmatrix}\&\text{We can represent that equation as follows }\begin{bmatrix}- \lambda - 20&-7\-14&12 - \lambda\end{bmatrix}\&\text{For a square matrix with dimensions of }2\&det\begin{vmatrix} a&b \ c&d \end{vmatrix}=ad-bc\&\text{From that, we can find the eigenvalues:}\&8\lambda + \lambda ^{2} - 338\&\lambda_{1}=-22.81;\lambda_{2}=14.81\end{align}$

← Didn't Know|Knew It →

Question

$\begin{align}&\text{Find any and all eigenvalues for the matrix }\&A=\begin{bmatrix}-2&-17&-11\17&-14&13\2&20&-17\end{bmatrix}\end{align}$

Tap to reveal answer

Answer

$\begin{align}&\text{Eigenvalues of a matrix, typically noted as }\lambda\text{, are those values which satisfy}\&det(A-\lambda I)\text{(I being the identity matrix). For the matrix }A=\begin{bmatrix}-2&-17&-11\17&-14&13\2&20&-17\end{bmatrix}\&\text{We can represent that equation as follows }\begin{bmatrix}- \lambda - 2&-17&-11\17&- \lambda - 14&13\2&20&- \lambda - 17\end{bmatrix}\&\text{For a square matrix with dimensions of }3\&det\begin{vmatrix} a&b&c \ d&e&f\g&h&i \end{vmatrix}=a(ei-fh)-b(di-fg)+c(dh-eg)\&\text{From that, we can find the eigenvalues:}\&(-1)\cdot (351\lambda + 33\lambda ^{2} + \lambda ^{3} + 9359)\&\lambda_{1}=-0.83+17.26i;\lambda_{2}=-0.83-17.26i;\lambda_{3}=-31.33\end{align}$

← Didn't Know|Knew It →

Question

$\begin{align}&\text{Calculate the eigenvalues for the matrix }\&\begin{bmatrix}-4&-10\12&-3\end{bmatrix}\end{align}$

Tap to reveal answer

Answer

$\begin{align}&\text{Eigenvalues of a matrix A, often denoted as }\lambda\text{, are values which satisfy}\&det(A-\lambda I)\text{, where I represents the identity matrix. For the given matrix }\begin{bmatrix}-4&-10\12&-3\end{bmatrix}\&\text{We can write a matrix of the form }\begin{bmatrix}- \lambda - 4&-10\12&- \lambda - 3\end{bmatrix}\&\text{For a square matrix with dimensions of }2\&det\begin{vmatrix} a&b \ c&d \end{vmatrix}=ad-bc\&\text{Knowing this, we can expand and solve:}\&7\lambda + \lambda ^{2} + 132\&\lambda_{1}=-3.50+10.94i;\lambda_{2}=-3.50-10.94i\end{align}$

← Didn't Know|Knew It →

Question

$\begin{align}&\text{Calculate the eigenvalues for the matrix }\&\begin{bmatrix}-15&14&5\-6&1&-4\-17&-11&-15\end{bmatrix}\end{align}$

Tap to reveal answer

Answer

$\begin{align}&\text{Eigenvalues of a matrix A, often denoted as }\lambda\text{, are values which satisfy}\&det(A-\lambda I)\text{, where I represents the identity matrix. For the given matrix }\begin{bmatrix}-15&14&5\-6&1&-4\-17&-11&-15\end{bmatrix}\&\text{We can write a matrix of the form }\begin{bmatrix}- \lambda - 15&14&5\-6&1 - \lambda&-4\-17&-11&- \lambda - 15\end{bmatrix}\&\text{For a square matrix with dimensions of }3\&det\begin{vmatrix} a&b&c \ d&e&f\g&h&i \end{vmatrix}=a(ei-fh)-b(di-fg)+c(dh-eg)\&\text{Knowing this, we can expand and solve:}\&(-1)\cdot (320\lambda + 29\lambda ^{2} + \lambda ^{3} - 992)\&\lambda_{1}=-15.74+12.27i;\lambda_{2}=-15.74-12.27i;\lambda_{3}=2.49\end{align}$

← Didn't Know|Knew It →

Question

$\begin{align}&\text{Find any and all eigenvalues for the matrix }\&A=\begin{bmatrix}-7&16\-5&-16\end{bmatrix}\end{align}$

Tap to reveal answer

Answer

$\begin{align}&\text{Eigenvalues of a matrix, typically noted as }\lambda\text{, are those values which satisfy}\&det(A-\lambda I)\text{(I being the identity matrix). For the matrix }A=\begin{bmatrix}-7&16\-5&-16\end{bmatrix}\&\text{We can represent that equation as follows }\begin{bmatrix}- \lambda - 7&16\-5&- \lambda - 16\end{bmatrix}\&\text{For a square matrix with dimensions of }2\&det\begin{vmatrix} a&b \ c&d \end{vmatrix}=ad-bc\&\text{From that, we can find the eigenvalues:}\&23\lambda + \lambda ^{2} + 192\&\lambda_{1}=-11.50+7.73i;\lambda_{2}=-11.50-7.73i\end{align}$

