The rank of the matrix \(A = \left[ {\begin{array}{*{20}{c}} 1&1&2\\ 1&2&3\\ 0&{ - 1}&1 \end{array}} \right]\)
3
The rank of a matrix is a fundamental concept in linear algebra. It represents the maximum number of linearly independent rows or, equivalently, the maximum number of linearly independent columns in the matrix. Another way to think about it is the dimension of the vector space spanned by its columns (or rows).
We are given the matrix:
$$A = \left[ {\begin{array}{ccc} 1 & 1 & 2 \\ 1 & 2 & 3 \\ 0 & -1 & 1 \end{array}} \right]$$We can find the rank by transforming the matrix into its Row Echelon Form (REF) using elementary row operations. The rank will be the number of non-zero rows in the REF.
Let's perform row operations:
The matrix is now in Row Echelon Form.
The resulting matrix in Row Echelon Form is:
$$ \left[ {\begin{array}{ccc} 1 & 1 & 2 \\ 0 & 1 & 1 \\ 0 & 0 & 2 \end{array}} \right] $$We count the number of non-zero rows in this matrix. All three rows have non-zero elements.
Since there are 3 non-zero rows, the rank of the matrix A is 3.
For a square matrix, if its determinant is non-zero, its rank is equal to its order (number of rows/columns). Let's calculate the determinant of A:
$$ \det(A) = 1 \cdot \left| {\begin{array}{cc} 2 & 3 \\ -1 & 1 \end{array}} \right| - 1 \cdot \left| {\begin{array}{cc} 1 & 3 \\ 0 & 1 \end{array}} \right| + 2 \cdot \left| {\begin{array}{cc} 1 & 2 \\ 0 & -1 \end{array}} \right| $$ $$ \det(A) = 1 \cdot ( (2)(1) - (3)(-1) ) - 1 \cdot ( (1)(1) - (3)(0) ) + 2 \cdot ( (1)(-1) - (2)(0) ) $$ $$ \det(A) = 1 \cdot ( 2 + 3 ) - 1 \cdot ( 1 - 0 ) + 2 \cdot ( -1 - 0 ) $$ $$ \det(A) = 1 \cdot (5) - 1 \cdot (1) + 2 \cdot (-1) $$ $$ \det(A) = 5 - 1 - 2 $$ $$ \det(A) = 2 $$Since the determinant is 2, which is not equal to 0, the rank of the 3x3 matrix A is 3.
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