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If A = $\begin{bmatrix} -1 & 1 & 0 \\ 0 & 2 &-2 \\ 0& 0 & 3 \end{bmatrix}$ then trace of the matrix 3A+ adj A is ____

asked in Linear Algebra by (125 points) | 57 views
+2

you can solve this question quickly without any matrix multiplication.

Since,  given matrix 'A' is a lower triangular matrix. 

So, |A| = -6 (multiplication all principal diagonals)

and eigen values of A will be -1 , 2 , 3 (i.e. all principal diagonals)

Now , since , Adj(A) = |A| * A-1 and eigen values of A-1  are the reverse of eigen values of A.

So, eigenvalues of A-1 will be $\frac{1}{\lambda _{i}}$ for i = 1,2,3...

So, Eigen values of Adj(A) = $\frac{\left | A \right |}{Eigen \;values\; of \; matrix\; A}$

So, here , Eigen values of Adj(A) will be :- $\frac{-6}{-1}$ , $\frac{-6}{2}$ , $\frac{-6}{3}$ . ie. 6,-3,-2.

Since , AX =  $\lambda$X and A2X =  $\lambda$2X

So, if $\lambda$ is an eigen value of A then $\lambda$2 will be an eigen value of A2

So, Eigen values of A2 will be :-  1, 4, 9

Now , Eigen values fro matrix 3A+ adj (A) will be :-

1) 3*1 + 6 = 9

2) 3*4 + (-3) = 9

3) 3*9 + (-2) = 25

Since , Trace of a matrix = Summation of its eigen values

So here ,  Trace = 9 + 9 + 25 = 43

 

1 Answer

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

$A^2 =$ $\begin{bmatrix} 1 & 1 & -2 \\ 0 & 4 &-10 \\ 0& 0 & 9 \end{bmatrix}$

$adjA = $$\begin{bmatrix} 6 & 0 & 0 \\ -3 & -3 & 0 \\ -2& -2 & -2 \end{bmatrix}^T = $$\begin{bmatrix} 6 & -3 & -2 \\ 0 & -3 & -2 \\  0& 0 & -2 \end{bmatrix} $

$3A^2 + adj A = $ = $\begin{bmatrix} 3 & 3 & -6 \\ 0 & 12 &-30 \\ 0& 0 & 27 \end{bmatrix} + $$\begin{bmatrix} 6 & -3 & -2\\ 0 & -3 & -2 \\  0& 0 & -2 \end{bmatrix} $ = $\begin{bmatrix} 9 & 0 & -8 \\ 0 & 9 & -32 \\  0& 0 & 25 \end{bmatrix} $

Trace = $9 + 9 + 25 = 43$
answered by Boss (35.4k points)

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