- Explicit X from the following equation:
$[adj(A)\cdot A - I](A - X)^{T}=I|A|$ Considering that $|A|\neq 1$.
- A is an invertible matrix and $|A|\neq 1$ , show that:
$[adj(A) - A^{-1}]^{-1} = \frac A{|A|-1} $
I know basic matrix properties such as $AA^{-1}=I$ , $A^{-1}=\frac 1{|A|}\cdot adj(A)$ , $(AB)^{-1}=B^{-1}A^{-1}$ , $(AB)^{T}=B^{T}A^{T}$ and $A\cdot adj(A)=adj(A)\cdot A=|A|I$ , which are probably required to solve this.
Thanks in advance!
You just need combine the basic matrix properties to solve your questions.
As you noticed, $adj(A) = |A| \cdot A^{-1}$. Hence, $$ adj(A) \cdot A = |A| \cdot A^{-1} \cdot A = |A| \cdot I $$ Note that $|A|$ is a scalar. So, after transposing both sides of your equation, you simply get: $$ \left(|A| - 1\right) A - \left(|A| - 1\right) X = |A|\cdot I. $$ Note here that transpose of a scalar is the same scalar and $I^T = I$. Now, divide both sides by $|A| - 1$ and you get an expression for $X$: $$ X = A - \frac{|A|}{|A| - 1}I $$
Same here, $adj(A) = |A| \cdot A^{-1}$ gives you right away the following: $$ \left(|A| \cdot A^{-1} - A^{-1}\right)^{-1} = \frac{1}{|A| - 1} \left(A^{-1}\right)^{-1} = \frac{1}{|A| - 1} A $$ Note that here we used the rule for scalar-matrix multiplication: Let $\alpha$ and $y$ are scalars and $X$ is a (invertible) matrix. Then $$ (\alpha \cdot X)^{y} = \alpha^y \cdot X^y $$
Note that in both questions condition $|A| - 1$ ensures that denominators in your expressions do not turn into zeros.