Question

Question 42: Let \(A = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]\)and \(B = \left[ {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right]\). Show that \(\det \left( {A + B} \right) = \det A + \det B\) if and only if \(a + d = 0\).

Short answer

It is proved that \(\det \left( {A + B} \right) = \det A + \det B\) if and only if \(a + d = 0\).

Step-by-step solution

Determine the matrix \(A + B\)

Compute the matrix \(A + B\) as shown below:

\(\begin{array}{c}A + B = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{1 + a}&{0 + b}\\{0 + c}&{1 + d}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{1 + a}&b\\c&{1 + d}\end{array}} \right]\end{array}\)

Show that \(\det \left( {A + B} \right) = \det A + \det B\) if and only if \(a + d = 0\)

The determinant of the matrix \(A + B\)is shown below:

\(\begin{array}{c}\det \left( {A + B} \right) = \left| {\begin{array}{*{20}{c}}{1 + a}&b\\c&{1 + d}\end{array}} \right|\\ = \left( {1 + a} \right)\left( {1 + d} \right) - cb\\ = 1 + a + d + ad - cb\\ = \det A + a + d + \det B\end{array}\)

Therefore, \(\det \left( {A + B} \right) = \det A + \det B\)if \(a + d = 0\).

Thus, it is proved that \(\det \left( {A + B} \right) = \det A + \det B\) if and only if \(a + d = 0\).