Let \(A\) be an \(m \times n\) matrix. Delete any \(m-k\) rows and \(n-l\) columns of \(A\text{.}\) The resulting matrix is called a \(k\times l\) sub-matrix of \(A\text{.}\) If \(k = l\text{,}\) then it is called a square sub-matrix of \(A\) of order \(k\text{.}\)
The rank of an \(m \times n\) matrix \(A\) is the order of the largest square sub-matrix of \(A\) whose determinant is non-zero. We denote the rank of a matrix \(A\) by \(r(A)\text{.}\)
We can see that \(\mid A\mid = 0\text{.}\) Hence \(r(A) \leq 2\text{.}\) Now we look for the square sub-matrix of \(A\) of order 2 whose determinant is non-zero. Consider the square sub-matrix \(B=
\begin{pmatrix}
1\amp 1 \\
2\amp 3 \\
\end{pmatrix}\text{,}\)\(\mid B\mid = 1\text{.}\) Hence \(r(A) = 2\text{.}\)
Suppose \(A\) is an \(m \times n\) matrix. The rank of \(A\) is equal to the number of non-zero rows in an echelon equivalent form of \(A\text{.}\) In other words, the rank of matrix \(A\) is the number of leading 1’s in any row-echelon matrix to which \(A\) can be carried by row operations.
Consider a matrix \(\begin{bmatrix} 1 \amp 1 \amp 2 \amp a^2\\1 \amp 1-a \amp 2\amp 0 \\2 \amp 2-a \amp 6-a \amp 4\end{bmatrix}\text{.}\) Find the rank of the matrix.
If \(r(A) = r(A^*) \) and \(r(A) = r \) whihc is strictly less than \(n\text{,}\) then number of unknowns, then there are infinite number of solutions and \(n-r\) variables can be chosen freely.
Suppose \(AX = B\) is a system of \(m\) linear equations in \(n\) unknowns with \(m \lt n\text{.}\) If \(r(A)\neq r(A^*)\text{,}\) then the system \(AX=B\) has no solution.
Suppose \(AX = B\) is a system of \(m\) linear equations in \(n\) unknowns with \(m \lt n\text{.}\) Then the system either has no solution or has infinitely many solutions.
Let us consider the system \(AX=B\) where \(A=\begin{bmatrix}1 \amp -2\amp 0\amp 3\\3\amp 2\amp 8\amp 1\\2\amp 3\amp 7\amp 2\\-1\amp 2\amp 0\amp 4\end{bmatrix}\)\(B=\begin{bmatrix}-4\\4\\3\\-3\end{bmatrix}\) and \(X=\begin{bmatrix}x_1\\x_2\\x_3\\x_4\end{bmatrix}.\)
