Question

find the eigen value and eigen vectors of the matrix.
(11 -4 -7
A= 7 -2 -5
10 -4 -6)​

Answer 1

Answer:

eigen value is 0,1,2

eigen vector at 0 is k(1 1 1)

eigen vector at 1 is (1 -1 2)

eigen vector at 2 is (1 ½ 1)

Answer 2

Tip:

• Eigenvalues are the special set of scalar values that is associated with the set of linear equations most probably in the matrix equation.
• Eigenvectors are the vectors (non-zero) that do not change the direction when any linear transformation is applied.

Explanation:

• We have $A=\left[\begin{array}{ccc}11&-4&-7\\7&-2&-5\\10&-4&-6\end{array}\right]$.
• We have to find the eigen value and eigen vector.

Step

Step 1 of 2:

For eigen values:

Characteristic equation : $|A-\lambda I|=0$

$|A-\lambda I|=\left|\begin{array}{ccc}11&-4&-7\\7&-2&-5\\10&-4&-6\end{array}\right|-\left|\begin{array}{ccc}\lambda&0&0\\0&\lambda&0\\0&0&\lambda\end{array}\right|$

             $=\left|\begin{array}{ccc}11-\lambda&-4&-7\\7&-2-\lambda&-5\\10&-4&-6-\lambda\end{array}\right|$

$|A-\lambda I|=(11-\lambda)[(-2-\lambda)(-6-\lambda)-20]+4[7(-6-\lambda)+50]-7[-28-10(-2-\lambda)]$

            $=(11-\lambda)[12+2\lambda+6\lambda+\lambda^2-20]+4[-42-7\lambda+50]-7[-28+20+10\lambda]\\\\=(11-\lambda)[\lambda^2+8\lambda-8]+4[8-7\lambda]-7[10\lambda-8]\\\\=3\lambda^2+96\lambda-88-\lambda^3+32-28\lambda-70\lambda+56\\\\=3\lambda^2-\lambda^3-2\lambda$

$|A-\lambda I|=0$

$\lambda^3-3\lambda^2+2\lambda=0$

$\lambda(\lambda^2-3\lambda+2)=0\\\\\lambda(\lambda^2-2\lambda-\lambda+2)=0\\\\\lambda(\lambda-1)(\lambda-2)=0$

So, $\lambda=0,1,2$

Eigen values are $0,1,2$            

Step 2 of 2:

Eigen Vector corresponding to Eigen Values:

(i) $\lambda=0$

$(A-\lambda I)X=0$

$\left[\begin{array}{ccc}11&-4&-7\\7&-2&-5\\10&-4&-6\end{array}\right]\left[\begin{array}{ccc}x\\y\\z\end{array}\right]=\left[\begin{array}{ccc}0\\0\\0\end{array}\right]$

$11x-4y-7z=0\\7x-2y-5z=0\\10x-4y-6z=0$

From solving the above equations we get,

$x=y=z=k$

The eigen vector is $\left[\begin{array}{ccc}1\\1\\1\end{array}\right]k$

(ii) $\lambda=1$

$(A-\lambda I)X=0$

$\left[\begin{array}{ccc}10&-4&-7\\7&-3&-5\\10&-4&-7\end{array}\right]\left[\begin{array}{ccc}x\\y\\z\end{array}\right]=\left[\begin{array}{ccc}0\\0\\0\end{array}\right]$

$10x-4y-7z=0\\7x-3y-5z=0\\10x-4y-7z=0$

Solving them we get,

$x=1,~y=-1,~z=2$

The eigen vector is $\left[\begin{array}{ccc}1\\-1\\2\end{array}\right]$

(iii) $\lambda=2$

$(A-\lambda I)X=0$

$\left[\begin{array}{ccc}9&-4&-7\\7&-4&-5\\10&-4&-8\end{array}\right]\left[\begin{array}{ccc}x\\y\\z\end{array}\right]=\left[\begin{array}{ccc}0\\0\\0\end{array}\right]$

$9x-4y-7z=0\\7x-4y-5z=0\\10x-4y-8z=0$

Solving them we get,

$x=1,~y=\frac{1}{2},~z=1$

The eigen vector is $\left[\begin{array}{ccc}1\\\frac{1}{2}\\1\end{array}\right]$

Final Answer:

Hence, the Eigen values are  $0,1,2$ and the Eigen vectors are  $\left[\begin{array}{ccc}1\\1\\1\end{array}\right]k$, $\left[\begin{array}{ccc}1\\-1\\2\end{array}\right]$, $\left[\begin{array}{ccc}1\\\frac{1}{2}\\1\end{array}\right]$