Math / Algebra

QuestionIf $A = \begin{bmatrix} b & 2 & -1 \\ 3 & 1 & -1 \\ -1 & 0 & 3 \end{bmatrix}$ and $\begin{bmatrix} 1 \\ \alpha \\ 1 \end{bmatrix}$ is an eigenvector of the matrix $A$ , then $b =$
Answer:

Studdy Solution

STEP 1

What is this asking? Given a matrix $A$ and one of its eigenvectors, we need to find the value of $b$ in the matrix. Watch out! Remember that if a vector $v$ is an eigenvector of a matrix $M$ , then $Mv = \lambda v$ for some scalar $\lambda$ (called the eigenvalue).
Don't mix up eigenvectors and eigenvalues!

STEP 2

1. Set up the eigenvector equation
2. Expand the matrix multiplication
3. Create a system of equations
4. Solve for $b$

STEP 3

We know that if $\begin{bmatrix} 1 \\ \alpha \\ 1 \end{bmatrix}$ is an eigenvector of matrix $A$ , then $A$ times the eigenvector equals the eigenvector scaled by some eigenvalue $\lambda$ .
Let's write that down: $A \begin{bmatrix} 1 \\ \alpha \\ 1 \end{bmatrix} = \lambda \begin{bmatrix} 1 \\ \alpha \\ 1 \end{bmatrix}$

STEP 4

Let's multiply the matrix $A$ by the eigenvector: $\begin{bmatrix} b & 2 & -1 \\ 3 & 1 & -1 \\ -1 & 0 & 3 \end{bmatrix} \begin{bmatrix} 1 \\ \alpha \\ 1 \end{bmatrix} = \begin{bmatrix} b + 2\alpha - 1 \\ 2 + \alpha \\ 2 \end{bmatrix}.$

STEP 5

So, our equation now looks like this: $\begin{bmatrix} b + 2\alpha - 1 \\ 2 + \alpha \\ 2 \end{bmatrix} = \begin{bmatrix} \lambda \\ \lambda \alpha \\ \lambda \end{bmatrix}.$

STEP 6

This gives us a system of equations by equating the corresponding components of the vectors: \begin{align} \label{eq:1} b + 2\alpha - 1 &= \lambda \\ 2 + \alpha &= \lambda \alpha \\ 2 &= \lambda\end{align}

STEP 7

From the third equation, we see that $\lambda = \mathbf{2}$ .
Let's substitute this into the second equation: $2 + \alpha = 2\alpha.$

STEP 8

Subtracting $\alpha$ from both sides gives us $\alpha = \mathbf{2}$ .

STEP 9

Now, let's substitute $\lambda = \mathbf{2}$ and $\alpha = \mathbf{2}$ into the first equation: $b + 2(\mathbf{2}) - 1 = \mathbf{2}.$

STEP 10

$b + 4 - 1 = 2$ $b + 3 = 2$ $b = 2 - 3$ $b = \mathbf{-1}$

STEP 11

Therefore, $b = -1$ .

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QuestionIf $A = \begin{bmatrix} b & 2 & -1 \\ 3 & 1 & -1 \\ -1 & 0 & 3 \end{bmatrix}$ and $\begin{bmatrix} 1 \\ \alpha \\ 1 \end{bmatrix}$ is an eigenvector of the matrix $A$ , then $b =$
Answer:

Studdy Solution

STEP 1

STEP 2

1. Set up the eigenvector equation
2. Expand the matrix multiplication
3. Create a system of equations
4. Solve for $b$

STEP 3

STEP 4

Let's multiply the matrix $A$ by the eigenvector: $\begin{bmatrix} b & 2 & -1 \\ 3 & 1 & -1 \\ -1 & 0 & 3 \end{bmatrix} \begin{bmatrix} 1 \\ \alpha \\ 1 \end{bmatrix} = \begin{bmatrix} b + 2\alpha - 1 \\ 2 + \alpha \\ 2 \end{bmatrix}.$

STEP 5

So, our equation now looks like this: $\begin{bmatrix} b + 2\alpha - 1 \\ 2 + \alpha \\ 2 \end{bmatrix} = \begin{bmatrix} \lambda \\ \lambda \alpha \\ \lambda \end{bmatrix}.$

STEP 6

This gives us a system of equations by equating the corresponding components of the vectors: \begin{align} \label{eq:1} b + 2\alpha - 1 &= \lambda \\ 2 + \alpha &= \lambda \alpha \\ 2 &= \lambda\end{align}

STEP 7

From the third equation, we see that $\lambda = \mathbf{2}$ .
Let's substitute this into the second equation: $2 + \alpha = 2\alpha.$

STEP 8

Subtracting $\alpha$ from both sides gives us $\alpha = \mathbf{2}$ .

