find the determinant of matrix 2

Question

if the determinant of A is given by $ \begin{vmatrix} a & c \\ b & d \\ \end{vmatrix}=5$

find\begin{vmatrix} 2a & 3d \\ 2b & 3c \\ \end{vmatrix}

Answer 1

You have $$ \det\begin{bmatrix} 2a & 3d \\ 2b & 3c \end{bmatrix}= 6\det\begin{bmatrix} a & d \\ b & c \end{bmatrix} $$ so the problem is whether we can deduce the last determinant from $$ \det\begin{bmatrix} a & c \\ b & d \end{bmatrix} $$ The answer is no.

You can't even predict whether the first matrix is invertible or not, given that the second matrix is invertible. Just consider the case $a=1$, $b=0$, $c=0$ and $d=5$.

I claim that we can suitably choose $a,b,c,d$ so that $$ \det\begin{bmatrix} a & d \\ b & c \end{bmatrix}=k $$ where $k$ is arbitrary. The equations are $$ \begin{cases} ad-bc=5 \\[4px] ac-bd=k \end{cases} $$ Multiply the first equation by $d$ and the second one by $c$ and subtract: we get $$ ad^2-ac^2=5d-kc $$ that is $$ a=\frac{5d-kc}{c^2-d^2} $$ so, as long as $c\ne d$ (which would make the problem trivial) and $c\ne -d$, we have the thesis.

If $d=-c\ne0$, one matrix is obtained from the other by multiplying the second column by $-1$, so the determinant changes sign.