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Linear Algebra - Solved Assignments - Fall 2004 Semester

LINEAR ALGEBRA SOLVED ASSIGNMENTS SEMESTER FALL-2004 Solution of Assignment # 1 of MTH501 (Fall2004) Q1. Consider the system of equations 22 3x y z ax z bx y z c+ + =+ =+ + = Show that the system to be consistent , the constant a, b and c must satisfy c = a + b. Answer: The augmented matrix correspond to the above system is 1 1 21 0 12 1 3abc . We will find out the echelon form of this matrix which can be obtained by the following steps. 2 1 3 12 3 21 1 21 0 12 1 31 1 20 1 1 , 20 1 1 21 1 20 1 1 ,0 0 0abcab a by R R R Rc aaa b by R R Rc a b + Now as you know that if augmented matrix of a linear system in echelon form have any row of the form [0 0 0 b] where b is non zero then that system will be inconsistent. Now in the above matrix we have third row as [0, 0, 0, c-a-b] and in order to system be consistent we must have c a b =0 which is equivalent to c = a + b. Q2. Is there a value of r so that the x = r, y = 2, z =1 is a solution to the following linear system? If there is a value find it. 3 2 44 52 3 2 9x zx y zx y z = + = + + = Answer: We will try to find out the solution of the above system, so the augmented matrix for the above system is3 0 2 41 4 1 52 3 2 9 and the echelon form of the augmented matrix can be obtained by the following steps. 1 22 1 3 1223 0 2 4 1 4 1 51 4 1 5 3 0 2 4 Re2 3 2 9 2 3 2 91 4 1 50 12 5 19 3 , 20 5 4 11 4 1 55 19 10 112 12 120 5 4 11 4 1 55 190 1 512 1223 830 012 12by placing R by Rby R R R RRR + + 331 4 1 55 190 1 1212 120 0 23 83RR 23 z = 83 z = 83/23 This doesnt match with the value of z given in the question. So we cant find the value of r from the given system of linear equations. Note: There are more then one method to solve that question as you can also put the given values x=r y=2 and z=1 in the system of linear equation and then you can conclude the same. But concept used in this question is that If system of linear equations is consistent then it will have unique solution or infinite many solutions. This is not possible that a system of linear equations have 2 or 3 or 10 solutions. Q3. An oil refinery produces low-sulfur and high-sulfur fuel. Each ton of low-sulfur requires 5 minutes in the blending plant and 4 minutes in the refining plant; each ton of high sulfur fuel requires 4 minutes in the blending plant and 2 minutes in the refining plant. If the blending plant is available for 3 hours and the refining plant is available for 2 hours, how many tons of each type of fuel should be manufactured so that the plants are fully utilized? Answer: The data given in the question can be formed into system of linear equations as Low-sulfur High-sulfur Blending plant 5 4 Refining Plant 4 2 As we are given in the question that we have blending plant and refining plant available for 3 and 2 hours respectively. Let x tons of low-sulfur and y tons of high sulfur be the amount should be manufactured so that plants are fully utilized. Then from the above data we must have the system, 5x + 4y = 180 4x + 2y = 120 Augmented matrix for the above system is1 25 4 180 1 2 604 2 120 4 2 1201 2 600 6 120by R R So we have x + 2y = 60 and -6y = - 120, thus we have y= 20 tons and x = 20tons are the required manufactured tons of each low-sulfur and high-sulfur so that we can utilize both plants for the given time. Q4. (a) Find a linear equation in the variables x and y that has the general solution x = 5 + 2t, y= t. (b) Show that x=t, 1 52 2y t = is also the general solution of the equation in part (a). Answer: (a) We have to find out linear equation for which the given parametric equations x = 5 + 2t, y= t define the coordinates of any point on that line. And in order to get the required linear equation we will simply find an equation independent of the parameter t Suppose x = 5 + 2t ---------- (1) y = t--------------- (2) Put t=y in eq. (1) we get x= 5 + 2y This is the required linear equation. (b) We have a linear equation x= 5 + 2y ----------- ( A) Put x = t in eq.(A) we get t =5 + 2y --------------- (B) From (B) y = ( t 5)/2 Y = t/2 5/2 1 52 2y t = This shows that x=t, 1 52 2y t = is also the general solution of the equation in part (a). Q5. Consider5 12 , 2 27 3 3kx y and wk = = = find the value(s) of k such that the vector w is in the span of x and y. Answer: We have to find the value(s) of k (if possible) under the condition that w is in the span of x and y W is in the span of x and y so we have to check whether x1x + x2y = w have solution? Here x1 and x2 are constants. To answer this, row reduce the augmented matrix [ x y w]: 1211 2 1 322 35 12 2 27 3 32 2 25 17 3 31 1 115 127 3 31 1 10 4 5 5 , 70 4 101 1 15 10 14 40 4 101 1 150 1 440 0 5kkk Rkk Rkk R R R RkkRkkR R + + + + + + + + The third equation is 0x2 = -5 this is not possible which shows that the system has no solution. The vector equation x1x + x2y = w has no solution and so w is not in the span of x and y. Thus we can not able to find the value(s) of k. Solution of Assignment # 2 of MTH501 (Fall2004) Q1. If the columns of the matrix 12 7 11 9 59 4 8 7 36 11 7 3 94 6 10 5 12 span R4 then write the vector1457 as linear combination of the columns of the above matrix. Solution: We have to find out the values of the unknowns such that we can write the given vector1457 as1 2 3 4 512 7 11 9 5 19 4 8 7 3 46 11 7 3 9 54 6 10 5 12 7c c c c c