How do I solve such Matrix problems in Linear Algebra?

Question

I was given a problem in which a matrix $A$ was specified along with its determinant value, now the determinant of another matrix $B$ was asked to be found out whose indices were scalar multiplies of $A$. What I did was to factor out the scalar from the matrix and substitute the value of matrix $A$. However to my great surprise I found out that each row had to be factored out and the scalars would be indexed to its power by the number of rows and then the determinant operation be applied. This is something that took me aback and left me absolutely confused. I am looking for a clear cut and elimination by contradiction (yet trivial) explanation to this.

Answer 1

As a brief example, consider

$$\begin{align}A&=\begin{pmatrix}a & b\\ c& d\end{pmatrix}\\[10 pt] \det(A)&=ad-bc \\[10pt] B&=nA \\[10 pt] &=\begin{pmatrix}na & nb\\ nc& nd\end{pmatrix} \\[10 pt] \det(B)&=nand-nbnc \\[10pt] &=n^2ad-n^2bc\\[10 pt] &=n^2(ad-bc)\\[10 pt] &=n^2\det(A) \end{align} $$ In general this is true: for an $p$-dimensional matrix, $\det(nA)=n^p\det(A)$. This is because we multiply together $p$ values from the matrix when computing the determinant, and thus we pick up $p$ copies of $n$. I suggest playing around with this idea if you are still confused.

Answer 2

One of the basic properties of determinants that you should have learned is that they are multilinear functions of the columns (or rows) of a matrix. Indeed, some treatments start with this and a few additional properties as the definition of a determinant and then derive from them the formulas that you’re likely familiar with.

Specifically, if $A_i$ denotes the $i$th column of the $n\times n$ matrix $A$, then (with a slight abuse of notation) we can write $\det A = \det(A_1,\dots,A_n)$. The multilinearity property says among other things that for every $1\le k\le n$, $$\det(A_1,\dots,cA_k,\dots,A_n)=c\det(A_1,\dots,A_k,\dots,A_n).$$ That is, multiplying a single column of $A$ by $c$ multiplies its determinant by $c$ as well. With $cA$, you’ve multiplied every column of $A$ by $c$, therefore $\det(cA)=c^n\det(A)$.