in Linear Algebra edited by
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4 votes
4 votes

​​​​​Which of the following is/are the eigenvector(s) for the matrix given below?

$$\begin{pmatrix} – 9 & – 6 & – 2 & – 4 \\ – 8 & – 6 & – 3 & – 1 \\ 20 & 15 & 8 & 5 \\ 32 & 21 & 7 & 12 \end{pmatrix}$$

  1. $\begin{pmatrix} – 1 \\ 1 \\ 0 \\ 1 \end{pmatrix}$
  2. $\begin{pmatrix} 1 \\ 0 \\ – 1 \\ 0 \end{pmatrix}$
  3. $\begin{pmatrix} – 1 \\ 0 \\ 2 \\ 2 \end{pmatrix}$
  4. $\begin{pmatrix} 0 \\ 1 \\ – 3 \\ 0 \end{pmatrix}$
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1 Answer

15 votes
15 votes
Best answer
Answer A,C, D

$Ax = \lambda x$
Learn very powerful and important way to multiply matrix and vector. $Ax$ is just linear combination of columns of $A$.. if you can understand this then solution of system of linear equation is just cakewalk for you. And complete linear algebra will come very intutive.

A.

$  \Bigg[\begin{smallmatrix}
-9 & -6 & -2 & -4
 \\ -8 & -6 & -3 & -1
 \\ 20 & 15 & 8 & 5
\\ 32 & 21 & 7 & 12
\end{smallmatrix} \Bigg]
\Bigg[\begin{smallmatrix}
-1\\ 1\\ 0 \\1
\end{smallmatrix} \Bigg] =
-1 \Bigg[\begin{smallmatrix}
-9\\ -8\\ 20 \\32
\end{smallmatrix} \Bigg] + 1
\Bigg[\begin{smallmatrix}
-6\\ -6\\ 15 \\21
\end{smallmatrix} \Bigg]
+ 0 \Bigg[\begin{smallmatrix}
-2\\ -3\\ 8 \\7 \end{smallmatrix} \Bigg]
+ 1 \Bigg[\begin{smallmatrix}
-4\\ -1\\ 5 \\12
\end{smallmatrix} \Bigg] =
\Bigg[\begin{smallmatrix}  -1\\1 \\ 0 \\1 \end{smallmatrix} \Bigg]$

B.

$  \Bigg[\begin{smallmatrix}
-9 & -6 & -2 & -4
 \\ -8 & -6 & -3 & -1
 \\ 20 & 15 & 8 & 5
\\ 32 & 21 & 7 & 12
\end{smallmatrix} \Bigg]
\Bigg[\begin{smallmatrix}
1\\ 0\\ -1 \\0
\end{smallmatrix} \Bigg] =
1 \Bigg[\begin{smallmatrix}
-9\\ -8\\ 20 \\32
\end{smallmatrix} \Bigg] + 0
\Bigg[\begin{smallmatrix}
-6\\ -6\\ 15 \\21
\end{smallmatrix} \Bigg]
-1 \Bigg[\begin{smallmatrix}
-2\\ -3\\ 8 \\7 \end{smallmatrix} \Bigg]
+ 0 \Bigg[\begin{smallmatrix}
-4\\ -1\\ 5 \\12
\end{smallmatrix} \Bigg] =
\Bigg[\begin{smallmatrix}  -7\\-5 \\ 12 \\25 \end{smallmatrix} \Bigg]$

C.

$  \Bigg[\begin{smallmatrix}
-9 & -6 & -2 & -4
 \\ -8 & -6 & -3 & -1
 \\ 20 & 15 & 8 & 5
\\ 32 & 21 & 7 & 12
\end{smallmatrix} \Bigg]
\Bigg[\begin{smallmatrix}
-1\\ 0\\ 2 \\2
\end{smallmatrix} \Bigg] =
-1 \Bigg[\begin{smallmatrix}
-9\\ -8\\ 20 \\32
\end{smallmatrix} \Bigg] + 0
\Bigg[\begin{smallmatrix}
-6\\ -6\\ 15 \\21
\end{smallmatrix} \Bigg]
+ 2 \Bigg[\begin{smallmatrix}
-2\\ -3\\ 8 \\7 \end{smallmatrix} \Bigg]
+ 2 \Bigg[\begin{smallmatrix}
-4\\ -1\\ 5 \\12
\end{smallmatrix} \Bigg] =
3\Bigg[\begin{smallmatrix}  -1\\ 0\\ 2\\2 \end{smallmatrix} \Bigg]
 $

 

D.

$  \Bigg[\begin{smallmatrix}
-9 & -6 & -2 & -4
 \\ -8 & -6 & -3 & -1
 \\ 20 & 15 & 8 & 5
\\ 32 & 21 & 7 & 12
\end{smallmatrix} \Bigg]
\Bigg[\begin{smallmatrix}
0\\ 1\\ -3 \\0
\end{smallmatrix} \Bigg] =
0 \Bigg[\begin{smallmatrix}
-9\\ -8\\ 20 \\32
\end{smallmatrix} \Bigg] + 1
\Bigg[\begin{smallmatrix}
-6\\ -6\\ 15 \\21
\end{smallmatrix} \Bigg]
-3 \Bigg[\begin{smallmatrix}
-2\\ -3\\ 8 \\7 \end{smallmatrix} \Bigg]
+ 0 \Bigg[\begin{smallmatrix}
-4\\ -1\\ 5 \\12
\end{smallmatrix} \Bigg]
= 3\Bigg[\begin{smallmatrix}  0\\ 1\\ -3 \\0 \end{smallmatrix} \Bigg]$
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4 Comments

Doing it traditionally takes more time, at least happened to me during the exam! :(

Wish I knew this column picture before the exam, super-duper intuitive & cool
0
0
as in option C and D both eigen values are same but the eigen vectors are different ,as per rule every eigen value possess an unique eigen vector but how the both option c and d are correct simultaneously
0
0

@Nalluri.Sasikanth Good observation.

BUT, You are wrong in saying that “ every eigen value possess an unique eigen vector ”.

Two different eigenvalues has two different eigenvector  -TRUE

But the  eigen value may also have different eigenvectors. 

Please watch this video.

I have given all PROOFS in the video. You will definitely like this.

2
2
Same eigen values can may or may not have same eigen vectors.

For different eigen values we will definitely have different eigen vectors
0
0