UNIT-3 LINEAR ALGEBRA (21MA1BSCDE) 2021-22 PDF

BMS COLLEGE OF ENGINEERING, BENGALURU DEPARTMENT OF MATHEMATICS Dept. of Math., BMSCE Unit 3: Linear Algebra UNIT 3: LINEAR ALGEBRA I. RANK OF MATRICES 1. Find t...

BMS COLLEGE OF ENGINEERING, BENGALURU DEPARTMENT OF MATHEMATICS Dept. of Math., BMSCE Unit 3: Linear Algebra UNIT 3: LINEAR ALGEBRA I. RANK OF MATRICES 1. Find the rank of the following matrices by reducing them into echelon form. 1 2 3  1 3 4 3  1 −2  1 4 2  3 9 12 3  3 −6  a)  . e)  . i)    2 6 5  1 3 4 1  7 −1   Ans:  =2 Ans:  =2 4 5  0 1 −3 −1 1 1 −1 1  5 6 7 8 1 0 1 1  1 −1 2 −1 6 7 8 9    f)  .  b)  3 1 0 2 . 3 1 0 1  j) 11 12 13 14 .     1 1 −2 0  Ans:  =2 16 17 18 19  Ans:  =2 1 2 3 2  Ans:  =2  2 3 −1 −1 2 3 5 1  1 2 −2 3  1 −1 −2 −4  g)  .  2 5 −4 6    1 3 4 5    c) 3 1 3 −2 . k)  −1 −3 2 −2 .   Ans:  =2   6 3 0 −7   2 4 −1 6  1 2 3 0 Ans:  =3 2 Ans:  =4  4 3 2  1 4 5  2 6 8  h)  3 2 1 3. d)  .    3 7 22  6 8 7 5 Ans:  =3 Ans:  =3 1 1 −1 0   4 4 −3 1  2. Find ‘ b ’ if the rank of   is 3. Ans: b = −6 b 2 2 2    9 9 b 3  II. Consistency and solution of linear system of equations: Page 1 of 9 Dept. of Mathematics, BMSCE Unit 3: Linear Algebra n = Number of Unknowns 1. Discuss the solution of the following system of linear equations x + 2 y + 3z = 0 x + 2y + z = 3 1. 3 x + 4 y + 4 z = 0. 2x + 3y + 2z = 5 7 x + 10 y + 12 z = 0 7. 3 x − 5 y + 5 z = 2. 3x + 9 y − z = 4 2 x1 + 3 x2 − 4 x3 + x4 = 0 Ans : x = −1, y = 1, z = 2 x1 − x2 + x3 + 2 x4 = 0 5 x1 − x3 + 7 x4 = 0 2 x + 6 y + 11 = 0 2.. 7 x1 + 8 x2 − 11x3 + 5 x4 = 0 6 x + 20 y − 6 z + 3 = 0 8.. Ans : x1 = x2 = 6 y − 18 z + 1 = 0 x3 = x4 = Ans: inconsistent 5x + 3 y + 7 z = 4 2 x1 − 2 x2 + 4 x3 + 3 x4 = 9 3 x + 26 y + 2 z = 9 x1 − x2 + 2 x3 + 2 x4 = 6 3. 7 x + 2 y + 10 z = 5. 9. 2 x1 − 2 x2 + x3 + 2 x4 = 3. 7 − 16k k +3 x1 − x2 + x4 = 2 Ans : x = y= z=k 11 11 Ans:Inconsistent 4x − 2 y + 6z = 8 2x + y − z = 0 x + y − 3z = −1 2 x + 5 y + 7 z = 52 4.. 10.. 15 x − 3 y + 9 z = 21 x+ y+z =9 Ans : x = 1, y = 3k − 2, z = k Ans : x = 1, y = 3, z = 5 2x − 3y + 7 z = 5 3x + 2 y + 2 z = 1 3x + y − 3z = 13 x + 2y = 4 5.. 2 x + 19 y − 47 z = 32 11. 10 y + 3 z = −2. Ans: inconsistent 2x − 3y − z = 5 2 x1 + 3x2 − x3 = 1 Ans : x = 2, y = 1, z = −4 3 x1 − 4 x2 + 3 x3 = −1 x1 + x2 − 2 x3 + x4 + 3x5 = 1 6. 2 x1 − x2 + 2 x3 = −3. 