← Didn't Know|Knew It →

Question

$\begin{align}&\text{Find the eigenvalues for the matrix }\&A=\begin{bmatrix}3&-18&-11\-6&13&-20\-19&-14&6\end{bmatrix}\end{align}$

Tap to reveal answer

Answer

$\begin{align}&\text{Eigenvalues of a matrix A, usually denoted as }\lambda\text{, are values which satisfy}\&det(A-\lambda I)\text{, where I is the identity matrix. For our matrix }A=\begin{bmatrix}3&-18&-11\-6&13&-20\-19&-14&6\end{bmatrix}\&\text{We can define a matrix of the form }\begin{bmatrix}3 - \lambda&-18&-11\-6&13 - \lambda&-20\-19&-14&6 - \lambda\end{bmatrix}\&\text{For a square matrix with dimensions of }3\&det\begin{vmatrix} a&b&c \ d&e&f\g&h&i \end{vmatrix}=a(ei-fh)-b(di-fg)+c(dh-eg)\&\text{Using this, we can solve for our eigenvalues:}\&(-1)\cdot (\lambda ^{3} - 22\lambda ^{2} - 462\lambda + 11735)\&\lambda_{1}=-22.30;\lambda_{2}=22.15+5.95i;\lambda_{3}=22.15-5.95i\end{align}$

← Didn't Know|Knew It →

Question

$\begin{align}&\text{Calculate the eigenvalues for the matrix }\&\begin{bmatrix}-13&-5&5\11&-17&18\11&-1&-3\end{bmatrix}\end{align}$

Tap to reveal answer

Answer

$\begin{align}&\text{Eigenvalues of a matrix A, often denoted as }\lambda\text{, are values which satisfy}\&det(A-\lambda I)\text{, where I represents the identity matrix. For the given matrix }\begin{bmatrix}-13&-5&5\11&-17&18\11&-1&-3\end{bmatrix}\&\text{We can write a matrix of the form }\begin{bmatrix}- \lambda - 13&-5&5\11&- \lambda - 17&18\11&-1&- \lambda - 3\end{bmatrix}\&\text{For a square matrix with dimensions of }3\&det\begin{vmatrix} a&b&c \ d&e&f\g&h&i \end{vmatrix}=a(ei-fh)-b(di-fg)+c(dh-eg)\&\text{Knowing this, we can expand and solve:}\&(-1)\cdot (329\lambda + 33\lambda ^{2} + \lambda ^{3} + 1172)\&\lambda_{1}=-7.08+3.47i;\lambda_{2}=-7.08-3.47i;\lambda_{3}=-18.84\end{align}$

← Didn't Know|Knew It →

Question

Give the characteristic polynomial of the matrix

$A = \begin{bmatrix} \frac{1}{2}& \frac{2}{3}\ \ -\frac{1}{4}& \frac{1}{3} \end{bmatrix}$

Tap to reveal answer

Answer

The characteristic polynomial of a square matrix can be derived as follows:

Determine $\det( \lambda I - A)$ , using the identity matrix with the same dimensions as (two by two):

$\lambda I - A$

$= \lambda \begin{bmatrix} 1& 0\ 0& 1 \end{bmatrix} - \begin{bmatrix} \frac{1}{2}& \frac{2}{3}\ \ -\frac{1}{4}& \frac{1}{3} \end{bmatrix}$

$= \begin{bmatrix} \lambda& 0\ 0& \lambda \end{bmatrix} - \begin{bmatrix} \frac{1}{2}& \frac{2}{3}\ \ -\frac{1}{4}& \frac{1}{3} \end{bmatrix}$

Subtract the matrices by subtracting elementwise:

$=\begin{bmatrix} \lambda - \frac{1}{2}& 0- \frac{2}{3}\ \ 0 - \left (-\frac{1}{4} \right )& \lambda - \frac{1}{3} \end{bmatrix}$

$=\begin{bmatrix} \lambda - \frac{1}{2}& - \frac{2}{3}\ \ \frac{1}{4} & \lambda - \frac{1}{3} \end{bmatrix}$

Find the determinant of this matrix by taking the product of the upper left-to lower right diagonal and subtracting the product of the upper right-to-lower left diagonal:

$\det( \lambda I - A)$

$=\begin{vmatrix} \lambda - \frac{1}{2}& - \frac{2}{3}\ \ \frac{1}{4} & \lambda - \frac{1}{3} \end{vmatrix}$

$=\left ( \lambda - \frac{1}{2} \right )\left ( \lambda - \frac{1}{3} \right ) - \left ( - \frac{2}{3}\right ) \left ( \frac{1}{4} \right )$

$= \lambda ^{2} - \frac{5}{6} \lambda + \frac{1}{6} - \left ( - \frac{1}{6}\right )$

$= \lambda ^{2} - \frac{5}{6} \lambda + \frac{1}{3}$ ,

the correct choice.

← Didn't Know|Knew It →

0

Knew It