Question

Let \(A = \left[ {\begin{aligned}{*{20}{c}}a&b\\c&d\end{aligned}} \right]\) and let \(k\) be a scalar. Find a formula that relates \(\det kA\) to \(k\) and \(\det A\).

Short answer

The formula that relates \(\det kA\) to \(k\) and \(\det A\) is \(\det kA = {k^2}\det A\).

Step-by-step solution

Determine matrix \(kA\)

Let \(A = \left[ {\begin{aligned}{*{20}{c}}a&b\\c&d\end{aligned}} \right]\).

Compute matrix \(kA\) as shown below:

\(\begin{aligned}{c}kA = k\left[ {\begin{aligned}{*{20}{c}}a&b\\c&d\end{aligned}} \right]\\ = \left[ {\begin{aligned}{*{20}{c}}{ka}&{kb}\\{kc}&{kd}\end{aligned}} \right]\end{aligned}\)

Determine the formula that relates \(\det kA\) to k and \(\det A\)

The determinant of matrix A is shown below:

\(\begin{aligned}{c}\det A = \left| {\begin{aligned}{*{20}{c}}a&b\\c&d\end{aligned}} \right|\\ = ad - bc\end{aligned}\)

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

\[\begin{aligned}{c}\det kA = \left| {\begin{aligned}{*{20}{c}}{ka}&{kb}\\{kc}&{kd}\end{aligned}} \right|\\ = \left( {ka} \right)\left( {kd} \right) - \left( {kb} \right)\left( {kc} \right)\\ = {k^2}\left( {ad - bc} \right)\\ = {k^2}\det A\end{aligned}\]

Thus, the formula that relates \(\det kA\) to \(k\) and \(\det A\) is \(\det kA = {k^2}\det A\).