1
$\begingroup$

How does the inverse diagonal matrix looks like - D$(3,3)$?

If I have diagonal matrix like this:

$$\begin{bmatrix} 5 & 0 & 0 \\ 0 & 7 & 0 \\ 0 & 0 & 6\end{bmatrix}$$

Is the inverse of this matrix is all non zero element raised by power of $-1$?

$$\begin{bmatrix} \frac15 & 0 & 0 \\ 0 & \frac17 & 0 \\ 0 & 0 & \frac16\end{bmatrix}$$

$\endgroup$

closed as off-topic by José Carlos Santos, 5xum, Ben, Claude Leibovici, mechanodroid Nov 2 '17 at 12:23

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – José Carlos Santos, 5xum, Ben, Claude Leibovici, mechanodroid
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ Yes.$~~~~~~~~~$ $\endgroup$ – JMoravitz Nov 1 '17 at 19:14
  • $\begingroup$ As the others have said, yes. However, remember that this does not hold for all matrices. In general, you cannot just raise each element to the power of $-1$. But it is true for diagonal matrices. $\endgroup$ – Eff Nov 1 '17 at 19:20
1
$\begingroup$

Yes.

We can check that it is indeed true by multiplying the two matrices together to see if we can get the identity matrix.

Multiplication of diagonal matrices $A$ and $B$ gives us another diagonal matrix $C$ where $$C_{ii}=A_{ii}B_{ii}.$$

$\endgroup$
0
$\begingroup$

Yes, the multiplicative inverse of a diagonal matrix is a diagonal matrix with the reciprocal of the diagonal numbers on its diagonal. The multiplicative inverse of $\begin{pmatrix}a & 0 & 0 \\ 0 & b & 0 \\0 & 0 & c\end{pmatrix}$ is $\begin{pmatrix}\frac{1}{a} & 0 & 0 \\ 0 & \frac{1}{b} & 0 \\0 & 0 & \frac{1}{c}\end{pmatrix}$

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.