# Find the inverse of this diagonal 4 x 4 matrix

$\begin{bmatrix} k_1 & 0 & 0 & 0 \\ 0 & k_2 & 0 & 0 \\ 0 & 0 & k_3 & 0 \\ 0 & 0 & 0 & k_4 \\ \end{bmatrix}$

I'm thinking you just have to convert this into a reduced-row echelon form like this:

$\left[ \begin{array}{cccc|cccc} k_1 & 0 & 0 & 0 & 1 & 0 & 0 & 0\\ 0 & k_2 & 0 & 0 & 0 & 1 & 0 & 0\\ 0 & 0 & k_3 & 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & k_4 & 0 & 0 & 0 & 1\\ \end{array} \right]$

So the result for the inverse would be this:
$\begin{bmatrix} \frac1{k_1} & 0 & 0 & 0 \\ 0 & \frac1{k_2} & 0 & 0 \\ 0 & 0 & \frac1{k_3} & 0 \\ 0 & 0 & 0 & \frac1{k_4} \\ \end{bmatrix}$

Is this right, or is there something I'm missing?

• It is right. To check, just multiply the matrices, you get the identity matrix. Well done. – user405743 Jun 25 '17 at 4:26
• Ok, thanks @Dragon – Bucephalus Jun 25 '17 at 4:26
• Thanks for the edit @Daichi – Bucephalus Jun 25 '17 at 4:34
• Further exercise: generalize this result to an $n \times n$ matrix. – Duncan Ramage Jun 25 '17 at 4:52
• As Duncan was saying, this is a general result that the inverse of a diagonal matrix (it it is invertible) is another diagonal matrix having entries reciprocal of the entries of the given matrix. – StubbornAtom Jun 25 '17 at 7:29

Because the matrix is invertible we can assume that all the $k_i$'s are non-zero. Your answer is correct. Elegant and sweet.