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Let uequals Start 3 By 1 Table 1st Row 1st Column 8 2nd Row 1st Column negative 4 3rd Row 1st Column 8 EndTable
and AequalsStart 3 By 2 Table 1st Row 1st Column 3 2nd Column negative 5 2nd Row 1st Column negative 2 2nd Column 4 3rd Row 1st Column 1 2nd Column 1 EndTable
. Is u in the plane in set of real numbers Rcubed spanned by the columns of​ A? Why or why​ not?

Let Uequals Start 3 By 1 Table 1st Row 1st Column 8 2nd Row 1st Column Negative 4 3rd Row 1st Column 8 EndTable And AequalsStart 3 By 2 Table 1st Row 1st Column class=

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dhunganas642

Answer:

To determine if the vector u is in the plane spanned by the columns of matrix A, we can check if u can be written as a linear combination of the columns of A. Let's calculate the linear combination:

u = 8 * [3, -2, 1] + (-4) * [-5, 4, 1] + 8 * [0, 1, 1]

Simplifying this expression, we get:

u = [24, -16, 8] + [20, -16, -4] + [0, 8, 8]

u = [44, -24, 12]

Now, we compare this result with the vector u given in the question:

u = [8, -4, 8]

Since [44, -24, 12] is not equal to [8, -4, 8], u is not in the plane spanned by the columns of A.

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