o µ ] } v W z x I s F t F x · 2020. 3. 16. · ^ } o µ ] } v W A B y F t F u s r w z r s C U A B...

10
Solution: A = 7 −2 −3 10 5 8 0 1 , B = 8 −1 4 −2 , C = −6 3 0 4 9 −2 8 5 , D = 5 −3 2 0 8 6 1 −2 −6 a) A2 x 4, B2 x 2; number of columns of A (4) is not equal number of rows of B (2). Therefore, AB does not exist. b) A T = 7 −3 −2 10 5 0 8 1 , A T B = 7 −3 −2 10 5 0 8 1 8 −1 4 −2 ቃ= 7 ∗ 8 + (−3) ∗ (4) 7 ∗ (−1) + (−3) ∗ (−2) −2 ∗ 8 + 10 ∗ 4 −2 ∗ (−1) + 10 ∗ (−2) 5∗8+0∗4 5 ∗ (−1) + 0 ∗ (−2) 8∗8+1∗4 8 ∗ (−1) + 1 ∗ (−2) = 44 −1 24 −18 40 −5 68 −10 . c) AC= 7 −2 −3 10 5 8 0 1 −6 3 0 4 9 −2 8 5 = 7 ∗ (−6) + (−2) ∗ 0 + 5 ∗ 9 + 8 ∗ 8 7 ∗ 3 + (−2) ∗ 4 + 5 ∗ (−2) + 8 ∗ 5 (−3) ∗ (−6) + 10 ∗ 0 + 0 ∗ 9 + 1 ∗ 8 (−3) ∗ (3) + 10 ∗ 4 + 0 ∗ (−2) + 1 ∗ 5 ൨= =ቂ 67 43 26 36 d) 2B +3I2 = 2 8 −1 4 −2 +3 1 0 0 1 ቃ= 2∗8+3∗1 2 ∗ (−1) + 0 2∗4+3∗0 2 ∗ (−2) + 3 ∗ 1 ൨ = ቂ 19 −2 8 −1 e) det 5 −3 2 0 8 6 1 −2 −6 = 5 −3 2 0 8 6 1 −2 −6 5 −3 2 0 1 −2 = 5*0*(-6)+(-3)*6*1+8*2*(-2) – 8*0*1- 5*6*(-2)–(-3)*2*(-6) = -26. f) det8 −1 4 −2 ቃ = 8 ∗ (−2) − (−1) ∗ 4 = −12. B -1 = 1/(-12) −2 1 −4 8 = 1/6 −1/12 1/3 −2/3 .

Transcript of o µ ] } v W z x I s F t F x · 2020. 3. 16. · ^ } o µ ] } v W A B y F t F u s r w z r s C U A B...

  • Solution:

    A = 7 −2 −3 10

    5 80 1

    , B = 8 −1 4 −2

    , C =

    −6 30 49 −28 5

    , D = 5 −32 0

    86

    1 −2 −6

    a) A2 x 4, B2 x 2; number of columns of A (4) is not equal number of rows of B (2). Therefore, AB does not exist.

    b) AT =

    7 −3−2 105 08 1

    , ATB =

    7 −3−2 105 08 1

    8 −1 4 −2

    =

    ⎣⎢⎢⎡7 ∗ 8 + (−3) ∗ (4) 7 ∗ (−1) + (−3) ∗ (−2)

    −2 ∗ 8 + 10 ∗ 4 −2 ∗ (−1) + 10 ∗ (−2)

    5 ∗ 8 + 0 ∗ 4 5 ∗ (−1) + 0 ∗ (−2)

    8 ∗ 8 + 1 ∗ 4 8 ∗ (−1) + 1 ∗ (−2) ⎦⎥⎥⎤

    =

    44 −124 −1840 −568 −10

    .

    c) AC= 7 −2 −3 10

    5 80 1

    −6 30 49 −28 5

    =

    7 ∗ (−6) + (−2) ∗ 0 + 5 ∗ 9 + 8 ∗ 8 7 ∗ 3 + (−2) ∗ 4 + 5 ∗ (−2) + 8 ∗ 5

    (−3) ∗ (−6) + 10 ∗ 0 + 0 ∗ 9 + 1 ∗ 8 (−3) ∗ (3) + 10 ∗ 4 + 0 ∗ (−2) + 1 ∗ 5 =

    =67 43 26 36

    d) 2B +3I2 = 2 8 −1 4 −2

    +3 1 0 0 1

    = 2 ∗ 8 + 3 ∗ 1 2 ∗ (−1) + 0 2 ∗ 4 + 3 ∗ 0 2 ∗ (−2) + 3 ∗ 1

    = 19 −2 8 −1

    e) det 5 −32 0

    86

    1 −2 −6=

    5 −32 0

    86

    1 −2 −6

    5 −32 01 −2

    = 5*0*(-6)+(-3)*6*1+8*2*(-2) – 8*0*1- 5*6*(-2)–(-3)*2*(-6) = -26.

    f) det 8 −1 4 −2

    = 8 ∗ (−2) − (−1) ∗ 4 = −12.

    B-1 = 1/(-12) −2 1 −4 8

    = 1/6 −1/12 1/3 −2/3

    .