I am given the following matrix A and I need to find a nullspace of this matrix.
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Answered question
2022-06-02
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I am given the following matrix A and I need to find a nullspace of this matrix.
I have found a row reduced form of this matrix, which is:
And then I used the formula , which gave me:A′=(120023100010131000012210)(x1x2x3x4x5)=(000) Hence I obtained the following system of linear equations:
How should I proceed from this point? Thanks
Answer & Explanation
Asher Swanson
Beginner2022-06-03Added 2 answers
Therefore,basis of null space=
Aliana Kaufman
Beginner2022-06-04Added 13 answers
The null space of a matrix is basically a solution of the following: Ax = 0 x is a linear combination of all the independent matrices that satisfy the above equation. You have already multiplied the echelon form of A with x and have equated it to 0, so if any of the resulting equations is true, Ax = 0 We see that x5 is a variable in all the equations, so for the sake of simplicity, we take x5 = c. (a constant) Thus, x3 = -10c/13 and x4 = -10c/22 x1, however, is still a problem because of the x2 for which we have no value. So we assign x2 another value, d (also a constant) and x1 = -2d - 10c/23 The two matrices which satisfy the equation Ax = 0 come out to be (-2d 1d 0 0 0)^T and (-10c/23 0 -10c/13 -10c/22 c)^T Let's call them y and z respectively. If Ay = 0 and Az = 0 then k(Ay) + t(Az) = 0; where k and t are constants Therefore, A (ky + tz) = 0 k * (-2d 1d 0 0 0)^T + t * (-10c/23 0 -10c/13 -10c/22 c)^T is equivalent to (if we take d and c common in their matrices) a * (-2 1 0 0 0)^T + b * (-10/23 0 -10/13 -10/22 1)^T In short, Ax = 0 is satisfied by any linear combination of the following two matrices: (-2 1 0 0 0)^T (-10/23 0 -10/13 -10/22 1)^T Together, they form the null space of A. The null space of A belongs to: (-2 1 0 0 0)^T; (-10/23 0 -10/13 -10/22 1)^T