Math

QuestionGiven matrix AA and vector b\mathbf{b}, determine if b\mathbf{b} is in the span of columns of AA. How many vectors are in this span?

Studdy Solution

STEP 1

Assumptions1. The matrix AA is given by [1050368]\left[\begin{array}{rrr}1 &0 & -5 \\0 & & -3 \\ -6 &8 &\end{array}\right] . The vector b\mathbf{b} is given by [19626]\left[\begin{array}{r}19 \\6 \\ -26\end{array}\right]
3. The columns of AA are denoted by a1,a,a3a_{1}, a_{}, a_{3}
4. The set WW is the span of {a1,a,a3}\left\{a_{1}, a_{}, a_{3}\right\}

STEP 2

First, we need to identify the vectors a1,a2,aa_{1}, a_{2}, a_{} from the matrix AA. These are the column vectors of the matrix.
a1=[106],a2=[028],a=[52]a_{1} = \left[\begin{array}{r}1 \\0 \\ -6\end{array}\right], a_{2} = \left[\begin{array}{r}0 \\2 \\8\end{array}\right], a_{} = \left[\begin{array}{r}-5 \\ - \\2\end{array}\right]

STEP 3

Now, we compare the vector b\mathbf{b} with the vectors a1,a2,a3a_{1}, a_{2}, a_{3}.
b=[19626]\mathbf{b} = \left[\begin{array}{r}19 \\6 \\ -26\end{array}\right]

STEP 4

We can see that b\mathbf{b} is not equal to any of the vectors a1,a2,a3a_{1}, a_{2}, a_{3}. Therefore, b\mathbf{b} is not in {a1,a2,a3}\left\{a_{1}, a_{2}, a_{3}\right\}.

STEP 5

The number of vectors in {a1,a2,a3}\left\{a_{1}, a_{2}, a_{3}\right\} is3, as there are three column vectors in the matrix AA.

STEP 6

Now, we need to check if b\mathbf{b} is in the span of {a1,a2,a3}\left\{a_{1}, a_{2}, a_{3}\right\}, which is WW. This means we need to find if there are scalars x1,x2,x3x_{1}, x_{2}, x_{3} such that b=x1a1+x2a2+x3a3\mathbf{b} = x_{1}a_{1} + x_{2}a_{2} + x_{3}a_{3}.

STEP 7

We can set up a system of linear equations to solve for the scalars x1,x2,x3x_{1}, x_{2}, x_{3}.
[10502362][x1x2x3]=[19626]\left[\begin{array}{rrr}1 &0 & -5 \\0 &2 & -3 \\ -6 & &2\end{array}\right] \left[\begin{array}{r}x_{1} \\ x_{2} \\ x_{3}\end{array}\right] = \left[\begin{array}{r}19 \\6 \\ -26\end{array}\right]

STEP 8

olving this system of linear equations, we find that x1=1,x2=2,x3=3x_{1} =1, x_{2} =2, x_{3} = -3.

STEP 9

Since we found a solution for the scalars x,x2,x3x_{}, x_{2}, x_{3}, it means that b\mathbf{b} can be expressed as a linear combination of a,a2,a3a_{}, a_{2}, a_{3}. Therefore, b\mathbf{b} is in WW.

STEP 10

The number of vectors in WW is infinite, as WW is the span of {a,a2,a3}\left\{a_{}, a_{2}, a_{3}\right\}, which means it includes all linear combinations of a,a2,a3a_{}, a_{2}, a_{3}.

STEP 11

To show that aa_{} is in WW, we need to express aa_{} as a linear combination of a,a,a3a_{}, a_{}, a_{3}.

STEP 12

We can do this by setting x=,x2=0,x=0x_{} =, x_{2} =0, x_{} =0 in the equation b=xa+x2a2+xa\mathbf{b} = x_{}a_{} + x_{2}a_{2} + x_{}a_{}.

STEP 13

Substituting these values, we get a=a+0a2+0a3=aa_{} =a_{} +0a_{2} +0a_{3} = a_{}, which shows that aa_{} is in WW.
So, the answers area. No, bb is not in {a,a2,a3}\left\{a_{}, a_{2}, a_{3}\right\} since bb is not equal to a,a2a_{}, a_{2}, or a3a_{3} and there are3 vectors in {a,a2,a3}\left\{a_{}, a_{2}, a_{3}\right\}. b. Yes, b\mathbf{b} is in WW and there are infinite vectors in WW. c. Yes, a\mathrm{a}_{} is in W.

Was this helpful?

Studdy solves anything!

banner

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

ParentsInfluencer programContactPolicyTerms
TwitterInstagramFacebookTikTokDiscord