1. ## Find characteristic root.

Find the Characteristic root of the matrix
$
\begin{bmatrix}
2 & 2 & 1 \\
1 & 3 & 1 \\
1 & 2 & 2 \\
\end{bmatrix}$

2. Originally Posted by varunnayudu
Find the Characteristic root of the matrix
$
\begin{bmatrix}
2 & 2 & 1 \\
1 & 3 & 1 \\
1 & 2 & 2 \\
\end{bmatrix}$
Write the characteristic polynomial of the matrix. Call the matrix A, then the characteristic polynomial is $\text{det $$A - tI$$}$.

I take it that you are worried about the cubic nature of the polynomial and you dont know how to solve cubics.
In this particular case, observe that if you expand the determinant as:

$\text{det (A - tI)} = (2 - t)[(3 - t)(2 - t) - 2] - 2[(2 - t)- 1] + 1[2 - (3 - t)]$ $= (2 - t)[t^2 - 5t + 4] - 2[1 - t] + 1[t - 1]$

Now observe that $t - 1$ is a factor of all the three [] terms.
$(2 - t)[t^2 - 5t + 4] - 2[1 - t] + 1[t - 1] = (2 - t)[(t-1)(t-4)] + 2[t - 1] + 1[t - 1]$

I hope you can finish it now

3. Originally Posted by varunnayudu
Find the Characteristic root of the matrix
$
\begin{bmatrix}
2 & 2 & 1 \\
1 & 3 & 1 \\
1 & 2 & 2 \\
\end{bmatrix}$
Solve $\det(A - \lambda I) = 0$ where A is the above matrix.

If you're stuck doing this, please show what you've done and where you get stuck.

4. ## Is this right answer....

i have solved it but dont know if it is right or wrong

$
3 \lambda^2 -7 \lambda + 5
$

$
(3 \lambda^2 -7 \lambda + 4) + 1
$

$
(\lambda - 1)(3 \lambda - 4) +1
$

5. Originally Posted by varunnayudu
i have solved it but dont know if it is right or wrong

$
3 \lambda^2 -7 \lambda + 5
$

$
(3 \lambda^2 -7 \lambda + 4) + 1
$

$
(\lambda - 1)(3 \lambda - 4) +1
$
I dont know how you got the above equation. Anyway I have worked it out completely below. But I have to make a comment about the way you did the algebra above.

Why did you write $3 \lambda^2 -7 \lambda + 5$ as $(\lambda - 1)(3 \lambda - 4) + 1$? When you want to find a root for a polynomial, you want to factor it completely. You cant leave terms like + 1 hanging in the equality.

Note: You want factor it completely because you know that the product of 2 real numbers is 0 means at least one of them is 0.

Originally Posted by Isomorphism
Now observe that $t - 1$ is a factor of all the three [] terms.
$(2 - t)[t^2 - 5t + 4] - 2[1 - t] + 1[t - 1] = (2 - t)[(t-1)(t-4)] + 2[t - 1] + 1[t - 1]$
I will continue from here,

$(2 - t)[(t-1)(t-4)] + 2[t - 1] + 1[t - 1] =$ $(t-1)[(2 - t)(t-4) + 3] = (t-1)[-t^2 + 6t - 5] = (t - 1)[-(t-1)(t-5)]$

Thus $\det(A - tI) = 0 \Rightarrow (t - 1)[-(t-1)(t-5)] = 0 \Rightarrow t = 1,1,5$

6. ## yes it is correct

Originally Posted by Isomorphism
Write the characteristic polynomial of the matrix. Call the matrix A, then the characteristic polynomial is $\text{det $$A - tI$$}$.

I take it that you are worried about the cubic nature of the polynomial and you dont know how to solve cubics.
In this particular case, observe that if you expand the determinant as:

$\text{det (A - tI)} = (2 - t)[(3 - t)(2 - t) - 2] - 2[(2 - t)- 1] + 1[2 - (3 - t)]$ $= (2 - t)[t^2 - 5t + 4] - 2[1 - t] + 1[t - 1]$

Now observe that $t - 1$ is a factor of all the three [] terms.
$(2 - t)[t^2 - 5t + 4] - 2[1 - t] + 1[t - 1] = (2 - t)[(t-1)(t-4)] + 2[t - 1] + 1[t - 1]$

I hope you can finish it now

After the above operation ......

Taking the commom terms $(\lambda -1)$ out of the equation we get that answer

7. ## Thanks all

Thanks all for helping me
Cheers