Answered

Welcome to Westonci.ca, the ultimate question and answer platform. Get expert answers to your questions quickly and accurately. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals.

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=

Sagot :

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.

We hope this was helpful. Please come back whenever you need more information or answers to your queries. Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. Get the answers you need at Westonci.ca. Stay informed by returning for our latest expert advice.