Sure! Let's find the determinant of the matrix
[tex]\[
\begin{vmatrix}
a & b & c \\
d & e & f \\
5g & 5h & 5i
\end{vmatrix}
\][/tex]
given that
[tex]\[
\begin{vmatrix}
a & b & c \\
d & e & f \\
g & h & i
\end{vmatrix} = 7.
\][/tex]
To start, let's note that the third row of the matrix [tex]\(\begin{vmatrix} a & b & c \\ d & e & f \\ 5g & 5h & 5i \end{vmatrix}\)[/tex] has a common factor of 5 in each element. We can factor out the 5 from each element in the third row.
This gives us:
[tex]\[
\begin{vmatrix}
a & b & c \\
d & e & f \\
5g & 5h & 5i
\end{vmatrix} = 5 \cdot \begin{vmatrix}
a & b & c \\
d & e & f \\
g & h & i
\end{vmatrix}.
\][/tex]
Next, recognize that the remaining [tex]\( \begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix} = 7 \)[/tex].
So, extracting the factor [tex]\(5\)[/tex] from each element in the third row again, multiplying by [tex]\(5\)[/tex] gives the overall multiplicative effect as [tex]\(5^3\)[/tex]:
[tex]\[
\begin{vmatrix}
a & b & c \\
d & e & f \\
5g & 5h & 5i
\end{vmatrix} = 5 \cdot 5 \cdot 5 \cdot \begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix},
\][/tex]
which simplifies to:
[tex]\[
5^3 \cdot 7.
\][/tex]
Now, calculating [tex]\(5^3\)[/tex] results in:
[tex]\[
5^3 = 5 \times 5 \times 5 = 125.
\][/tex]
Therefore:
[tex]\[
\begin{vmatrix}
a & b & c \\
d & e & f \\
5g & 5h & 5i
\end{vmatrix}= 125 \times 7 = 875.
\][/tex]
Thus, the determinant of the matrix [tex]\(\begin{vmatrix} a & b & c \\ d & e & f \\ 5g & 5h & 5i \end{vmatrix}\)[/tex] is:
[tex]\[
\boxed{875}.
\][/tex]
