Question

Question: Let \(A = \left( {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right)\). Use formula (1) for a determinant (given before Example 2) to show that \(\det A = ad - bc\). Consider two cases: \(a \ne 0\) and \(a = 0\).

Short answer

It is proved that \({\rm{det}}A = ad - cb\) when \(a \ne 0\) and \(a = 0\).

Step-by-step solution

Consider \(a \ne 0\) and determine the determinant

Consider \(a \ne 0\), apply the row operations on the matrix \(A\).

\[\begin{array}A = \left( {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}a&b\\0&{d - c{a^{ - 1}}b}\end{array}} \right)\;\;\;\;\;\;\;\;\;\;{R_2} \to {R_2} - \frac{1}{a}{R_1}\\ = U\end{array}\]

Find the determinant of \(A\).

\(\begin{array}{\rm{det}}A = a\left( {d - c{a^{ - 1}}b} \right) - b\left( 0 \right)\\ = ad - cb\end{array}\)

Thus, it is proved that \({\rm{det}}A = ad - cb\) when \(a \ne 0\).

Consider \(a = 0\) and determine the determinant

Consider \(a = 0\), apply the row operations on the matrix \(A\).

\[\begin{array}A = \left( {\begin{array}{*{20}{c}}0&b\\c&d\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}c&d\\0&b\end{array}} \right)\;\;\;\;\;\;\;\;\;\;{R_2} \leftrightarrow {R_1}\\ = U\end{array}\]

Find the determinant of \(A\).

\[\begin{array}{\rm{det}}A = {\left( { - 1} \right)^1}\left( {cb} \right)\\ = 0 - bc\\ = ad - bc\end{array}\]

Thus, it is proved that \({\rm{det}}A = ad - cb\) when \(a = 0\).