Question

Exercises 25 and 26 prove Theorem 4 for \(A = \left( {\begin{aligned}{*{20}{c}}a&b\\c&d\end{aligned}} \right)\).

25. Show that if \(ad - bc = {\bf{0}}\), then the equation \(A{\bf{x}} = {\bf{0}}\) arhas more than one solution. Why does this imply that Ais not invertible? (Hint:First, consider \(a = b = {\bf{0}}\). Then, if aand bare not both zero, consider the vector \(x = \left( {\begin{aligned}{*{20}{c}}{ - b}\\a\end{aligned}} \right)\).)

Short answer

Matrix A is not invertible because the solution set isnot unique.

Step-by-step solution

Check whether A is invertible or not

The matrix equation can be represented as shown below:

\(\begin{aligned}{c}A{\bf{x}} = 0\\\left( {\begin{aligned}{*{20}{c}}a&b\\c&d\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}{{x_1}}\\{{x_2}}\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}0\\0\end{aligned}} \right)\end{aligned}\)

It is given that\(ad - bc = 0\). For the equation\(A{\bf{x}} = 0\), consider the case when\(a = b = 0\). Then, the equation becomes

\(\begin{aligned}{c}\left( {\begin{aligned}{*{20}{c}}0&0\\c&d\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}{{x_1}}\\{{x_2}}\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}0\\0\end{aligned}} \right)\\\left( {\begin{aligned}{*{20}{c}}0\\{c{x_1} + d{x_2}}\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}0\\0\end{aligned}} \right)\end{aligned}\).

So,\(c{x_1} + d{x_2} = 0\). It is observed that \({x_2} = - \frac{{c{x_1}}}{d}\). It means, for \({x_1} = d\), the value is \({x_2} = - c\).

In this case, the solution set is \({\bf{x}} = \left( {\begin{aligned}{*{20}{c}}d\\{ - c}\end{aligned}} \right)\).

The columns are linearly independent because one column is a scalar multiple of other columns.

Check whether ­A is invertible or not

Now, consider the case when all the entries are non-zero, that is,\(a \ne 0\),\(b \ne 0\),\(c \ne 0\), and\(d \ne 0\).

The equation becomes as shown below:

\(\begin{aligned}{c}A{\bf{x}} = 0\\\left( {\begin{aligned}{*{20}{c}}a&b\\c&d\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}{{x_1}}\\{{x_2}}\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}0\\0\end{aligned}} \right)\end{aligned}\)

Let,\(\left( {\begin{aligned}{*{20}{c}}{{x_1}}\\{{x_2}}\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}{ - b}\\a\end{aligned}} \right)\).

It can be represented as shown below:

\(\begin{aligned}{c}A{\bf{x}} = 0\\\left( {\begin{aligned}{*{20}{c}}a&b\\c&d\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}{ - b}\\a\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}0\\0\end{aligned}} \right)\\\left( {\begin{aligned}{*{20}{c}}{ - ab + ab}\\{ - bc + ad}\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}0\\0\end{aligned}} \right)\end{aligned}\)

As\( - bc + ad = 0\), x has anon-trivial solution in the equation \(A{\bf{x}} = 0\).

State whether ­A is invertible or not

From the above cases, the equation \(A{\bf{x}} = 0\)does not have a unique solution (more than one solution).

According to theorem 5, the matrix can't be invertible for more than one solution.

Thus, A isnot invertible.