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Question 20 of 20 View Policies Current Attempt in Progress Find a 3 \times 3 matrix $A$ that has eigenvalues $\lambda = 0, 18, -18$ with corresponding eigenvectors $\begin{bmatrix} 0\\1\\-1 \end{bmatrix}, \begin{bmatrix} 1\\-1\\1 \end{bmatrix}, \begin{bmatrix} 0\\1\\1 \end{bmatrix}$ respectively.

          Question 20 of 20
View Policies
Current Attempt in Progress
Find a 3 \times 3 matrix $A$ that has eigenvalues $\lambda = 0, 18, -18$ with corresponding eigenvectors
$\begin{bmatrix} 0\\1\\-1 \end{bmatrix}, \begin{bmatrix} 1\\-1\\1 \end{bmatrix}, \begin{bmatrix} 0\\1\\1 \end{bmatrix}$
respectively.
        
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Question 20 of 20
View Policies
Current Attempt in Progress
Find a 3 ×3 matrix A that has eigenvalues λ = 0, 18, -18 with corresponding eigenvectors
< b m a t r i x >
, 
    < b m a t r i x >
, 
    < b m a t r i x >
respectively.

Added by Cory C.

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Elementary and Intermediate Algebra
Elementary and Intermediate Algebra
Alan S. Tussy, R. David Gustafson 5th Edition
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Question 20 of 20 View Policies Current Attempt in Progress Find a 3x3 matrix A that has eigenvalues 2, -1, and 4, respectively. Textbook and Media Save for Later Attempts: 0 of 3 used Submit Answer
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Transcript

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00:01 Hello students, in this question we can see that 4 charged particle form a square figure.
00:06 So here we have 1, 2, 3 and 4 charges with q1 is equal to plus q charge, q2 is equal to q3 is equal to q and we have q4 is equal to minus 9q.
00:22 So according to the question we have to make the force acting on the particle 1 should be 0.
00:27 So looking at the force acting on this particle as we say that f3 and f2 will be cancelling with each other having opposite direction.
00:39 So there will be the force acting from the particle so we have f4 and there will be f2.
00:50 So we can write f1 is equal to f14 minus f12.
00:56 So as this is related to this angle so 45 degree we have 1 by 4 pi epsilon 0 q into 9q divided by r14 square cos 45 minus 1 by 4 pi epsilon 0 q, q by r12 square...
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