2 x1 − x2 + 2 x3 + 2 x4 + 6 x5 = 2 3 x1 + 1x2 − 2 x3 = 5 12. 3 x1 + 2 x2 − 4 x3 − 3 x4 − 9 x5 = 3. Ans: Inconsistent Ans: x1 = 1, x2 = 2a, x3 = a x4 = −3b, x5 = b 2x + 3 y + 5z = 9 2. Investigate the values of  and  so that the equations 7 x + 3 y − 2 z = 8 have (i) no solution (ii) 2x + 3y +  z =  unique solution (iii) infinite number of solutions. Ans: (i) If  = 5 and   9 (ii)   5 and  can be any value (iii)  = 5 and  = 9. x + 2 y + 3z = 6 3. Determine the values of a and b for which the system x + 3 y + 5 z = 9 have (i) no solution (ii) 2 x + 5 y + az = b unique solution (iii) infinite number of solutions. Page 2 of 9  Dept. of Mathematics, BMSCE Unit 3: Linear Algebra x + ay + z = 3 4. Find the values of a and b for which the system of equations x + 2 y + 2 z = b is consistent. x + 5 y + 3z = 9 Ans: If a = −1 and b =6 equations will be consistent and have infinite number of solutions. If a  −1 and b has any value, equations will be consistent and have a unique solution. 3x − y + 4 z = 3 5. Show that if   −5 the system of equations x + 2 y − 3z = −2 have a unique solution. If  = −5 6 x + 5 y +  z = −3 show that the equations are consistent. Determine the solution in each case. 4 −9 4 − 5k 13k − 9 Ans: when   −5, x = , y = , z = 0 , when  = −5, x = , y= , z=k. 7 7 7 7 3x + 4 y + 5 z = a 6. Show that the equations 4 x + 5 y + 6 z = b do not have a solution unless a + c = 2b. 5x + 6 y + 7 z = c 5 x + 3 y + 2 z = 12 7. Prove that the equations 2 x + 4 y + 5 z = 2 are incompatible unless c = 74 ; and in that case the 39 x + 43 y + 45 z = c equations are satisfied by x = 2 + t , y = 2 − 3t , z = −2 + 2t , where t is any arbitrary quantity. x + y + z =1 8. For what values of k the equations 2 x + y + 4 z = k have a solution and solve them completely in 4 x + y + 10 z = k 2 each case. Ans: when k = 1, x = −3 z , y = 2 z + 1 , when k = 2 , x = 1 − 3 z , y = 2 z. −2 x + y + z = a 9. Test for consistency x − 2 y + z = b where a, b and c are constants. x + y − 2z = c Ans: if a + b + c  0 inconsistent , if a + b + c = 0 , then infinitely many solution. III. Solution of linear system of equations by Gauss elimination method: 1. Solve the following system of equations by Gauss elimination method 2 x + y + z = 10 10 x + 2 y + z = 9 3x + 2 y + 3z = 18 2 x + 20 y − 2 z = −44 a). d). x + 4 y + 9 z = 16 −2 x + 3 y + 10 z = 22 Ans: x = 7, y = −9, z = 5 Ans: x = 1, y = −2, z = 3 2 x + 2 y + z = 12 2 x1 + x2 + 4 x3 = 12 3x + 2 y + 2 z = 8 8 x1 − 3x2 + 2 x3 = 20 e). b) 5 x + 10 y − 8 z = 10. 