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If $A$ is $3 \times 3$ invertible matrix, then

show that for any scalar $k\neq 0$, $kA$ is also invertible and that $$(kA)^{-1}=\frac{A^{-1}}{k}$$

Chinny84
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Dhruv Raghunath
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  • To prove, just show that $$\left(\frac 1kA^{-1}\right)(kA) = I$$ –  Feb 21 '17 at 18:00
  • how does it show that they are equal – Dhruv Raghunath Feb 21 '17 at 18:05
  • Square matrix inverses are two-sided and unique. Proving the result in my above comment would show that $\frac 1kA^{-1}$ has the defining property of the inverse. Thus it is the two-sided, unique inverse. Thus $kA$ must be invertible. –  Feb 21 '17 at 18:08
  • what I'm not able to understand is how does the inverse of kA leads to k coming in the denominator – Dhruv Raghunath Feb 21 '17 at 18:20
  • @DhruvRaghunath Another way is to look this problem element-by-element. If $a_{ij}$ and $b_{ij}$ are the elements of $A$ and $A^{-1}$, then $\Sigma_j a_{ij}b_{jl} = \delta_{il}$. On the other hand, $ka_{ij}$ and $b_{ij}/k$ are the elements of $kA$ and $\frac1kA^{-1}$. $\Sigma_j ka_{ij}\frac1kb_{jl} = \delta_{il}$. – Guangliang Feb 21 '17 at 18:20
  • @DhruvRaghunath Because you need $k\cdot \frac 1k = 1$. –  Feb 21 '17 at 18:21
  • thnx for the help i just realized wt @Bye_World said XD. Thanks for the help to both of u – Dhruv Raghunath Feb 21 '17 at 18:24

1 Answers1

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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 $\left(\frac 1k {A^{-1}}\right)\cdot (kA)=Id=(kA)\cdot \left(\frac 1k {A^{-1}}\right)$.

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

So by the uniqueness of the inverse matrix we have that $\frac1k A^{-1}$ is the inverse of the matrix $kA$.

Giulio
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