Question

find latent roots and vectors for the matrix
[1 2 3
0 2 3
0 0 2]​

Answer 1

Answer:

This implies that

x2+2ax=4x−4a−13

or

x2+2ax−4x+4a+13=0

or

x2+(2a−4)x+(4a+13)=0

Since the equation has just one solution instead of the usual two distinct solutions, then the two solutions must be same i.e. discriminant = 0.

Hence we get that

(2a−4)2=4⋅1⋅(4a+13)

or

4a2−16a+16=16a+52

or

4a2−32a−36=0

or

a2−8a−9=0

or

(a−9)(a+1)=0

So the values of a are −1 and 9.

Answer 2

Given: A matrix is given

$A=\left[\begin{array}{ccc}1&2&3\\0&2&3\\0&0&2\end{array}\right]$

To Find:

Latent roots of matrix?

Step-by-step explanation:

Given matrix A.

• Firstly we find characteristic eqation of matrix A=

                   $\begin{vmatrix}A-\lambda I\end{vmatrix}=0\\\Rightarrow\begin{vmatrix}\left[\begin{array}{ccc}1&2&3\\0&2&3\\0&0&2\end{array}\right] -\left[\begin{array}{ccc}\lambda&0&0\\0&\lambda&0\\0&0&\lambda\end{array}\right] \end{vmatrix} =0\\\\\Rightarrow\begin{vmatrix}\left[\begin{array}{ccc}1-\lambda&2&3\\0&2-\lambda&3\\0&0&2-\lambda\end{array}\right] \end{vmatrix}=0\\\\\Rightarrow 1-\lambda\begin{vmatrix}2-\lambda &3 &\\0&2-\lambda \end{vmatrix}-2\begin{vmatrix}0&3\\0&2-\lambda\end{vmatrix}$

            $\Rightarrow (1-\lambda)(2-\lambda)^2-2(0)+3(0)\\\\\Rightarrow-\lambda^3+5\lambda^2-8\lambda+4$

• Now we need to find root of above equation and it willbe our latent roots.

      After solving we get roots of euation are $\lambda=1 ,\lambda=2,\lambda=2$

Thus, lantent roots of given Matrix is 1,2.