retagged by
422 views

1 Answer

2 votes
2 votes

WRONG APPROACH

we know that $A^{-1} = \frac{1}{|A|} (adj(A))$

Multiply by $A$ on both sides

$AA^{-1} = \frac{1}{|A|}A(adj(A))$ $\Leftrightarrow$ $|A|I = A(adj(A))$

from the matrix given, we conclude that $|A|= \frac{1}{5^{3}}$

We want to find the value of $\vert(5A)^{-1}\vert$.


$\vert(5A)^{-1}\vert =\frac{1}{\vert 5A\vert}=   $  $ \frac{1}{5^{3}\vert A \vert}$ chip in the value of $\vert A\vert$ from above.

Therefore
$ \vert (5A)^{-1} \vert = 1$

CORRECT APPROACH

There is a slight in the question , in RHS that is not a determinant but a matrix. So from now on in this solution we will consider it as a matrix.

$A^{-1} = \frac{1}{|A|} (adj(A))$

Multiply by $A^{-1}$ on both sides

$AA^{-1} = \frac{1}{|A|}A(adj(A))$ $\Leftrightarrow$ $|A|I = A(adj(A))$

From the given matrix, we conclude that$|A|I=\frac{1}{5}I \Rightarrow$ $|A|=\frac{1}{5}$

We want to find the value of $\vert(5A)^{-1}\vert$.

$\vert(5A)^{-1}\vert =\frac{1}{\vert 5A\vert}=   $  $ \frac{1}{5^{3}\vert A \vert}$ chip in the value of $\vert A\vert$ from above.

Therefore $ \vert (5A)^{-1} \vert = \frac{1}{5^2}$

Thank you @AGNIDEB+MUKHERJEE

edited by

Related questions

1 votes
1 votes
0 answers
1
0 votes
0 votes
1 answer
2
srestha asked May 27, 2019
1,236 views
$1)$ How to find a matrix is diagonalizable or not?Suppose a matrix is $A=\begin{bmatrix} \cos \Theta &\sin \Theta \\ \sin\Theta & -\cos\Theta \end{bmatrix}$Is it diagona...