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Question

Let A be a 3*3 matrix whose characteristics roots are 3,2,-1. If $B=A^2-A$ then |B|=?

a)24

b)-2

c)12

d)-12

Please explain in detail.

Comments

24

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Characterstics Roots means Eigen values

Therefore for A Eigen values are : 3 ,2,-1

for $A^{2}$ eigen values will be: $3^{2}$ ,$2^{2}$ ,$(-1)^{2}$

Eigen values for $A^{2}$ = 9 ,4 , 1

Eigen values for A = 3 ,2, -1

Eigen values for B(i.e $A^{2}$-A) =Eigen values for $A^{2}$ –  Eigen values for A

                                                     =(9-3),(4-2),(1-(-1))

                                                     =6,2,2

Determinant value of B = Product of Eigen values

                                      = (6*2*2)

                                      =24

Answer 1

Let

A = $\begin{bmatrix} 3 &0 &0 \\ 0&2 &0 \\ 0&0 &-1 \end{bmatrix}$

A$^{2}$ = $\begin{bmatrix} 9 &0 &0 \\ 0&4 &0 \\ 0&0 &1 \end{bmatrix}$

B = A$^{2}$ - A

    =$\begin{bmatrix} 6 &0 &0 \\ 0&2 &0 \\ 0&0 &2 \end{bmatrix}$

Thus |B| = 24

Answer 2

Characteristic roots are nothing but eigenvalues. So the eigenvalues of matrix A are $3,2,-1$.

Also, $determinant\ of\ matrix\ B = Product\ of\ its\ eigenvalues.$

Now eigenvalues of matrix $B$ can be found by substituting the corresponding eigenvalues of matrix $A$.

$B= A^2-A$

$1.\ B = 3^2-3=9-3=6$

$2.\ B=2^2-2=4-2=2$

$3.\ B=-1^2-(-1)=1+1=2$

$|B|=6*2*2=24\ (Answer)$