4 x1 + 11x2 − x3 = 33 −51 115 35 Ans: x1 = 3, x2 = 2, x3 = 1. Ans: x = ,y= ,z = 4 8 4 x1 + 4 x2 − x3 = −5 2 x1 + 4 x2 + x3 = 3 x1 + x2 − 6 x3 = −12 3x1 + 2 x2 − 2 x3 = −2 f) 3 x1 − x2 − x3 = 4. c). x1 − x2 + x3 = 6 117 81 148 Ans: x1 = 2, x2 = −1, x3 = 3. Ans: x1 = , x2 = − , x3 =. 71 71 71 Page 3 of 9 Dept. of Mathematics, BMSCE Unit 3: Linear Algebra 2 x1 + x2 + 3x3 = 1 5 x1 + x2 + x3 + x4 = 4 4 x1 + 4 x2 + 7 x3 = 1 x1 + 7 x2 + x3 + x4 = 12 g). 2 x1 + 5 x2 + 9 x3 = 3 x1 + x2 + 6 x3 + x4 = −5 Ans: x1 = −1 / 2, x2 = −1, x3 = 1. j). x1 + x2 + x3 + 4 x4 = −6 2 x1 − 7 x2 + 4 x3 = 9 Ans: x1 = , x2 = , x1 + 9 x2 − 6 x3 = 1 x3 = , x4 =. h). −3 x1 + 8 x2 + 5 x3 = 6 2 x1 + 2 x2 + x3 + 2 x4 = 7 Ans : x1 = 4, x2 = 1, x3 = 2 − x1 + 2 x2 + x4 = −2 k) −3 x1 + x2 + 2 x3 + x4 = −3 2 x1 + x2 + x3 = 10 − x1 + 2 x4 = 0 3x1 + 2 x2 + 3x3 = 18 i). 80 17 53 40 x1 + 4 x2 + 9 x3 = 16 Ans: ,− , , 37 37 37 37 Ans: x1 = 7, x2 = −9, x3 = 5. 2. Balance the chemical equation x1C3 H 8 + x2O2 → x3CO2 + x4 H 2O finding x1, x2 , x3 , x4 using Gauss elimination method 3. Find the traffic flow in the net of one-way streets directions shown in the figure 4. Using Kirchhoff’s voltage and current law, find the system of equations in current from the following circuit and then find the current using Gauss elimination method:1) Page 4 of 9  Dept. of Mathematics, BMSCE Unit 3: Linear Algebra i) ii) iii) iv) IV. Gauss-Seidel Iteration method The solution converges if the system is diagonally dominant. Page 5 of 9 Dept. of Mathematics, BMSCE Unit 3: Linear Algebra ( ) Suppose AX = B is diagonally dominant with A = aij , X = ( xi ) and B = ( bi ). x1n +1 = 1 a11 ( b1 − a12 x2n − a13 x3n ) Then the iterative formula is x2n +1 = 1 a22 ( b2 − a21 x1n +1 − a23 x3n ) x3n +1 = 1 a33 ( b3 − a31 x1n +1 − a32 x2n +1 ) 20 x + y − 2 z = 17 8 x1 + x2 − x3 = 8 3x + 20 y − z = −18 2 x1 + x2 + 9 x3 = 12 a.. i.. 2 x − 3 y + 20 z = 25 x1 − 7 x2 + 2 x3 = −4 Ans: x = 1, y = −1, z = 1 Ans: x1 = 1 x2 = 1, x3 = 1 5 x + 2 y + z = 12 4 x1 + 2 x2 + x3 = 11 x + 4 y + 2 z = 15 − x1 + 2 x2 = 3 b.. x + 2 y + 5 z = 20 j. 2 x1 + x2 + 4 x3 = 16. Ans: x = 0.996, y = 2, z = 3 Ans: