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Consider an$n脳m$ matrix $A$with rank$\left(A\right)$<.$n$

We need to show that there exists vector$\stackrel{鈫�}{b}鈭�{\mathrm{鈩�}}^n$ such that the system$A\stackrel{鈫�}{x}=\stackrel{鈫�}{b}$ is inconsistent.

Let$E=$ rref.$\left(A\right)$

Also, rank$\left(A\right)$ =rank$\left(E\right)$ .

Since rank$\left(A\right)$<$n$, there exists a row in $E=$rref $\left(A\right)$with all entries zero.

Since$A$ an $n脳m$matrix, without loss of generality, assume that$i$ th row

( $1鈮�i鈮�n$) of$E$ is zero.

Now, consider the vector$\stackrel{鈫�}{d}=\left[\begin{array}{c}0\\ .\\ .\\ .\\ 1\\ 0\\ 0\end{array}\right]$ , non-zero entry 1, is at $i$th position.

The equation $E\stackrel{鈫�}{x}=\stackrel{鈫�}{d}鈬�0=1$.

This is not possible.

Hence, the given system is inconsistent.

Here, the systems$E\stackrel{鈫�}{x}=\stackrel{鈫�}{d}$ and$EA=\stackrel{鈫�}{b}$ are equivalent since$E=$ rref$\left(A\right)$.

Hence, the system$A\stackrel{鈫�}{x}=\stackrel{鈫�}{b}$ is inconsistent.

Consider an$n脳m$ matrix $A$with$m$ <.$n$

We need to show that there exists vector$\stackrel{鈫�}{b}鈭�{\mathrm{鈩�}}^n$ such that the system $A\stackrel{鈫�}{x}=\stackrel{鈫�}{b}$is inconsistent.

Since$A$ an $n脳m$matrix with$m$ <,$n$rank $\left(A\right)$$鈮�m$.

Consider $E=$rref such that there exists a row in $E$ with all entries zero.

Since $A$an$n脳m$ matrix, without loss of generality, assume that$i$ th row

( $1鈮�i鈮�n$) of $E$is zero.

Now, consider the vector$\stackrel{鈫�}{d}=\left[\begin{array}{c}0\\ .\\ .\\ .\\ 1\\ 0\\ 0\end{array}\right]$ , non-zero entry 1, is at$i$ th position.

The equation$E\stackrel{鈫�}{x}=\stackrel{鈫�}{d}鈬�0=1$ .

This is not possible.

Hence, the given system is inconsistent.

Here, the system$E\stackrel{鈫�}{x}=\stackrel{鈫�}{d}$and$EA=\stackrel{鈫�}{b}$ are equivalent since$E=$ rref$\left(A\right)$.

Hence, the system$A\stackrel{鈫�}{x}=\stackrel{鈫�}{b}$ is inconsistent.