Answer to Question #147781 in Linear Algebra for Saranya

Question

Find the sum and product of the matrix of the given eigen values of the matrix
A=2 1 2
1 3 1
2 2 -6

Answer 1

We have matrix: A=

$$\begin{pmatrix} 2 & 1 & 2 \\ 1 & 3 & 1 \\ 2 & 2 & -6 \end{pmatrix}$$

Find the eigenvalues of the matrix:

Associate matrix Vlasna vectors values and introduce a variable.

$\begin{pmatrix} 2- \lambda & 1 & 2 \\ 1 & 3-\lambda & 1 \\ 2 & 2 & -6 - \lambda \end{pmatrix}$

To do this, find the determinant of the matrix and equate this expression to zero.

$(2-\lambda)*((3 - \lambda)*(-6-\lambda)-2*1)-1*(1*(-6=\lambda)-2*2)+2*(1*1-(3-\lambda)*2)=0$

After transformations, we get:

$\lambda$³$+\lambda$² $-31\lambda +40 = 0$

After applying the rules, it is clear that the roots of the equation are not integers. Let's calculate the approximate $\lambda$ .

$\lambda$₁ $=-6,60$

$\lambda$₂ $=4,15$

$\lambda$₃ $=1,45$

Multiply the matrix A by the matrix of its eigenvalues:

$\begin{pmatrix} 2 & 1 & 2 \\ 1 & 3 & 1 \\ 2 & 2 & -6 \end{pmatrix}$ $*$ $\begin{pmatrix} -6,6 \\ 4,15 \\ 1,45 \end{pmatrix}$ $=$ $\begin{pmatrix} -6,15 \\ 7,3 \\ -13,6 \end{pmatrix}$

Maybe, if I misunderstood something, I add to the report: the sum and product of the matrix A itself:

The sum of the matrix: A + A = 2A

$2\begin{pmatrix} 2 & 1 & 2\\ 1 & 3 & 1\\ 2 & 2 & -6 \end{pmatrix}=\begin{pmatrix} 4 & 2 & 4 \\ 2 & 6 & 2\\ 4 & 4 & -12 \end{pmatrix}$

And product of the matrix: $A * A = A^2$

$A*A=\begin{pmatrix} 2 & 1 & 2 \\ 1 & 3 & 1 \\ 2 & 2 & -6 \end{pmatrix}*\begin{pmatrix} 2 & 1 & 2 \\ 1 & 3 & 1 \\ 2 & 2 & -6 \end{pmatrix}=\begin{pmatrix} 9 & 9 & -7 \\ 7 & 12 & -1 \\ -6 & -4 & 42 \end{pmatrix}$

Multiplication is performed for each component of the matrix separately:

$a^2_11=a_11*a_11 +a_12*a_21+a_13*a_31= 2*2+1*1+2*2=9$

$a^2_12=a_11*a_12 +a_12*a_22+a_13*a_32=2*1+1*3+2*2=2+3+4=9$

$a^2_13=a_11*a_13 +a_12*a_23+a_13*a_33= 2*2+1*1+2*(-6)=-7$

$a^2_21=a_21*a_11 +a_22*a_21+a_23*a_31=1*2+3*1+1*2=7$

$a^2_22=a_21*a_12 +a_22*a_22+a_23*a_32=1*1+3*3+1*2=12$

$a^2_23=a_21*a_13 +a_22*a_23+a_23*a_33=1*2+3*1+1*(-6)=-1$

$a^2_31=a_31*a_11 +a_32*a_21+a_33*a_31= 2 *2 +2*1+(-6)*2=-6$

$a^2_32=a_31*a_12 +a_32*a_22+a_33*a_32=2*1+2*3+(-6)*2=-4$

$a^2_33=a_31*a_13 +a_32*a_23+a_33*a_33=2*2+2*1+(-6)*(-6)=42$