Question-75

Consider \(\displaystyle k\) linear equations in \(\displaystyle n\) variables expressed in the form \(\displaystyle Ax=b\). Which of the following are true?

• If \(\displaystyle n=k\), there is always at most one solution

• If \(\displaystyle n >k\), you can always solve the system

• If \(\displaystyle n >k\), nullity of \(\displaystyle A\) is non-zero

• If \(\displaystyle n< k\), then for some \(\displaystyle b\), there is no solution to \(\displaystyle Ax=b\)

• If \(\displaystyle n< k\), then the only solution of \(\displaystyle Ax=0\) is \(\displaystyle x=0\)

Answer

• If \(\displaystyle n=k\), there is always at most one solution

• If \(\displaystyle n >k\), you can always solve the system

• If \(\displaystyle n >k\), nullity of \(\displaystyle A\) is non-zero

• If \(\displaystyle n< k\), then for some \(\displaystyle b\), there is no solution to \(\displaystyle Ax=b\)

• If \(\displaystyle n< k\), then the only solution of \(\displaystyle Ax=0\) is \(\displaystyle x=0\)

Solution

Option-1

This is not true. Consider:

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

This system has infinitely many solutions.

Option-2

This is not true. Let \(\displaystyle n=4,k=2\):

\[ \begin{equation*} A=\begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{bmatrix} ,\ x=\begin{bmatrix} x_{1}\\ x_{2}\\ x_{3}\\ x_{4} \end{bmatrix} ,\ b=\begin{bmatrix} 0\\ 1 \end{bmatrix} \end{equation*} \]

This system has no solutions.

Option-3

This is true. If \(\displaystyle n >k\), the rank of \(\displaystyle A\) is at most \(\displaystyle k\). Therefore, the nullity of \(\displaystyle A\) is at least \(\displaystyle n-k >0\).

Option-4

This is true. If \(\displaystyle n< k\), the matrix looks tall. Let \(\displaystyle c_{1} ,\cdots ,c_{n}\) be the \(\displaystyle n\) columns. The span of these columns, the column space, would have dimension at most \(\displaystyle n\). This is a subspace of \(\displaystyle \mathbb{R}^{k}\). Since \(\displaystyle n< k\), we can find a vector \(\displaystyle b\in \mathbb{R}^{k}\) which is not in the column space. For this choice, \(\displaystyle Ax=b\) has no solutions.

Option-5

This is false. Consider \(\displaystyle n=2,k=3\):

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

This system has infinitely many solutions.