The rank, basis for row space, basis for column space of the given matrix. The rank of the given matrix is given by rank A = 1 _ . The basis for the row space is [ 2 − 1 3 ] _ . The basis for the column space is [ − 2 1 ] T _ . Definition used: A set of vectors is a basis for a vector space V if the vectors in the set are linearly independent, and if any vector in V can be expressed as a linear combination of the vectors in the set. Consider p vectors that each have n components. Then these vectors are linearly independent if the matrix formed, with these vectors as row vectors, has rank p. Calculation: The given matrix is [ 4 − 2 6 − 2 1 − 3 ] . Let A = [ 4 − 2 6 − 2 1 − 3 ] Reduce the given matrix as shown below: A = [ 4 − 2 6 − 2 1 − 3 ] ∼ [ 4 − 2 6 0 0 0 ] ( Row 2 + ( 0.5 ) Row 1 ) ∼ [ 2 − 1 3 0 0 0 ] ( ( 0.5 ) Row 1 ) Thus, the row reduced form of the given matrix is [ 2 − 1 3 0 0 0 ] . From the row reduced matrix form of the given matrix, it can be observed that there is only one non-zero row. Hence, rank A = 1 _ . Also, note that there is only one linearly independent row and hence, the basis for the row space is [ 2 − 1 3 ] _ . Find the transpose of the given matrix as shown below: A = [ 4 − 2 6 − 2 1 − 3 ] A T = [ 4 − 2 − 2 1 6 − 3 ] Reduce the transpose of the given matrix as shown below: A T = [ 4 − 2 − 2 1 6 − 3 ] ∼ [ 0 0 − 2 1 6 − 3 ] ( Row 1 + ( 2 ) Row 2 ) ∼ [ 0 0 − 2 1 0 0 ] ( Row 3 + ( 3 ) Row 2 ) ∼ [ − 2 1 0 0 0 0 ] ( Interchange Row 1 and Row 2 ) Thus, the row reduced form of transpose of the given matrix is [ − 2 1 0 0 0 0 ] . From the row reduced matrix form of transpose matrix, it can be observed that there is only one non-zero column which is linearly independent. Hence, the basis for the column space is [ − 2 1 ] T _ .

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