Problem-3

Question

Is the following statement true or false: If the system \(Ax=b\) has a solution, say \(x^{*}\), then we can express it as \(x^{*} =A^{-1} b\).

Hint

Think about what happens when there are \(2\) equations and \(3\) unknowns.

Solution

This statement is false. Generating a counter example. Let \(x^{*}\) be a solution to a system \(Ax = b\) as given below:

\[ \begin{equation*} x^{*} =\begin{bmatrix} 1\\ 0\\ 1 \end{bmatrix} \end{equation*} \]

\[ \begin{equation*} \begin{aligned} 2x_{1} +x_{2} -x_{3} & =1\\ x_{1} -x_{3} & =0 \end{aligned} \end{equation*} \]

\[ \begin{equation*} A=\begin{bmatrix} 2 & 1 & -1\\ 1 & 0 & -1 \end{bmatrix} ,\ \ x=\begin{bmatrix} x_{1}\\ x_{2}\\ x_{3} \end{bmatrix} ,\ \ b=\begin{bmatrix} 1\\ 0 \end{bmatrix} \end{equation*} \]

\(A\) is not a square matrix. Inverse exists only for square matrices. Even if \(A\) is a square matrix, it need not be invertible. For example, consider the following example:

\[ \begin{equation*} A=\begin{bmatrix} 1 & -1\\ -2 & 2 \end{bmatrix} ,\ b=\begin{bmatrix} 0\\ 0 \end{bmatrix} \ \rightarrow \ x^{*} =\begin{bmatrix} 1\\ 1 \end{bmatrix} \end{equation*} \]

Here the system \(Ax=b\) has a solution \(x^{*}\), but it cannot be written as \(A^{-1} b\) as \(A\) is not invertible. Note that the above system has infinitely many solutions and only one of them was listed.