MATH 240 Midterm: MATH 240 NIU Test3sample Solution
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Solutions to the sample questions for exam 3 on 11/30/2007. Prof. john beachy: let l : r5 r4 be the linear transformation de ned by l(x) = ax, for the matrix. 0 0 1 1 (a) nd a basis for ker l; Since ker l is the solution space of the equation l(x) = 0, we need to nd the solution space of the system of equations ax = 0. After row-reducing the matrix, we have the corresponding equations x1 = 2x4; x3 = x4; x5 = 0. Choose x2 = 1, x4 = 0 and then x2 = 0, x4 = 1 to obtain the basis. 0 (b) nd a basis for range l; Since range l is the column space of a, as basis we take the columns of a that correspond to the leading 1"s in the reduced matrix, giving us the basis. 0 (c) nd dim(ker l) and dim(range l).
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