Mat1503 May/June 2018 Solutions

Page 1 of 4 MAT1503
May/June 2018
Memorandum

QUESTION 1
Let
0 0 2 3
0 1 −2 2
𝐴=[ ]
−1 2 5 0
3 1 2 0
(a) Evaluate the determinant, det( 𝐴) by expanding along the
last column. (6)

0 1 −2 0 0 2
det( 𝐴) = −3 |−1 2 5 | + 2 |−1 2 5|
3 1 2 3 1 2
−1 5 −1 2 −1 2
= −3 [−1 | | − 2| |] + 2 [2 | |]
3 2 3 1 3 1
= −3(17 + 14) + 4(−7)
= −93 − 28
= −121 [by expanding along the last column]


0 1 2 0 1 −2
det( 𝐴) = 2 |−1 2 0| − 3 |−1 2 Always look for the row or
5|
3 1 0 3 1 2 column with most zeros when
applying cofactor expansion
−1 2 1 −2 1 −2
= 2 [2 | |] − 3 [1 | |+3| |] along a row or column.
3 1 1 2 2 5
= 4(−7) − 3(4 + 27)
= −28 − 93
= −121 [by expanding along the first row]
0 2 3 0 2 3
det( 𝐴) = −1 | 1 −2 2| − 3 | 1 −2 2|
1 2 0 2 5 0
0 2 3 0 2 0 2 3 0 2
= −1 [| 1 −2 2| | 1 −2|] − 3 [| 1 −2 2| | 1 −2|]
1 2 0 1 2 2 5 0 2 5
= −1((10) − (−6)) − 3((23) − (−12))

= −16 − 105
= −121 [by expanding along the first column and applying Sarrul’s Rule
or Basket-Weave Method]
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James Training Solutions
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