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 \(k\text{.}\)
Definition1.4.2.
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 \(r(A) = 2\text{.}\)
Observation1.4.4.
We list the following results without proof.
An \(m \times n\) matrix is of rank 0 if and only it is a zero matrix.
An \(n \times n\) square matrix \(A\) has rank \(n\) if and only if \(|A| \neq 0\text{.}\)
An \(n \times n\) square matrix \(A\) has rank strictly less than \(n\) if and only if \(|A| = 0\text{.}\)
For an \(m \times n\) matrix \(A\text{,}\) rank of \(A \leq \min (m, n).\)
The rank of a matrix is not affected if we insert zero column or a zero row (of appropriate size) to it.
If \(A\) is an \(m \times n\) matrix, then \(r(A) = r(A^T)\text{,}\) where \(A^T\) is the transpose of\(A\text{.}\)
Theorem1.4.5.
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.
Example1.4.6.
Let us find the ranl of \(A=\begin{bmatrix}1 \amp -2\amp 0\amp 3\amp -4\\3\amp 2\amp 8\amp 1\amp 4\\2\amp 3\amp 7\amp 2\amp 3\\-1\amp 2\amp 0\amp 4\amp -3\end{bmatrix}\text{.}\)
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.
Solution.
Let us apply elementary row operations of \(A\text{.}\) We have
Clearly the rank of \(A\) is 3 if \(a\neq 0, 2\text{.}\) If \(a=0\) or \(a=2\text{,}\) then it is easy to check that rank of \(A\) is 2.
Theorem1.4.8.Number of Solution of a System.
Let \(AX = B\) be a system of \(m\) linear equations in \(n\) unknowns and \(A^* = [A~|~B]\text{,}\) the augmented matrix of the system. Then
The system has a solution i.e. the system is consistent if and only if \(r(A) = r(A^*)\text{.}\)
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.
If \(r(A) = r(A^*)\) and \(r(A) = n\text{,}\) the number of unknowns, then there is only one (unique) solution for the system.
Corollary1.4.9.
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.
Corollary1.4.10.
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.
Thus we have \(r(A)=r(A^*)=3\text{,}\) hence this system has a unique solution.
Example1.4.12.
` Let us consider the system \(AX=B\) where \(A=\begin{bmatrix}1 \amp -2\amp 0\amp 3\\3\amp 2\amp 8\amp 1\\2\amp 3\amp 7\amp 2\\-1\amp 2\amp 0\amp 4\end{bmatrix}\quad B=\begin{bmatrix}-4\\4\\3\\-3\end{bmatrix}\quad X=\begin{bmatrix}x_1\\x_2\\x_3\\x_4\end{bmatrix}.\)Solution.
