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\( \{ \frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ -1 \end{bmatrix}, \frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ 1 \end{bmatrix} \} \) is an orthanormal basis of \( \mathbb{R}^2 \) (you do not need to verify this). Calculate weights \( c_1 \) and \( c_2 \) so that \( \frac{c_1}{\sqrt{2}} \begin{bmatrix} 1 \\ -1 \end{bmatrix} + \frac{c_2}{\sqrt{2}} \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \end{bmatrix} \) 11. (5 points) (a) \( \{ \begin{bmatrix} -2 \\ 1 \end{bmatrix}, \begin{bmatrix} 1 \\ 2 \end{bmatrix} \} \) is an orthogonal basis of \( \mathbb{R}^2 \) (you do not need to verify this). Calculate weights \( c_1 \) and \( c_2 \) so that \( c_1 \begin{bmatrix} -2 \\ 1 \end{bmatrix} + c_2 \begin{bmatrix} 1 \\ 2 \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \end{bmatrix} \)

          \( \{ \frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ -1 \end{bmatrix}, \frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ 1 \end{bmatrix} \} \) is an orthanormal basis of \( \mathbb{R}^2 \) (you do not need to verify this). Calculate weights \( c_1 \) and \( c_2 \) so that
\( \frac{c_1}{\sqrt{2}} \begin{bmatrix} 1 \\ -1 \end{bmatrix} + \frac{c_2}{\sqrt{2}} \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \end{bmatrix} \)

11. (5 points) (a) \( \{ \begin{bmatrix} -2 \\ 1 \end{bmatrix}, \begin{bmatrix} 1 \\ 2 \end{bmatrix} \} \) is an orthogonal basis of \( \mathbb{R}^2 \) (you do not need to verify this). Calculate weights \( c_1 \) and \( c_2 \) so that
\( c_1 \begin{bmatrix} -2 \\ 1 \end{bmatrix} + c_2 \begin{bmatrix} 1 \\ 2 \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \end{bmatrix} \)

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${(1)/(√(2)) < b m a t r i x > , (1)/(√(2)) < b m a t r i x > } is an orthanormal basis of ℝ^2 (you do not need to verify this). Calculate weights c1 and c2 so that (c1)/(√(2)) < b m a t r i x > + (c2)/(√(2)) < b m a t r i x > = < b m a t r i x > 11. (5 points) (a) { < b m a t r i x > , < b m a t r i x > } is an orthogonal basis of ℝ^2 (you do not need to verify this). Calculate weights c1 and c2 so that c1 < b m a t r i x > + c2 < b m a t r i x > = < b m a t r i x >$

Added by James H.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Ben Blakesley

06/16/2022

Step 1

Step 1: Since {[i][l]} is an orthonormal basis of R2, we can express the vector [l] as a linear combination of the basis vectors: [l] = c1[i] + c2[l] Show more…

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({[i][l]} } is an orthonormal basis of R2 (you do not need to verify this). Calculate weights c1 and c2 so that []+8[l]-[l] 11. (5 points) (a) {[3][I]} is an orthogonal basis of R2 (you do not need to verify this). Calculate weights c1 and c2 so that c[]+cr[]=[i]

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00:01 Okay, so we have these two vectors u1 and u2.

00:03 So the first thing we want to do is check that this forms an orthogonal basis for r2.

00:08 So we just need to check that they're orthogonal.

00:11 So we need to check the dot product is equal to 0.

00:15 So u1 .u2, so the vector 41 .1 .1 .1, 12.

00:22 So this is going to be 4 times minus 3 plus 1 times 12, which is minus 12, plus 12, which is minus 12, which is 0, so they're orthogonal, and there's two of them.

00:34 So orthogonal implies linearly independent, and there are two of them, so they span r2.

00:40 So we've shown that this is an orthogonal basis.

00:43 The next thing we want to do is take the vector x to be 0 minus 51, and write this as a linear combination of u1 and u2.

00:56 So we want to write 0 minus 51 equals c times u1, plus c times u2 so we could do this in a few ways we could just solve the um the easiest way probably here is to solve the pair of simultaneous equations so c times u1 is 4 c1 c1 plus c2 u2 gives us minus 3 c2 12 c2 and then we have two equations we have 0 equals 4 c1 minus 3 c2 and we have minus 51 equals c1 plus 12 c2.

01:40 So then we have two equations we can solve for c1 and c2.

01:46 So this first equation, if we rearrange, gives us 3c2 equals 4c1.

01:52 Or in other words, we could write this as c1 is 3 over 4c2, for example...

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