\(\Delta =\left| \begin{matrix} 5 & 3 \\ 3 & 7 \end{matrix} \right| =5(7)-3(3)\)
= 119 - 93 = 26
\(\Delta x=\left| \begin{matrix} 17 & 3 \\ 31 & 7 \end{matrix} \right| =17(7)-31(3)\)
= 119 - 93 = 2
\(\Delta y=\left| \begin{matrix} 5 & 17 \\ 3 & 31 \end{matrix} \right| =5(31)-17(3)\)
= 155 - 51 = 104

\(\therefore\) Solution set is (1, 4)

A =\(\left( \begin{matrix} 1 & -2 & 3 \\ -2 & 4 & -1 \\ -1 & 2 & 7 \end{matrix}\begin{matrix} 4 \\ -3 \\ 6 \end{matrix} \right) \)
The order of A is 3 x 4
\(\therefore \rho (A)\le \text{minimum} \ of(3,4)\)
\(\rho (A)\le 3\)
Let us transform the matrix A to an echelon form

The matrix is in echelon form and the number of non- zero rows is 2
\(\therefore \rho (A)=2\)

Let A =\(\left( \begin{matrix} 1 & 2 & -1 \\ 2 & 4 & 1 \\ 3 & 6 & 3 \end{matrix}\begin{matrix} 3 \\ -2 \\ -7 \end{matrix} \right) \)
The order of A is 3 x 4
\(\therefore \rho \left( A \right) \le 3\) [Since minimum of (3, 3) is 3]
Let us transform the matrix to an echelon form.

The matrix is in echelon form and the number of non- zero rows is 2.
\(\therefore \rho (A)=2\)

Let A =  \(\left( \begin{matrix} -1 & 2 & -2 \\ 4 & -3 & 4 \\ -2 & 4 & -4 \end{matrix} \right) \)
The order of A is 3 x 3
\(\therefore \rho (A)\le 3\) [Since minimum of (3,3) is 3]
Let us transform the matrix to an echelon form

The matrix is in echelon form and the number of non-zero rows is 2.
\(\therefore \rho (A)=2\)

Let \(A=\left( \begin{matrix} 2 & -1 & 1 \\ 3 & 1 & -5 \\ 1 & 1 & 1 \end{matrix} \right) \)
The order of A is 3 x 3
\(\therefore \rho (A)\le 3\) [Since minimum of (3,3) is 3]
Let us transform the matrix A to an echelon form

This matrix is in echelon from and number of non zero rows is 3.
\(\therefore \rho (A)=3\)