STEP 9

Now, let's substitute $\lambda = \mathbf{2}$ and $\alpha = \mathbf{2}$ into the first equation: $b + 2(\mathbf{2}) - 1 = \mathbf{2}.$

STEP 10

$b + 4 - 1 = 2$ $b + 3 = 2$ $b = 2 - 3$ $b = \mathbf{-1}$

STEP 11

Therefore, $b = -1$ .

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QuestionIf A=[b2−131−1−103]A = \begin{bmatrix} b & 2 & -1 \\ 3 & 1 & -1 \\ -1 & 0 & 3 \end{bmatrix}A=⎣⎡​b3−1​210​−1−13​⎦⎤​ and [1α1]\begin{bmatrix} 1 \\ \alpha \\ 1 \end{bmatrix}⎣⎡​1α1​⎦⎤​ is an eigenvector of the matrix AAA, then b=b =b=Answer:

Studdy Solution

STEP 1

STEP 2

1. Set up the eigenvector equation2. Expand the matrix multiplication3. Create a system of equations4. Solve for bbb

STEP 3

STEP 4

STEP 5

So, our equation now looks like this: [b+2α−12+α2]=[λλαλ]. \begin{bmatrix} b + 2\alpha - 1 \\ 2 + \alpha \\ 2 \end{bmatrix} = \begin{bmatrix} \lambda \\ \lambda \alpha \\ \lambda \end{bmatrix}. ⎣⎡​b+2α−12+α2​⎦⎤​=⎣⎡​λλαλ​⎦⎤​.

STEP 6

This gives us a system of equations by **equating** the corresponding components of the vectors: \begin{align*} \label{eq:1} b + 2\alpha - 1 &= \lambda \\ 2 + \alpha &= \lambda \alpha \\ 2 &= \lambda\end{align*}

STEP 7

From the **third equation**, we see that λ=2\lambda = \mathbf{2}λ=2.Let's **substitute** this into the **second equation**: 2+α=2α. 2 + \alpha = 2\alpha. 2+α=2α.

STEP 8

Subtracting α\alphaα from both sides gives us α=2\alpha = \mathbf{2}α=2.

STEP 9

Now, let's **substitute** λ=2\lambda = \mathbf{2}λ=2 and α=2\alpha = \mathbf{2}α=2 into the **first equation**: b+2(2)−1=2. b + 2(\mathbf{2}) - 1 = \mathbf{2}. b+2(2)−1=2.

STEP 10

b+4−1=2 b + 4 - 1 = 2 b+4−1=2 b+3=2 b + 3 = 2 b+3=2 b=2−3 b = 2 - 3 b=2−3b=−1 b = \mathbf{-1} b=−1

STEP 11

Therefore, b=−1b = -1b=−1.

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QuestionIf $A = \begin{bmatrix} b & 2 & -1 \\ 3 & 1 & -1 \\ -1 & 0 & 3 \end{bmatrix}$ and $\begin{bmatrix} 1 \\ \alpha \\ 1 \end{bmatrix}$ is an eigenvector of the matrix $A$ , then $b =$
Answer:

1. Set up the eigenvector equation
2. Expand the matrix multiplication
3. Create a system of equations
4. Solve for $b$

So, our equation now looks like this: $\begin{bmatrix} b + 2\alpha - 1 \\ 2 + \alpha \\ 2 \end{bmatrix} = \begin{bmatrix} \lambda \\ \lambda \alpha \\ \lambda \end{bmatrix}.$

This gives us a system of equations by equating the corresponding components of the vectors: \begin{align} \label{eq:1} b + 2\alpha - 1 &= \lambda \\ 2 + \alpha &= \lambda \alpha \\ 2 &= \lambda\end{align}

From the third equation, we see that $\lambda = \mathbf{2}$ .
Let's substitute this into the second equation: $2 + \alpha = 2\alpha.$

Subtracting $\alpha$ from both sides gives us $\alpha = \mathbf{2}$ .

Now, let's substitute $\lambda = \mathbf{2}$ and $\alpha = \mathbf{2}$ into the first equation: $b + 2(\mathbf{2}) - 1 = \mathbf{2}.$

$b + 4 - 1 = 2$ $b + 3 = 2$ $b = 2 - 3$ $b = \mathbf{-1}$

Therefore, $b = -1$ .