Looking for answer to part c c. Find the change of coordinates matrix P from B to C. [x]C = P [x]B

Looking for answer to part c c. Find the change of coordinates matrix P from B to C. [x]C = P [x]B

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3a. Given the vector x and the basis B = [b1, b2, b3] of a subpace of R4 determine if x is in the span of B, and if so find the coordinates of x with respect to B and give the coordinate vector [x]B. x:[1,1,1,-1] b1: [1,0,2,0] b2:[0,1,3,0] b3:[0,0,4,1] **Part A Answered** (Credits to abtrehearn) Part a: Vector x is in the span of B if and only if it is a linear combination of the vectors in B. Then there are scalars x1,x2,x3 such that x1b1+x2b2+x3b3=x, or x1(1,0,2,0)+x2(0,1,3,0)+x3(0,0,4,1)=(1,1,1,−1) a system of four simultaneous linear equations in unknowns x1,x2,x3, which has the unique solution (x1,x2,x3)=(1,1,−1), so the vector x can be expressed as a linear combination if the vectors in B, and x is in the span of b. The coordinate vector [x]B is [1,1,-1]. b. Given the basis C = [c1, c2, c3], determine if C is in the span of B. If so write the vector x using the C coordinates. and give the coordinate vector [x]C. c1:[1,2,-4,-3] c2:[1,3,3,-2] c3:[-1,3,-9,4] *Part B Answered* (Credit to Alchemista, explain further if you would like to) In order to determine if C is in the span of B, Use c1, c2, and c3 in a linear combination with B to determine if its in the span of B. Matrix B = c1 1 0 0 1 0 1 0 2 2 3 4 -4 0 0 1 -3 Reduced: 1 0 0 1 0 1 0 2 0 0 1 -3 0 0 0 0 Matrix B = c2 1 0 0 1 0 1 0 3 2 3 4 3 0 0 1 -2 Reduced: 1 0 0 1 0 1 0 3 0 0 1 -2 0 0 0 0 Matrix B = c3 1 0 0 -1 0 1 0 3 2 3 4 -9 0 0 1 4 Reduced: 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 You can express the first vector in C as a Linear Combination of B, the second, but not the third. As you can see the third has no solution (the system is inconsistent) The answer then is no, since there are vectors in the span of C that are not in B. c. Find the change of coordinates matrix P from B to C. [x]C = P [x]B

If you want, you can also perform the reduction on an augmented matrix with all three vectors at once. \[ \left[ \begin {array}{cccccc} 1&0&0&1&1&-1\\ 0&1&0& 2&3&3\\ 2&3&4&-4&3&-9\\ 0&0&1&-3&- 2&4\end {array} \right] \] Reduced: \[\left[ \begin {array}{cccccc} 1&0&0&1&1&0\\ 0&1&0&2 &3&0\\ 0&0&1&-3&-2&0\\ 0&0&0&0&0&1 \end {array} \right]\]

i dont think that way would work, if i remember correctly the way you wrote it implies that columns 1 and 2 of C would be part of B making it a 4x5 matrix wouldnt it?

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