x1 = 1, x2 = 2, x3 = 3 2x + y + 6z = 9 k. Start with ( 2, 2, −1) and solve 8 x + 3 y + 2 z = 13 c.. 5 x1 − x2 + x3 = 10 x + 5y + z = 7 2 x1 + 4 x2 = 12 Ans: x = 1, y = 1, z = 1 x1 + x2 + 5 x3 = −1 28 x + 4 y − z = 32 Ans : x1 = 2.5555, x2 = 1.7222, x3 = −1.0555 x + 3 y + 10 z = 24 d.. 10 x1 + x2 + x3 = 12 2 x + 17 y + 4 z = 35 Ans: x = 0.998, y = 1.723, z = 2.024 2 x1 + 10 x2 + x3 = 13 l.. 2 x1 + 2 x2 + 10 x3 = 14 10 x + 2 y + z = 9 Ans: x1 = x2 = x3 = 1 2 x + 20 y − 2 z = −44 e.. −2 x + 3 y + 10 z = 22 27 x1 + 6 x2 − x3 = 85 Ans: x = 1, y = −2, z = 3 6 x1 + 15 x2 + 2 x3 = 72 m.. x1 + x2 + 54 x3 = 110 83x + 11y − 4 z = 95 7 x + 52 y + 13z = 104 Ans: x1 = 2.4255, x2 = 3.573, x3 = 1.926 f.. 3 x + 8 y + 29 z = 71 x1 − 8 x2 + 3x3 = − 4 Ans: x = 1.06, y = 1.37, z = 1.96 2 x1 + x2 + 9 x3 = 12 n.. 54 x + y + z = 110 8 x1 + 2 x2 − 2 x3 = 8 2 x + 15 y + 6 z = 72 Ans : x1 = x2 = x3 = 1 g.. − x + 6 y + 27 z = 85  2 −1 0 0   x  0 Ans: x = 1.926, y = 3.573, z = 2.425  −1 2  −1 0   y  0  = 5 x1 − x2 = 9 o.  0 −1 2 −1  z  0 .      − x1 + 5 x2 − x3 = 4  0 0 −1 2   u  5  h.. − x2 + 5 x3 = −6 Ans: x = 1, y = 2, z = 3, u = 4 Ans: x1 = 1.99, x2 = 0.99, x3 = −1 Page 6 of 9  Dept. of Mathematics, BMSCE Unit 3: Linear Algebra V. Characteristic values (Eigen values) and characteristic vectors (Eigen vectors) If A is a square matrix, then  is said to be an eigen value of the matrix if there exists a non-zero vector X such that AX =  X. X is called the eigen vector corresponding to the eigen value .  X =  IX  ( A −  I ) X = 0. We seek non-trivial solution of ( A −  I ) X = 0. X is non-trivial if  ( A −  I )  n  A −  I = 0. If A is matrix of size 3  3 then − 3 + Tr ( A )  2 −  M ii22  + A = 0. 1 1 3  2 1 −1 1 5 1  1 1 −2      3 1 1  −1 −2 1  a. Ans:  = −2,3,6; x1 = [ − k ,0, k ],. g. Ans :  = 1, −1, 4; x1 = [2, −1,1],. x2 = [k , −k , k ], x3 = [k , 2k , k ] x2 = [0,1,1], x3 = [ −1, −1,1] 3 1 4  1 2 2 0 2 6   0 2 1     b. 0 0 5 .  −1 2 2  Ans:  = 2,3,5 x1 = k1[1, −1,0] h. Ans :  = 1, 2, 2; x1 = k1[1,1, −1],. x2 = k2 [1,0,0], x3 = k3 [3, 2,1] x2 = k2 [2,1,0], x3 = k3 [2,1,0]  8 −6 2  1 1 3  −6 7 −4  1 5 1     c.  2 −4 3 . 3 1 1 Ans :  = 0,3,15 x1 = [1, 2, 2], i. Ans:  = −2,3,6; x1 = [−1,0,1],. x2 = [2,1, −2], x3 = [2, −2,1] x2 = [1, −1,1], x3 = [1, 2,1] 2 0 1 2 2 1 0 2 0  1 3 1      d. 1 0 2 . 