Question

If $A$ is a scalar matrix with scalar $k\ne 0$, of order $3$, then $k{A}^{-1}$ is:

• A. $\frac{1}{k}I$
• B. $\frac{1}{k^2}I$
• C. $k^{2}I$
• D. $\frac{1}{k^3}I$

Answer 1

The correct option is C $k^{2}I$

It is well-known that if you find an inverse for a matrix, that inverse matrix will be unique.

So what we have to do is to show that $\left(k{A}^{-1}\right)\cdot \left(kA\right)=Id=\left(kA\right)\cdot \left(k{A}^{-1}\right)$.

We have $\left(k{A}^{-1}\right).\left(kA\right)=\left(kk\right)\cdot (A^{-1}.A)=Id=Id$ and $\left(kA\right)\cdot \left(k{A}^{-1}\right)=\left(kk\right)\cdot \left(A{A}^{-1}\right)=Id$.

So by the uniqueness of the inverse matrix we have that $k{A}^{-1}$ is the inverse of the matrix $kA$.