Question

If A is an idempotent matrix satisfying
$(I-0.4A{)}^{-1}=I-\alpha A$
where I is the unit matrix of the same order as that of A then the value of $\alpha$ is equal to

• A. $\alpha =-\frac{2}{3}$
• B. $\alpha =\frac{4}{3}$
• C. $\alpha =\frac{2}{3}$
• D. $\alpha =\frac{3}{2}$

Answer 1

The correct option is A $\alpha =-\frac{2}{3}$

$A$ is an idempotent i.e, $A^2=A$

Given $(I-0.4A{)}^{-1}=(I-\alpha A)$

$(I-0.4A)(I-0.4A{)}^{-1}=(I-0.4A\left)\right(I-\alpha A)$

$I=(I-0.4A)(I-\alpha A)$

$I=I-\alpha AI-0.4AI+0.4\alpha A^2$

$I=I-\alpha A-0.4A+0.4\alpha A$

$\Rightarrow \alpha A+0.4A=0.4\alpha A$

$\Rightarrow (\alpha +0.4)A=0.4\alpha A$

$\Rightarrow \alpha +0.4=0.4\alpha$

$\Rightarrow \alpha =\frac{-0.4}{0.6}=\frac{-2}{3}$