Section 1.3) p.32/13, 14, 18, 22, 25, 26 In 13 and 14, determine if b is a linea

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Section 1.3) p.32/13, 14, 18, 22, 25, 26 In 13 and 14, determine if b is a linear combination of the vectors formed from the columns of the matrix A 13) A = [1 -4 2 0 3 5 -2 8 -4 ] , b = [3 -7 -3 ] 14) A = [1 0 5 -2 1 -6 0 2 8 ] , b = [2 -1 6 ] 16) Let v1 = [1 0 -2 ], v2 = [-2 1 7 ], and y = v1 = [h -3 -5 ]. For what value(s) of h is y in the plane generated by v1 and v2? For 18, list five vectors in Span{ v1, v2}. For each vector show weights on v1 and v2 used to generate the vector and list the three entries of the vector. 18) v1 = [1 1 -2 ], v2 = [-2 3 0 ] 22) Construct a 3x3 matrix A, with nonzero entries, and a vector b in R3 such that b is not in the set spanned by the columns of A. 25) Let A = [1 0 -4 0 3 -2 -2 6 3 ] , b = [4 1 -4 ]. Denote the columns of A by a1, a2, a3, and let W = Span{a1, a2, a3}. a) Is b in {a1, a2, a3}? How many vectors are in {a1, a2, a3}? b) Is b in W? How many vectors are in W ? c) Show that a1 is in W. (Row operations are unnecessary.) 26) Let Let A = [2 0 6 -1 8 5 1 -2 1 ] , b = [10 3 7 ], and let W be the set of all linear combinations of the columns of A. a) Is b in W ? b) Show that the second column of A is in W. Section 1.4) p.40/8, 10, 16, 18, 22 8) Use definition of Ax to write the matrix equation as a vector equation or vice versa. z1[2 -4 ]+z2[-1 5 ]+z3[-4 3 ]+z4[0 2 ]=[5 12 ] 10) Write the system first as a vector equation and then as a matrix equation. 4x1-x2=85x1+3x2=2 3x1-x2=1 16) Let A = [1 -2 -1 -2 2 0 4 -1 3 ] , b = [b1 b2 b3 ] Show that the equation Ax = b does not have a solution for all possible b, and describe the set of all b for which Ax = b does have a solution. 18) Can every vector in R4 be written as a linear combination of the columns in the matrix A above? Do the columns of B span R3? A = [1 3 0 3 -1 -1 -1 1 0 -4 2 -8 2 0 3 -1 ] B = [1 4 1 2 0 1 3 -4 0 2 6 7 2 9 5 -7 ] 22) Let v1 = [0 0 -3 ], v2 = [0 -3 9 ], v3 = [4 -2 -6 ]. Does {v1, v2, v3} span R3? Why or why not?

Explanation / Answer

13) if b can be written as a linear combination of the columns of the matrix A, you will have, a[1 -4 2]+ c[0 3 5]+ d[-2 8 4]=[ 3 -7 3], where all a,c,d are not 0. This will give us three equations of the form, a-2d=3, -4a + 3c +8d= -7 and 2a+5c+4d=3. You can solve these equations using Cramer's rule, You can solve 14 similarly. For 16, y will lie in the span of {v1, v2} if y cab be written as a*v1+ b*v2. So you have a[1 0 -2] + b[-2 1 7]=[ h -3 -5], from this you get b=-3 and -2a+ 7b=-5, so a=13. From that we get h has to be a-2b=19. 18) you will get 5 vectors in span {v1, v2} by av1+bv2, for different values of(a,b). a is the weight of v1 and b is the weight of v2. 22The columns of a 3*3 matrix will span the whole of R3 if all of them linearly independent. So write two independent columns of A say [1 1 1], [1 2 1], make the 3 rd column linearly dependent on the 1st two, simplest way to do it is add the 1st two, so A=[ 1 1 1, 1 2 1, 2 3 2]. Now you have to construct a vector v such that v cannot be written as a[1 1 1]]+ c[1 2 1]. Take a to be 1, c to be 1, if b is [ 2 3 3] it cannot be written as a linear combination of these columns. Take b such that all the three equations involving a and c are not satisfied, I think you can solve the rest using these concepts

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