1 2 2  Ans:  = 1, 2,3 x1 = [1,0, −1] j. Ans:  = 5,1,1; x1 = [1,1,1],. x2 = [0,1,0], x3 = [1,0,1] x2 = [1,0, −1], x3 = [2, −1,0]  −2 2 −3  3 10 5   2 1 −6   −2 −3 −4       −1 −2 0   3 5 7  e. Ans :  = 5, −3, −3 x1 = k[1, 2, −1]. k. Ans :  = 2, 2,3. x2 = [3k1 − 2k2 , k1 , k2 ] x1 = x2 = [5, 2, −3], x3 = [1,1, −2] 6 −2 2  2 −2 2   −2 3 −1 1 1   1   2 −1 3  1 3 −1 f. Ans:  = 8, 2, 2; x1 = [2, −1,1],. l. Ans :  = 2, 2, −2. x2 = [1,0, −2], x3 = [1, 2,0] x1 = x2 = [0,1,1] x3 = [−4, −1,1] Page 7 of 9  Dept. of Mathematics, BMSCE Unit 3: Linear Algebra 3 −2 −5 4 −1 −5   −2 −1 −3 m. Ans:  = −5, 2, 2. x1 = [3, 2, 4], x2 = x3 = [1,3, −1] VI. Determine the largest (dominant) Eigen value and the corresponding Eigen vector of the following matrices using Rayleigh’s Power Method  2 −1 0   6 −2 2   −1 2 −1  −2 3 −1 1.  . 6.    0 −1 2   2 −1 3  Ans :  = 3.41 Ans :  = 8  4 1 1 6 1  A=  1 2 0  2.  1 3. 7.  . Ans :  = 4.418  0 0 3  3. Ans :  = 4 3 −2 −5 4  25 1 2   −1 −5 1 3 0  −2 −1 −3 8.  .  2 0 −4  Ans:  = −5  4 1 −1 Ans :  = 25.182  2 3 −1  1 3 −1 4.  . 3 2 4  −2 1 5  9.   Ans :  = 6  −1 4 10   −8 6 −2  Ans :  = 11  6 −7 4  5.    −2 4 −3 Ans :  = −15 SELF STUDY: VII. Solution of system of equations by Gauss Jacobi iterative Method 2x + y + 6z = 9 10 x + 2 y + z = 9 8 x + 3 y + 2 z = 13 2 x + 20 y − 2 z = −44 a.. c.. x + 5y + z = 7 −2 x + 3 y + 10 z = 22 Ans: x = 1, y = 1, z = 1 Ans: x = 1, y = −2, z = 3 28 x + 4 y − z = 32 83x + 11y − 4 z = 95 x + 3 y + 10 z = 24 7 x + 52 y + 13z = 104 b. d.. 2 x + 17 y + 4 z = 35 3 x + 8 y + 29 z = 71 Ans: x = 0.998, y = 1.723, z = 2.024 Ans: x = 1.06, y = 1.37, z = 1.96. Page 8 of 9  Dept. of Mathematics, BMSCE Unit 3: Linear Algebra 54 x + y + z = 110 8 x1 + x2 − x3 = 8 2 x + 15 y + 6 z = 72 2 x1 + x2 + 9 x3 = 12 e. g.. − x + 6 y + 27 z = 85 x1 − 7 x2 + 2 x3 = −4 Ans: x = 1.926, y = 3.573, z = 2.425 Ans: x1 = 1 x2 = 1, x3 = 1. 4 x1 + 2 x2 + x3 = 11 5 x1 − x2 = 9 − x1 + 2 x2 = 3 − x1 + 5 x2 − x3 = 4 f. 2 x1 + x2 + 4 x3 = 16 − x2 + 5 x3 = −6 Ans: x1 = 1, x2 = 2, x3 = 3 Ans: x1 = 1.99, x2 = 0.99, x3 = −1. VIII. Using Cayley-Hamilton Theorem find the inverse of the matrix 1 1 3 1. 1 5 1. 3 1 1 3 1 4 2. 0 2 6 .   0 0 5   8 −6 2  3.  −6 7 −4 .    2 −4 3  2 0 1 4. 0 2 0 .   1 0 2   −2 2 −3 5.  2 1 −6 .    −1 −2 0   6 −2 2  6.  −2 3 −1    2 −1 3   2 1 −1 7.  1 1 −2     −1 −2 1  Page 9 of 9