Math133McGillsolutions

# Math133McGillsolutions - MATH 133 Practice Exam(Solutions x...

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Unformatted text preview: MATH 133 - Practice Exam (Solutions) x = 3t + 2 y = −2t + 1 , and the point A(1, −1, 2), Find the 1. Given the following line L1 : z=t equation of the plane perpendicular to L1 and passing by A. Solution n = d1 = (3, −2, 1) and A(1, −1, 2) so the equation of the plane is given by 3(x − 1) − 2(y + 1) + (z − 2) = 0 =⇒ 3x − 2y + z = 7. x = 3t + 2 y = −2t + 1 , and the point A(1, −1, 2). 2. Find the distance between the line L1 : z=t Solution A general point P on the line has coordinates (3t + 2, −2t + 1, t) for this point to be the closest to point A we should have vector AP = (3t + 1, −2t − 2, t − 2) perpendicular to the directional vector of the line (3, −2, 1) thus 3(3t + 1) − 2(−2t − 2) + 1(t − 2) = 0 =⇒ t = −5/14 and the distance between point A and the line is the norm of the vector AP = ||AP || = (−1/14)2 + (−9/7)2 + (−33/14)2 3. Find the equation of line AB given that A(1, 2, 3) and B (−2, 0, 2). Solution The directional vector of the line is vector AB = (−3, −2, −1) and thus the parametric equations of the line are given by x = −3t + 1 y = −2t + 2 z = −t + 3 4. Find a vector perpendicular to the plane ABC given that A(1, 2, 3), B (−2, 0, 2), and C (2, 4, −1). Solution A vector perpendicular to the plane is the cross product of any of two vectors in the plane AB = (−3, −2, −1) and AC = (1, 2, −4) for example. n = AB × AC = (10, −13, −4). x = 3t + 2 y = −2t + 1 ﬁnd the intersection of P1 5. Given (P1 ) : 4x − 3y + z = 2 and L1 : z=t and L1. Solution To ﬁnd the intersection of L1 and P1 we replace the equations of the line into that of the plane to get 1 3 4(3t + 2) − 3(−2t + 1) + t = 2 =⇒ t = − 19 We now replace in the equations of the line to get the point (29/19, 25/19, −3/19). 6. Find the solution of the following system using Gauss-Jordan elimination 4x1 + x2 − 3x3 + x4 = 1 −x1 + 3x2 − x3 + 2x4 = 7 Solution 4 1 −3 1 1 −1 3 −1 2 7 4R2 + R1 → R1 0 13 −7 9 29 −1 3 −1 2 7 1 −3 1 −2 −7 0 13 −7 9 29 −R2 → R2 R2 ↔ R1 R2 /13 → R2 1 −3 1 −2 −7 0 1 −7/13 9/13 29/13 3R2 + R1 → R1 1 0 −8/13 1/13 −4/13 0 13 −7 9 29 So the solution of the system is given by x1 x2 x3 x4 = −4/13 + 8/13 t − 1/13 s = 29/13 + 7/13 t − 9/13 s =t =s k x − 2y = 3 have 2x − ky = 4 7. For what value(s) of k does the system a) No solution. b) An inﬁnite number of solutions. c) A unique solution. Solution k −2 3 2 −k 4 k −2 3 1 −k/2 2 1 −k/2 2 k −2 3 1 0 R2 /2 → R2 R1 ↔ R2 − kR1 + R2 → R2 −k/2 2 − 4) −2k + 3 1 (k 2 2 a) No solution is when k = 2 or k = −2. 2 b) There is no value of k for which the system has an inﬁnite number of solutions. c) We get a unique solution when k = 2 and k = −2. 8. Given A = 2 4 −1 −3 a) Write A−1 as a product of elementary matrices. b) Write A as a product of elementary matrices. Solution 2 4 −1 −3 −1 −3 2 4 13 24 13 01 − R1 → R1 =⇒ E2 = − 2R1 + R2 =⇒ E3 = 13 0 −2 01 10 R1 ↔ R2 =⇒ E1 = −1 0 01 10 −2 1 R2 /(−2) → R2 =⇒ E4 = − 3R2 + R1 → R1E5 = − =⇒ E1 1 = − =⇒ E2 1 = − =⇒ E3 1 = 1 0 0 −1/2 1 −3 01 − =⇒ E5 1 = a) A−1 = E5 E4 E3E2 E1 − − − − − b) A = E1 1 E2 1 E3 1 E4 1 E5 1 9. Find the dimension of the following subspaces of Rn 1 2 4 −1 a) span 0 , 1 , 1 , −1 −1 1 −1 −2 1 1 −1 b) span 1 , 0 , 1 1 2 0 1 0 −1 −2R1 + R2 → R2 2 1 1 −4R1 + R3 → R3 a) 4 1 −1 R1 + R4 → R4 −1 −1 −2 3 −1 0 01 10 21 − =⇒ E4 1 = 10 01 Solution 01 10 13 01 10 0 −2 0 −1 1 3 1 3 −1 −3 0 −1 1 3 0 0 0 0 =⇒ basis = b) 1 0 0 0 1 0 0 0 1 11 1 02 −1 1 0 111 0 −1 1 021 11 1 0 1 −1 02 1 10 2 0 1 −1 00 3 1 0 , 1 0 −1 3 −R1 + R2 → R2 R1 + R3 → R3 − R2 ↔ R2 −R2 + R1 → R1 −2R2 + R3 → R3 =⇒ dimension = 3. 