**WILL GIVE MEDALS, SUPER EASY QUESTION** https://gyazo.com/93f26f1944e2f6f171354d917159af95

Question

jim_thompson5910 · Answer

based on these properties http://math.kennesaw.edu/~plaval/math3260/det2.pdf if we multiply any column by a constant k, then det(A) = k*det(A)

anonymous · Answer

okay , how does this help me in this question?

anonymous · Answer

is the first one 12? I think i got the first one, second is tricky

jim_thompson5910 · Answer

$$
\det\left[
\begin{array}{ccc}
a & 6 & d\
b & 6 & e\
c & 6 & f\
\end{array}
\right]
=
\det\left[
\begin{array}{ccc}
a & 6*1 & d\
b & 6*1 & e\
c & 6*1 & f\
\end{array}
\right]
$$

$$
\det\left[
\begin{array}{ccc}
a & 6 & d\
b & 6 & e\
c & 6 & f\
\end{array}
\right]
=
6*\det\left[
\begin{array}{ccc}
a & 1 & d\
b & 1 & e\
c & 1 & f\
\end{array}
\right]
$$


$$
\det\left[
\begin{array}{ccc}
a & 6 & d\
b & 6 & e\
c & 6 & f\
\end{array}
\right]
=
6*2
$$


$$
\det\left[
\begin{array}{ccc}
a & 6 & d\
b & 6 & e\
c & 6 & f\
\end{array}
\right]
=
12
$$

anonymous · Answer

yes i got it! can you help me with the second one?

anonymous · Answer

i can't seem to get the second one for some reason

jim_thompson5910 · Answer

still thinking on that one

anonymous · Answer

i still can't figure it out lol

jim_thompson5910 · Answer

I think I might have something. One second

jim_thompson5910 · Answer

$$
\det\left[
\begin{array}{ccc}
a & -4 & d\
b & -7 & e\
c & -10 & f\
\end{array}
\right]
=
\det\left[
\begin{array}{ccc}
a & -3-1 & d\
b & -6-1 & e\
c & -9-1 & f\
\end{array}
\right]
$$


$$
\det\left[
\begin{array}{ccc}
a & -4 & d\
b & -7 & e\
c & -10 & f\
\end{array}
\right]
=
\det\left[
\begin{array}{ccc}
a & -3 & d\
b & -6 & e\
c & -9 & f\
\end{array}
\right]
-
\det\left[
\begin{array}{ccc}
a & 1 & d\
b & 1 & e\
c & 1 & f\
\end{array}
\right]
$$


$$
\det\left[
\begin{array}{ccc}
a & -4 & d\
b & -7 & e\
c & -10 & f\
\end{array}
\right]
=
\det\left[
\begin{array}{ccc}
a & -3*1 & d\
b & -3*2 & e\
c & -3*3 & f\
\end{array}
\right]
-
\det\left[
\begin{array}{ccc}
a & 1 & d\
b & 1 & e\
c & 1 & f\
\end{array}
\right]
$$


$$
\det\left[
\begin{array}{ccc}
a & -4 & d\
b & -7 & e\
c & -10 & f\
\end{array}
\right]
=
-3*\det\left[
\begin{array}{ccc}
a & 1 & d\
b & 2 & e\
c & 3 & f\
\end{array}
\right]
-
\det\left[
\begin{array}{ccc}
a & 1 & d\
b & 1 & e\
c & 1 & f\
\end{array}
\right]
$$


$$
\det\left[
\begin{array}{ccc}
a & -4 & d\
b & -7 & e\
c & -10 & f\
\end{array}
\right]
=
-3*(-5)-(2)
$$


$$
\det\left[
\begin{array}{ccc}
a & -4 & d\
b & -7 & e\
c & -10 & f\
\end{array}
\right]
=
13
$$

anonymous · Answer

perfect, thanks man

jim_thompson5910 · Answer

no problem