Is it true that $\det(2*B)=160$, if $\det(B)=5$

Question

I want to clarify (because I can't find the answer) that I have to multiply diagonals, not multiply B by another B.

B is a 5x5 matrix. So if I put ones and one 5 in middle , I have $2*2*2*2*10 = 160\\$ $[1 1 1 1 1: 1 1 1 1 1: 1 1 5 1 1:1 1 1 1 1: 1 1 1 1 1]$

Answer 1

$2B$ is the matrix whose entries are twice the corresponding entries in $B$.

Since det is a multilinear function of the columns (or rows), $\det(2B)=2^n \det(B)$, when $B$ is $n \times n$. When $n=5$, this gives $\det(2B)=2^5 \det(B)=32 \det(B)$.

Answer 2

When you multiply a matrix by a scalar, you multiply every entry of the matrix by that scalar. In finding determinants you can factor that scalar out of each row, so you get the scalar to the power of the matrix dimension factored out. In your 5 by 5 matrix you get $2^5=32$ times the original determinant which gives you $32 det (B) = 32*5 =160$

For a scalar of $3$ and dimension of four you get $3^4=81$ times the original determinant. That is $det (3B)= 81 det(B)$.

Answer 3

First notice that $2B$ is defined to her $(2I_5)B.$ Since $2I_5$ has $5$ entires, all of which are 2 along the diagonal, then $$\det(2B)=\det(2I_5B)=\det(2I_5)\det(B)=(2^5)\cdot 5=32*5=160.$$ This generalizes to: $B$ is $n\times n,$ then $\det(2B)=2^n \det(B)=5\cdot 2^n.$