Question

Let \(A = \left( {\begin{aligned}{*{20}{c}}{\bf{1}}&{\bf{2}}\\{\bf{1}}&{\bf{3}}\\{\bf{1}}&{\bf{5}}\end{aligned}} \right)\). Construct a \({\bf{2}} \times {\bf{3}}\) matrix \(C\) (by trial and error) using only \({\bf{1}}\), \( - {\bf{1}}\), and \({\bf{0}}\) as enteries, such that \(CA = {I_{\bf{2}}}\). Compute \(AC\) and note that \(AC \ne {I_{\bf{3}}}\).

Short answer

\(\left( {\begin{aligned}{*{20}{c}}1&1&{ - 1}\\{ - 1}&1&0\end{aligned}} \right)\)

Step-by-step solution

Construct matrix \(C\) using the elements \({\bf{1}}\), \( - {\bf{1}}\), and \({\bf{0}}\)

As the product \(CA\) is an identity matrix of \(2 \times 2\), the first row of \(C\) is \(\left( {\begin{aligned}{*{20}{c}}1&1&{ - 1}\end{aligned}} \right)\).

When \(A\) is multiplied by the first row of \(C\), the first row of \(CA\) becomes \(\left( {\begin{aligned}{*{20}{c}}1&0\end{aligned}} \right)\).

Construct matrix \(C\) using the elements \({\bf{1}}\), \( - {\bf{1}}\), and \({\bf{0}}\)

The second row of \(C\) is \(\left( {\begin{aligned}{*{20}{c}}{ - 1}&1&0\end{aligned}} \right)\).

When \(A\) is multiplied by the second row of \(C\), the second row of \(CA\) becomes \(\left( {\begin{aligned}{*{20}{c}}0&1\end{aligned}} \right)\).

So, matrix \(C\) is \(\left( {\begin{aligned}{*{20}{c}}1&1&{ - 1}\\{ - 1}&1&0\end{aligned}} \right)\).