Question

1. (15 points) Prove whether the following sets are linearly dependent or independent, and determine whether they form a basis of the vector space to which they belong. s 10110 -1 ) / -1 2) / 2 1 17 ) } in M2x2(R). "11-21 )'(1 1)'( 10 )'(2 –2 )S (b) {23 – X, 2x2 +4, -2x3 + 3x2 + 2x +6} in P3(R) (the set of polynomials of degree less than 3. (c) {æ4—23+5x2–8x+6, – x4+x2–5x2 +5x-3, x4+3x2 – 3x+5, 2x4+x3+4x2 +8x} in P4(R) (the set of polynomials of degree less than 4).

Answer 1

M2x2 -1 21 12 ol, 12 - 2 1. cay { (19), 1; 3), (12), (..)} in Mann 9 (44;) +-( )+('2 ) +d (3-4)=6) 1 -21) ", l Consider the relation, Then we have a C+2d = 0 -6 +20 + 0 =0 - 2a + b + c +2d = 0 a +b - 20 =0 Now consider the matrix A= 0 1 2 -2 1 1 2 1 - 2 apply elementary row operations on A. ? A R3 +(27R1, pio - 27 Trofl. 27 R2 EUR 0 - 2 1 Ret Rex 0 - 21 To -16 R4+ R2 0 0 1 7 Loo 3 -3] Rot R3 ri 0 0 9 11- Tio o 97 R2+(-27R3 0-1 0-13 0-1 0-13 R? +(-3)R3 Looi 7 Loo 01 LOO 0-24 Ri+(-9) Ry riooor - 0-100 R2+(13) Ry R3+(-7) Ru Loo o Them we havie a zo. -b=o or, b=o cao, d=0 lie. aso, b=0, C70, dao . go, the given set is linearly independent with 4 vectors. How dimension of Max2 is 4 and my linearly independent set with a vector is a basis of Manz so the given set { (-20), (: 7'), 1713), (3.,)} is a basis of Manel

16) { *?a, 2m'tu, -213 +3m++en+63 in 1P (IR); Consider, a(?-«) + 6(em44) + c(-2012+ 3+*+20+6=0 on, la – 26 ) m3 + (267367*"+ (a +2c)*+(4.6+6)=0 a -20 =0 =0. Them, 26+30=0 -a +20 = 0 - 3) 1:46+60 = 0 - ☺ a 20, 263-30 het m consider c=2.. Then a=4, b=-3, c=2 is a solution of The above system of equations a nd Them from definition of linearly dependent set { x'x, anith, :-2m² +n + 2x +6} is linearly dependent. O p love (C) { 4-23+50–84 +16, -*+m?- 5m*+5mg x"+3mY_34+5, 2n" + m² + un +8 ni} Consider the relation, al mm 375m"-en +6) + b(-x*+m3- 5n't Su-3) te(44+3+7+3m + 5) + d(204+3+40'+84) = 0

" (a-b+c+2d) 4+-a+b+d)*?+ (52-56 +3c +41d) * . + (-ea +56 -- 3c +8d)* + (ba-3b + 5e) =0 Then we have A-b+c+20 co - a+b + d = 0 50 - 5 b +30 +4d = 0 -8 a +56-36 +.8 d = 0 6a -3b + 50 =0 Consider the matrix A T - 1 17 and apply elementary 1 5 -8. 5. -5 5. 3 3 -3 5 4 8 - row operations on A. A. Retri T! Rz + (-5)R, 10: Rutier... To R5 +6R 0 - 3 3 1 17 12 +2 -11 5 16. -1 -6 -17 R1 + (-1) R2 - 1 0 (3) R, P 11, R3+ (2) R2 Ru+(-5)R2 0 0 0 3 R5 + Rż do 3.06

R5 + Ry ri -1 0 1 2 0 -3 0 _0 0 0 Then we have a -b -d=0 - C+2d = 0 3 d = 0 -3b +60=0 2d = 0 cod=0, b=0,C=0, a=o. . aso, bao, czo, dao so, the given set is linearly independent set with 4 vectors. How, dimension of IPu (IR) is 5 so borsis of Pu (R) centains 5 rectors, But falin²+5n-8x+6, ntn? sn't 5m-3, n44 347-3n+ 5, anh+23+4n+8x}. Contains four vectors. So the above set commot be a bensis of Pu (R).