10. Given that abc def ghi Find −a d 2g a) −b e 2h −c f 2i b) 3a 3b 3c g + 3a h + 3b i + 3c 2d 2e 2f c) abc 2d 2e 2f def Solution 4 = −2 −R1 → R1 a) We get the matrix by applying the operations 2R3 → R3 transposing A so the determiant is given by (−2)(−1)(2) = 4 3R1 → R1 R2 ↔ R3 b) We get the matrix by applying the operations 2R3 → R3 R2 + R1 → R2 so the determinant is given by (−2)(3)(−1)(2) = 12. c) The determinant is 0 since the matrix has 2 proportional rows. 11. Find the determinant of the following matrix 1 0 2 −3 0 2 1 −2 1 0 −1 0 1 1 0 A= 0 −2 1 −1 1 0 0 1 −1 0 Solution 1 0 2 −3 2 1 −2 1 −1 0 1 1 0 −2 1 −1 0 0 1 −1 0 0 0 1 0 1 2 =− −1 0 0 2 −3 1 −2 1 0 1 1 0 1 −1 = −(−2 − 2 + 3) = 1 1 2 −3 1 = − −1 1 0 1 −1 1 12. Find an orthogonal basis for R3 containing the vector −1 2 Solution 1 1 0 −1 , 0 , 1 span R3 we now use the Gram-Schmidt The three vectors 2 0 0 process to get an orthogonal basis. 1 u1 = v1 = −1 2 5/6 u2 = v2 − projv2 u1 = 1/6 . −1/3 u3 = v3 − projv3 u1 − projv3 u2 5 13. Find W⊥ given each of the following cases 1 2 −1 , 0 a) W = span 2 1 1 0 b) W = span 2 −1 Solution 12 a) W = col(A) where A = −1 0 21 1 −1 2 is 2 01 −1/2 =⇒ W⊥ = span −3/2 . 1 1 = col(A) where A = 0 2 −1 W⊥ = null(AT ) The RREF form of x = −1/2 t =⇒ y = −3/2 t z=t 1 0 b) W = span 2 −1 W⊥ = null(AT ) [1 0 2 − 1] =⇒ x = t − 2s y=r z=s w=t −2 0 s + 1 0 x1 1 x2 0 = t + x3 0 1 x4 1 0 −1 14. Given A = 0 1 0 −1 0 1 10 1/2 0 1 −3/2 0 1 r 0 0 a) Find an orthogonal matrix P and a diagonal matrix D such that D = P T AP . Solution 1−λ 0 −1 (A − λI ) = 0 1−λ 0 −1 0 1−λ 6 det(A − λI ) = λ(λ − 1)(λ − 2) =⇒ λ = 0, λ = 1, λ = 2 √ 1 1/ 2 λ = 0 =⇒ e1 = 0 =⇒ q1 = √0 1 1/ 2 0 0 1 =⇒ q1 = 1 λ = 1 =⇒ e2 = 0 0 √ 1 1/ 2 λ = 2 =⇒ e3 = 0 =⇒ q1 = √0 −1 −1/ 2 √ √ 1/ 2 0 1/ 2 0 0√ Q= √1 1/ 2 0 −1/ 2 000 D= 0 1 0 002 15. Find the matrix of the linear transformation for a) the counterclockwise rotation about the origin by and angle of π . 3 b) the relfection about the line 3x1 − 2x2 = 0. c) the projection on the line 4x1 − x2 = 0. Solution a) T = cos(π/3) − sin(π/3) sin(π/3) cos(π/3) = √ 1/2 − 3/2 √ 3/2 1/2 b) The image of (x, y ) under this transformation is given by 7x 24y 24x 57y , , ) (x, y ) − 2proj(3,−4)(x, y ) = ( − 25 25 25 25 7/25 24/25 so T is given by −24/25 57/25 c) The image of (x, y ) under this transformation is given by x 4y 4x 16y proj(1,4)(x, y ) = ( + , + ) 17 17 17 17 so T is given by 1 17 4 17 4 17 16 17 16. Are the vectors w,u, and v independant? 1 5 3 −2 , u = 6 , v = 2 . a) w = 3 −1 1 7 1 2 0 0 , u = −1 , v = 2 . b) w = 3 1 1 Solution To solve this problem we write the vectors as the rows (or columns) of a matrix and then either reduce the matrix to RREF or compute the determinant of the matrix • if the RREF of the matrix is the identity or the determinand is not zero then they are indepedant. • if the RREF of the matrix has at least one row of zeroes or the determinant is zero then the vectors are linearly dependant. 1 −2 3 a) det 5 6 −1 = 0 =⇒ dependant. 32 1 103 b) det 2 −1 1 = 9 = 0 =⇒ independant. 021 8 ...
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