Find the determinants of the given matrices Consider scalars $a,b,c,d,e,f$ such that
$\det\left( \begin{array}{ccc}
a & 1 & d\\
b & 1 & e\\
c & 1 & f\\ \end{array} \right) = 7$ and $\det\left( \begin{array}{ccc}
a & 1 & d\\
b & 2 & e\\
c & 3 & f\\ \end{array} \right) = 11$
What is $\det\left( \begin{array}{ccc}
a & 3 & d\\
b & 3 & e\\
c & 3 & f\\ \end{array} \right)$ and $\det\left( \begin{array}{ccc}
a & 3 & d\\
b & 4 & e\\
c & 5 & f\\ \end{array} \right)$?
For the first one, I was thinking that the determinant would be $3\cdot 7 = 21$, and I think the determinant of the second one would still just be $11$ (since it's just a multiple of the second determinant matrix). Does this seem right? How can I solve these kinds of problems?

Givan an $n \times n$ matrix $A = (a_1, a_2, \ldots, a_{n})$, where $a_{i}$ are column vectors. There are identities
$$det(a_1, \ldots, ca_i, \ldots, a_n) = cdet(A)$$ and
$$det(a_1, \ldots, a_{i-1}, a_i + v, a_{i+1} \ldots, a_n) = det(A) + det(a_1, \ldots, a_{i-1}, v, a_{i+1}, \ldots, a_n)$$, where $v$ is a column vector.
Therefore, returning to your questions,
$$det\begin{pmatrix}
a & 3 & d\\
b & 3 & e\\
c & 3 & f
\end{pmatrix} = 3 \cdot 7 = 21$$
and
$$det\begin{pmatrix}
a & 3 & d\\
b & 4 & e\\
c & 5 & f
\end{pmatrix} = 2 \cdot 7 + 11 = 25.$$
