Question

Question: Verify that \(\det AB = \left( {\det A} \right)\left( {\det B} \right)\) for the matrices in Exercises 37 and 38. (Do not use Theorem 6.)

37. \(A = \left[ {\begin{array}{*{20}{c}}3&0\\6&1\end{array}} \right],B = \left[ {\begin{array}{*{20}{c}}2&0\\5&4\end{array}} \right]\)

Short answer

It is verified that \(\det AB = \left( {\det A} \right)\left( {\det B} \right)\).

Step-by-step solution

Determine matrix \(AB\)

Compute matrix \(AB\) as shown below:

\(\begin{aligned}{}AB &= \left[ {\begin{aligned}{{}{}}3&0\\6&1\end{aligned}} \right]\left[ {\begin{aligned}{{}{}}2&0\\5&4\end{aligned}} \right]\\ &= \left[ {\begin{aligned}{{}{}}{6 + 0}&{0 + 0}\\{12 + 5}&{0 + 4}\end{aligned}} \right]\\ &= \left[ {\begin{aligned}{{}{}}6&0\\{17}&4\end{aligned}} \right]\end{aligned}\)

Verify whether \(\det AB = \left( {\det A} \right)\left( {\det B} \right)\)

If A is a triangular matrix, then according to theorem 2,\(\det A\) is the product of the entries on its main diagonal.

The determinants of matrices\(A\) and \(B\)are shown below:

\(\begin{aligned}{}\det A &= \left| {\begin{aligned}{{}{}}3&0\\6&1\end{aligned}} \right|\\ &= 3\\\det B &= \left| {\begin{aligned}{{}{}}2&0\\5&4\end{aligned}} \right|\\ &= 8\end{aligned}\)

The determinant of matrix \(AB\)is shown below:

\(\begin{aligned}{}\det AB &= \left| {\begin{aligned}{{}{}}6&0\\{17}&4\end{aligned}} \right|\\ &= 24\\ &= 3 \times 8\\ &= \left( {\det A} \right)\left( {\det B} \right)\end{aligned}\)

Hence, it is verified that\(\det AB = \left( {\det A} \right)\left( {\det B} \right)\).