Question

Exercise 1 Let 1 1 2 4 A= -16 2 5 1 2 - 1 0 2 3 loo-1/ and B 1 2 (1 1 -3 -1 2 2 0 / (1) Compute det A. 3 (2 .-1)-(3.0) = 1.(-2) - (0) = -2 (2) Evaluate : det (+(2(342)*)*") (3) Compute det B.

Answer 1

lo 2 1) Given A=[12 17 Lo 0 - 1 According to the formula TMT la b c eodet det d e f - Loh id h i. se po = aider (0.5) -todet (4 ) + c. det (59) 1. det (23) – 2. det (o 3) + (-). det (0 ) det ( a b) = ad-be det ( 2 ) = 2(-1)-3-0) – 2

I det (8 3) = 0.(-1) - 3.0 20. det (O 3) = 0.0-2.0=0, substitute all these obtained values in a 1.(-2)-2.0-1.0 -2. . det (A) = -2 det (462(3A%))))) To find det of the given first solve 2 - the brackets one by one 2342 0 a do 0 1.1+2.0-1.0 1.2 +2.2 -1.0 0.1 +2:0+30 0.2+ 2.2 +30 10.1 +0.07-10 0.2+0.2+0.0 1.4+2.3+(-)-(-) 0.(-) +2.3 + 3.(-1) 0.6-) +0.3+(-1).(-) :34²= 3 342 - 2616 61 13 18 187 d 43 =lo 12 a >2(342) (342) = 13 18) T (342) = 18 0 12 9 i 18 12 OM O Oo M 116 ool 436 24 o (36 18 6/7 36 24 o 36 18 6) To find the inverse of the matrix Augment with a 3x3 identity matrix =

Convert the matrix to identity matrix Reduce the matrix to row echelon form (36 24 olo 107 R R2 R2 R₂-t Ri (36 24 o o 1o7 O-4011-116 0 36 19 6 loo' R3 R3-R, 136 24 ololo = o 4 oli -1/6 o swap matrix R2 R3 To - 4o 1-1/6 ol R3 R3 - 2/R2 [36 24 oloiol > Loo 4 1/2-2/ R3 -1/4R3 (36 24 o o 1o7 -6 6 0 - 1 To o 1/4 / 46 R2 R2-6, R3 36 24 oo io lo -6 o 3/2-1/4 ol 1/6R2. (36 24 o o of o Io / 1/24 0 o o 1-/4-1/g 16

R R , -24.R2 R, 1/60R1 Inverse matrix is on the right of augmented matrix. cto 0 0 - 1 1/24 o 1-1/-% 1/6 (16 o ol -1/4 1/24 o 14 18 16 12/3 o ol = 14 -1/ 2/3) -1 V60 det ( ) - 0.det () +0.41 %) 물을 pidet (462 (39237) ") 3. det ( 1 2 12 124) 5 - 6 1 - 2 Joda echelon form to find det R2 Reduce matrix to row swap matrix Rit + 12 5 -1 67 Ti 2 4 1 2 1 - 2 1 -3 2

R₂ R2 - 1½.R, = 0 -3/2 5/2 R3 12 R3-R1 . - 6 s lo 4o 4 1-3 20 Ru R4 - 1/2 R (2 5 + 6 To -3/25/2 " lo 1/25/23 ) I R R4. 1,2 s 6 0 -4 0 -4 R3 R3 - 8/9, R2 12 s - 6 01/25/2-3 loo -20/1 - 2011 10 -3/25/2 ) Ru R4 - 23. R2 12 5 I 6) e MF T W g Fake o q şta a To o

Ru Rut R3 To o -20 - 20 The determinant of the matrix equals the diagonal product of the matrix Interchanging two rows negate the determinate, therefore multiply the result by (-1) ². i det B = 0