Solved: In Exercises 33–36, verify that det EA = (det E)

Step 1 of 2

Consider the elementary matrix \(E=\left[\begin{array}{ll}
1 & 0 \\
k & 1
\end{array}\right]\) and a \(2 \times 2\) matrix \(A=\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]\).

Objective is to verify the property det( EA ) = (detE)(detA)  

The product of EA.

\(\begin{aligned}
E A & =\left[\begin{array}{ll}
1 & 0 \\
k & 1
\end{array}\right]\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right] \\
& =\left[\begin{array}{ll}
(1)(a)+(0)(c) & (1)(b)+(0)(d) \\
(k)(a)+(1)(c) & (k)(b)+(1)(d)
\end{array}\right] \\
E A & =\left[\begin{array}{cc}
a & b \\
k a+c & k b+d
\end{array}\right]
\end{aligned}\)

To find the determinant of the elementary matrix \(E=\left[\begin{array}{ll}
1 & 0 \\
k & 1
\end{array}\right]\).

\(\begin{aligned}
\operatorname{det} E & =\left|\begin{array}{ll}
1 & 0 \\
k & 1
\end{array}\right| \\
& =1-0 \\
& =1
\end{aligned}\)