Q.If $\begin{vmatrix} 6i & -3i & 1 \\ 4 & 3i & -1 \\ 20 & 3 & i \end{vmatrix} = x+iy$ then
A) x=3, y=1
B) x=1, y=3
C) x=0, y=3
D) x=0, y=0

A standard property of determinants: $\det A=\det A'$ for any square matrix $A$.

One of the fundamental properties of determinants is that the determinant of a matrix and its transpose are always equal, i.e.

$\det A=\det A'$ …

Use the identity $\det(A^{-1}) = 1/\det(A)$ and check it against the claim $\det(A^{-1})=\det(A)$.

For an invertible matrix $A$, $A A^{-1} = I$, so taking determinants:

$\det(A)\det(A^{-1}) = \det(I) = 1 \implies \det(A^{-1}) = \dfrac{1}{\det A}$

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"Singular" is defined as $|A|=0$.

A square matrix $A$ is called singular if and only if $|A| = 0$ (equivalently, it has no inverse). This is the …

Scaling every entry of a $2\times2$ matrix by 5 scales its determinant by $5^2=25$.

For a square matrix $A$ of order $n$ and scalar $k$, the rule is $|kA|=k^n|A|$ (each of the $n$ rows contributes one factor of $k$) …

For an $n\times n$ matrix, $|kA| = k^n|A|$.

…

Multiplying every entry of an $n\times n$ matrix by $K$ scales the determinant by $K^n$.

If $A$ is a square matrix of order $n$, then $|KA| = K^n |A|$, because $K$ can be pulled out of each of the $n$ rows, contributing a factor $K$ each time. …

$\det(A^{-1})=\dfrac{1}{\det(A)}$.

For any invertible matrix $A$, $AA^{-1}=I$. Taking determinants of both sides and using the property $\det(PQ)=\det(P)\det(Q)$:

$\det(A)\det(A^{-1})=\det(I)=1.$

Since $A$ is invertible, $\det(A)\ne0$, so we can divide:

$\det(A^{-1})=\dfrac{1}{\det(A)}.$ …