Question

Let \(A = \left( {\begin{aligned}{*{20}{c}}{\bf{1}}&{\bf{1}}&{\bf{1}}&{\bf{0}}\\{\bf{0}}&{\bf{1}}&{\bf{1}}&{\bf{1}}\end{aligned}} \right)\). Construct a \({\bf{4}} \times {\bf{2}}\) matrix \(D\) using only \({\bf{1}}\) and \({\bf{0}}\) as enteries, such that \(AD = {I_{\bf{2}}}\). It is possible that \(CA = {I_{\bf{4}}}\) for some \({\bf{4}} \times {\bf{2}}\) matrix \(C\)? Why or why not?

Short answer

(D = \left( {\begin{aligned}{*{20}{c}}1&0\\0&0\\0&0\\0&1\end{aligned}} \right)\) and matrix \(C\) is not defined as the column vectors of \(A\) are linearly independent.

Step-by-step solution

Construct matrix \(D\) using the elements \({\bf{1}}\) and \({\bf{0}}\)

The first column of the matrix \(D\) is \(\left( {\begin{aligned}{*{20}{c}}1\\0\\0\\0\end{aligned}} \right)\).

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

Construct matrix \(D\) using the elements \({\bf{1}}\) and \({\bf{0}}\)

The second column of matrix \(D\) is \(\left( {\begin{aligned}{*{20}{c}}0\\0\\0\\1\end{aligned}} \right)\).

Find matrix \(C\)

If \(CA = {I_4}\), then \(CA{\bf{x}}\) is equal to \({\bf{x}}\) in \({\mathbb{R}^4}\). As the columns in \(A\) are linearly independent and \(A{\bf{x}} = 0\) for a non-zero vector \({\bf{x}}\), \(CA{\bf{x}}\) is not equal to \({\bf{x}}\).

So, \(D = \left( {\begin{aligned}{*{20}{c}}1&0\\0&0\\0&0\\0&1\end{aligned}} \right)\) and matrix \(C\) are not defined as the column vectors of \(A\) are linearly independent.