Invertible complex matrices

Question

I have 2 complex matrices:

$$A(p) = \pmatrix{ 1 & 0 & i\\ p & i & p\\ 1-i & p & 0\\ }$$

$$B(p) = \pmatrix{ p & -1 & -1\\ p-i & 1 & 1\\ p+i & 1 & 0\\ }$$

I have to decide if they are invertible for

$A(p), $

$B(p),$

$(A(p)B(p)),$

$((A(p))^{10}(B(p))^3).$

What is the best method to solve it? Calculate theirs determinant? And how to deal with $(A(p)B(p))$ and $((A(p))^{10}(B(p))^3).$

Answer 1

The determinants are, respectively

$$\det\left(A(p)\right)=(-1+i)(p^2-1)\\ \det\left(B(p)\right)=-2p+i$$

Now

• $A(p)$ is invertible iff $p^2\neq+1$

• $B(p)$ is invertible iff $p\neq \frac i2$

• $A(p)B(p)$ is invertible iff $0\neq\det(AB)=\det(A)\det(B)$. Thus, $AB$ is invertible iff $p\neq \frac i2\wedge p\neq \pm 1$

• Similarly for $A^{10}B^3$

Answer 2

$\det(A)=(1-i)(1-p^2)$ and $\det(B)=i-2p$, so if $p$ is real but not equal to $\pm 1$. Matrices $A$ and $B$ are non-singular so they are invertable. Since $\det (AB)= \det(A) \det(B)$, $AB$ is also invertible. Next, $\det A^m= (\det(A))^m$, so the other product is also invertable.