Determine if the columns of the matrix form a linearly independent set. Justify each answer. [14−30−2−751−4−575]

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SoosteethicU · Accepted Answer

The given matrix [14−30−2−751−4−575] The columns of the matrix form a linearly independent set if equation Ax=0 has only trivial solution Reduce matrix in reduced echelon form [14−30−2−751−4−575]R1↔R3R2−1/2R1→R2[−4−5750−9232−3200−1313] [−4−5750−9232−3200−1313]R3+1/4R1→R3R3+11/18R2→R3[−4−5750−9232−3200−1313] [−4−5750−9232−3200−1313]−3R3→R3R2−3/2R3→R2[−4−5750−9200001−1] [−4−5750−9200001−1]R1−7R3→R1−2/9R2→R2∼[−4−50120100001−1] [−4−50120100001−1]R1+5R2→R11/4R1→R1∼[100−30100001−1] Thus. we get equation Ax=0 has not only trivial solution, the columns of the matrix form a linearly dependent set. Hence, the columns of the matrix form a linearly dependent set Result: The columns of the matrix form a linearly dependet set

alenahelenash · Accepted Answer

To determine if the columns of the matrix form a linearly independent set, we need to check if the columns are linearly independent or not.Let&#039;s denote the given matrix as A:A=[14−30−2−751−4−575]To check for linear independence, we can perform row reduction (Gaussian elimination) on the matrix and see if we obtain any rows of zeros.Applying row reduction to matrix A:[14−30−2−751−4−575]→[14−3001−110−11195]&amp;nbsp;→[14−3001−1100816]The row-reduced echelon form of the matrix is:[14−3001−1100816]Since the row-reduced echelon form has no rows of zeros, we can conclude that the columns of matrix A form a linearly independent set.Therefore, the columns of the matrix [14−30−2−751−4−575] are linearly independent.

star233 · Accepted Answer

Result:The columns of the matrix form a linearly independent set.Solution:Let&#039;s denote the given matrix as A:A=[14−30−2−751−4−575]To check if the columns are linearly independent, we need to solve the equation Ax=0. This can be done by finding the reduced row-echelon form of the augmented matrix [A|0].Applying row operations, we can perform the following steps:[14−30−2−751−4−575]R2→R2+2R1R3→R3+4R1→[14−3001−11011−55]R3→R3−11R2→[14−3001−11006−6]R1→R1+16R3R2→R2+R3→[140−1010−5006−6]R1→R1−4R2R3→16R3→[10019010−5001−1]The reduced row-echelon form of the augmented matrix is:[10019010−5001−1]From this row-echelon form, we can see that the only solution to the equation Ax=0 is the trivial solution x=[000]. Therefore, the columns of the matrix are linearly independent.

Determine if the columns of the matrix form a linearly independent set. Justify

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