Consider the matrixA=⎡⎣⎢12−1370251−2−42⎤⎦⎥.Find a basis for the column space of A. A.⎧⎩⎨⎪⎪⎡⎣⎢100⎤⎦⎥,⎡⎣⎢310⎤⎦⎥⎫⎭⎬⎪⎪ B.⎧⎩⎨⎪⎪⎡⎣⎢12−1⎤⎦⎥,⎡⎣⎢370⎤⎦⎥,⎡⎣⎢251⎤⎦⎥⎫⎭⎬⎪⎪ C.⎧⎩⎨⎪⎪⎡⎣⎢12−1⎤⎦⎥,⎡⎣⎢370⎤⎦⎥⎫⎭⎬⎪⎪ D.⎧⎩⎨⎪⎪⎡⎣⎢12−1⎤⎦⎥⎫⎭⎬⎪⎪
Question
Consider the matrixA=⎡⎣⎢12−1370251−2−42⎤⎦⎥.Find a basis for the column space of A. A.⎧⎩⎨⎪⎪⎡⎣⎢100⎤⎦⎥,⎡⎣⎢310⎤⎦⎥⎫⎭⎬⎪⎪ B.⎧⎩⎨⎪⎪⎡⎣⎢12−1⎤⎦⎥,⎡⎣⎢370⎤⎦⎥,⎡⎣⎢251⎤⎦⎥⎫⎭⎬⎪⎪ C.⎧⎩⎨⎪⎪⎡⎣⎢12−1⎤⎦⎥,⎡⎣⎢370⎤⎦⎥⎫⎭⎬⎪⎪ D.⎧⎩⎨⎪⎪⎡⎣⎢12−1⎤⎦⎥⎫⎭⎬⎪⎪
Solution
To find a basis for the column space of matrix A, we need to find the linearly independent columns of A.
Let's write matrix A in column form: A = [12 -1; 37 0; 25 1; -2 -4].
To determine the linearly independent columns, we can perform row reduction on A.
Performing row reduction on A, we get the following row echelon form:
[1 0; 0 1; 0 0; 0 0].
The columns corresponding to the pivot positions in the row echelon form are the linearly independent columns of A.
Therefore, a basis for the column space of A is given by the columns of A that correspond to the pivot positions in the row echelon form.
In this case, the columns [12 -1] and [37 0] correspond to the pivot positions.
So, the basis for the column space of A is {[12 -1], [37 0]}.
Therefore, the correct answer is B.
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