Consider the matrixA=⎡⎣⎢212424−12042−5⎤⎦⎥.Find a basis for the column space of A. A.⎧⎩⎨⎪⎪⎡⎣⎢212⎤⎦⎥,⎡⎣⎢424⎤⎦⎥,⎡⎣⎢−120⎤⎦⎥⎫⎭⎬⎪⎪ B.⎧⎩⎨⎪⎪⎡⎣⎢212⎤⎦⎥,⎡⎣⎢−120⎤⎦⎥,⎡⎣⎢42−5⎤⎦⎥⎫⎭⎬⎪⎪ C.⎧⎩⎨⎪⎪⎡⎣⎢100⎤⎦⎥,⎡⎣⎢010⎤⎦⎥,⎡⎣⎢001⎤⎦⎥⎫⎭⎬⎪⎪ D.⎧⎩⎨⎪⎪⎡⎣⎢200⎤⎦⎥,⎡⎣⎢−110⎤⎦⎥,⎡⎣⎢401⎤⎦⎥⎫⎭⎬⎪⎪
Question
Consider the matrixA=⎡⎣⎢212424−12042−5⎤⎦⎥.Find a basis for the column space of A. A.⎧⎩⎨⎪⎪⎡⎣⎢212⎤⎦⎥,⎡⎣⎢424⎤⎦⎥,⎡⎣⎢−120⎤⎦⎥⎫⎭⎬⎪⎪ B.⎧⎩⎨⎪⎪⎡⎣⎢212⎤⎦⎥,⎡⎣⎢−120⎤⎦⎥,⎡⎣⎢42−5⎤⎦⎥⎫⎭⎬⎪⎪ C.⎧⎩⎨⎪⎪⎡⎣⎢100⎤⎦⎥,⎡⎣⎢010⎤⎦⎥,⎡⎣⎢001⎤⎦⎥⎫⎭⎬⎪⎪ D.⎧⎩⎨⎪⎪⎡⎣⎢200⎤⎦⎥,⎡⎣⎢−110⎤⎦⎥,⎡⎣⎢401⎤⎦⎥⎫⎭⎬⎪⎪
Solution
To find a basis for the column space of matrix A, we need to find the linearly independent columns of A.
Step 1: Write matrix A in column form: A = [2 1 2; 4 2 4; -1 2 0; 4 -5 -2]
Step 2: Reduce matrix A to echelon form or row-reduced echelon form.
Perform row operations to obtain echelon form: R2 = R2 - 2R1 R3 = R3 + (1/2)R1 R4 = R4 - 2R1
A = [2 1 2; 0 0 0; 0 2 1; 0 -7 -6]
Step 3: Identify the pivot columns in the echelon form of A.
The pivot columns are the columns with leading non-zero entries. In this case, the pivot columns are the first and third columns.
Step 4: Write the pivot columns as a basis for the column space of A.
The pivot columns are [2; 0; 0; 0] and [2; 0; 2; -7]. Therefore, a basis for the column space of A is: B = {[2; 0; 0; 0], [2; 0; 2; -7]}
So, the correct answer is B.
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