Homework answers / question archive / Consider the following system of equations 2x + y + 2 =7 ex - 2y + 2z = 3 3x + 4y - 2 = 8 (a) Write the above system in the form AX = B and identify A, X and B

Consider the following system of equations 2x + y + 2 =7 ex - 2y + 2z = 3 3x + 4y - 2 = 8 (a) Write the above system in the form AX = B and identify A, X and B

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Consider the following system of equations 2x + y + 2 =7 ex - 2y + 2z = 3 3x + 4y - 2 = 8 (a) Write the above system in the form AX = B and identify A, X and B. (b) Solve the system of equations using Gaussian Elimination (by hand). (c) Find A-1 (d) Solve the above system using the inverse, i.e find X = A B (e) Find the rank(A) (f) Find det(A4) (g) Find det(A-1 ) a b c 2.- Let A= d e f and assume that det(A) = -6 gh i Find the determinant of the following matrix.

2 2 1] 3.- Let A = 5 2 find det(A - XI) = 0 2 2 4.- Show that v1 = (1, -3, 2), v2 = (1, 0, -1), v3 = (1, 2, -4) span R and express v = (9, 8, 7) as a linear combination of V1, V2, V3. 5.- Determine whether the given set of vectors is a basis for: (a) S= {1 -3x3, 2x + 5x2,1-x + 3x2} P2 (b) S = {(1, -1, 1), (2, 5, -2), (3, 11, -5) } R3

# 1 [a) 1x+ y+2:7 X - Zy+ 22 - 3 d A = Ye BE 3x +4 4 -1 38 (b) Gaussian Elimination It's a bit easier ( Human Calculations ) to solve the's system: x - 24 +2 2 = 3 2xty + 2 =7 ( Exchange R , (s R2 ) 3 x + 4 4 - 2 - 8 - Z 2 4 2 -2 - 2 P1+ P z - PZ D 5 - 3 R. + RJ - RS O 10 - 2 2 +Rs - R 3 -1 Back ward Substitution 4 .* - 27 4 +2 2 =3 3 1-3(3) =1 54 =10 54 -32 21 3 2 =3 7 - 2(2) +2( 3) = 3 X- 4+6 =3 x +2-3 5 *=1

#2 h Let B= -20 -20 -20 - -6 -2 B is obtained from A by doing the following steps 1. Sump ( Permate ) R. As PJ 2. Multiply New Ry by (t ) 3. Multiply Rz by -2 = 12 #3 + A-AL - 4 - 2 = [-x) [*-)(1-2)-41 -2(4-12-2)+1 (4-5+2) = (12) (10 -#2+2" -4) -4+42 + 4-1 = (1-2) (23- 42 +4) 4 3 2 -5 =22- 142 + 12 - #3+7 A" -62 +FA-5 =-2'+4AL-/82+ 1 =0 1 -1 9 -19 7 3-2 +42-6 3 + 7 = ( x-1)(-2 + rx-1) - 1 1 -1

#4 Let ( x.y. 2) ERS and albic ERR + b + c Led A= N deff Al= 1(2) - 1 ( 8 )+1(3) =2 -1 +3 exist =) unique solution for each x.y, 2 Spans IRS At b + c = 9 -34 42c=[ 24-5-40=4 Z. Gaussian Elimination = C= -24 , 6-135

#5 aj n = 2, S = { ( - 3 x4 2 x + 5 x 2 1 - * + 3271 Linearly Indep. ? c, ( t - 3 x * ) + (z ( 1 * + 5 x2 ) + c, ( 4 - *+ 3x2 ) =0 C , = s c x' + z tz x + 5 ( x 6 + (, - 6, * + 3 6 *= 0 CI+ =0 61 = - C3 -36+ 562+ 363=0 3-3(-4)+5/83)+36=0 C1=0 => Linearly lad. Now, dim ( 5 ) = 3 S SPARS P, (x ) S is a basis.

# 5 ( 6) " ( ). . ( 2)+4 (;)=(:) 2 We have A= - 1 11 2 -5 |Notice that def(A) =0 Adoesn't exist A Lie = o has more then one solution ) Linearly dependent. 3 5 can not be a basis for /R Since dim[e' ]= 3

a b c 2.— Let A = d e f and assume that det(A) = —6 g h 1' Find the determinant of the following matrix. 9 h 3' —2d —23 —2f