Question

6. IfA= 3 X 3 invertible matrix, then show that for any scalar k (non-zero), kA is
invertible and (kA)^-1 = 1/kA^-1​

Answer 1

Answer:

Derivation.

Step-by-step explanation:

t 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 (1kA−1)⋅(kA)=Id=(kA)⋅(1/kA−1).

We have (1/kA−1)⋅(kA)=(1/kk)⋅(A−1A)=1⋅Id=Id and (k/A)⋅(1/kA−1)=(k1/k)⋅(AA−1)=1⋅Id=Id.

So by the uniqueness of the inverse matrix /we have that 1/kA−1 is the inverse of the matrix kA