how can we find the rank of a Matrix?
give an example?
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Rank of a square matrix is found by finding the determinant. If the determinant is non-zero, the dimension of the square matrix is the rank.
If the determinant of A of size n x n is zero, then it is possible that there are two identical rows or columns. Or, algebraic sum of some rows or columns is equal to a row or column. Then the rank cannot be n.
We remove one of the identical rows or columns and then find the determinant of n-1 x n-1 matrix. If its determinant is non zero, then n-1 will be the rank of the matrix A.
Determinant is found by the usual method.
![det| \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] | = det | \left[\begin{array}{ccc}1&2&3\\4-1&5-2&6-3\\7-1&8-2&9-3\end{array}\right] | \\ \\ \\ = det| \left[\begin{array}{ccc}1&2&3\\3&3&3\\6&6&6\end{array}\right] | = 0 \\ \\](https://tex.z-dn.net/?f=det%7C++%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%26amp%3B2%26amp%3B3%5C%5C4%26amp%3B5%26amp%3B6%5C%5C7%26amp%3B8%26amp%3B9%5Cend%7Barray%7D%5Cright%5D+%7C+%3D+det+%7C+%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%26amp%3B2%26amp%3B3%5C%5C4-1%26amp%3B5-2%26amp%3B6-3%5C%5C7-1%26amp%3B8-2%26amp%3B9-3%5Cend%7Barray%7D%5Cright%5D+%7C+%5C%5C+%5C%5C+%5C%5C+%3D+det%7C+%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%26amp%3B2%26amp%3B3%5C%5C3%26amp%3B3%26amp%3B3%5C%5C6%26amp%3B6%26amp%3B6%5Cend%7Barray%7D%5Cright%5D+%7C+%3D+0+%5C%5C+%5C%5C+ "det| \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] | = det | \left[\begin{array}{ccc}1&2&3\\4-1&5-2&6-3\\7-1&8-2&9-3\end{array}\right] | \\ \\ \\ = det| \left[\begin{array}{ccc}1&2&3\\3&3&3\\6&6&6\end{array}\right] | = 0 \\ \\")
Determinant is zero because 3nd row is a multiple of 2nd row.
Let us remove the third row and column. and find the determinant again of the 2x2 matrix.
![det| \left[\begin{array}{ccc}1&2\\4&5\end{array}\right] | = 1*5 - 4*2 = -3 \neq 0. \\ \\ So\ Rank = 2.\\](https://tex.z-dn.net/?f=det%7C++%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%26amp%3B2%5C%5C4%26amp%3B5%5Cend%7Barray%7D%5Cright%5D+%7C+%3D+1%2A5+-+4%2A2+%3D+-3++%5Cneq+0.+%5C%5C+%5C%5C+So%5C+Rank+%3D+2.%5C%5C "det| \left[\begin{array}{ccc}1&2\\4&5\end{array}\right] | = 1*5 - 4*2 = -3 \neq 0. \\ \\ So\ Rank = 2.\\")
If the determinant of A of size n x n is zero, then it is possible that there are two identical rows or columns. Or, algebraic sum of some rows or columns is equal to a row or column. Then the rank cannot be n.
We remove one of the identical rows or columns and then find the determinant of n-1 x n-1 matrix. If its determinant is non zero, then n-1 will be the rank of the matrix A.
Determinant is found by the usual method.
Determinant is zero because 3nd row is a multiple of 2nd row.
Let us remove the third row and column. and find the determinant again of the 2x2 matrix.
karthik4297:
then rank of example would be 1?
no no 3-1=2
power supply gone. i continue ans
oo k
u seem to know it. then why ask a question?
select as best answer.
ok see u.
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Step-by-step explanation:
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