Mat1503 Oct/Nov 2017 Solutions

MAT1503
October/November 2017


QUESTION 1

(a) (i) No solution: The 𝑟𝑟𝑒𝑓 matrix has a row with nonzero number
(leading entry) in the last (augmented) column of [𝐴|𝑏] and 0′𝑠 in
all of the columns of the coefficient part.
(ii) Infinitely many solutions: The number of variables in the
coefficient part of [𝐴|𝑏] is more than the number of nonzero rows
in the last (augmented) column of [𝐴|𝑏].
(iii) Unique solution: The number of variables in the coefficient part
of [𝐴|𝑏 is equal to the number of nonzero rows equal to the in the
last (augmented) column of [𝐴|𝑏].


(b) (i) To be discussed, note this for now: A determinant is calculated
only for square matrices (n x n). In essence 𝑑𝑒𝑡(𝐸) ≠ 0, since it
cannot be calculated.

(ii) A 2 𝑥 2 matrix is not invertible 𝑖𝑓𝑓 𝑎𝑑 − 𝑏𝑐 = 0.
𝑎+𝑏−1 0
𝐴=[ ] ⟹ det(𝐴) 𝑜𝑟 |𝐴| = 0
0 3
⟹ (𝑎 + 𝑏 − 1)(3) − (0)(0) = 0
3−3𝑏 3−3𝑎
3𝑎 + 3𝑏 − 3 = 0 ∴ 𝑎 = 𝑎𝑛𝑑 𝑏 = 𝑓𝑜𝑟 𝐴 𝑡𝑜 𝑏𝑒 𝑖𝑛𝑣𝑒𝑟𝑡𝑖𝑏𝑙𝑒.
3 3

5 0
𝐵=[ ] ⟹ det(𝐴) 𝑜𝑟 |𝐴| = 0
0 2𝑎 − 3𝑏 − 7
⟹ (5)(2𝑎 − 3𝑏 − 7) − (0)(0) = 0
10𝑎 − 15𝑏 − 35 = 0
15𝑏+35 10𝑎−35
∴𝑎= 𝑎𝑛𝑑 𝑏 = 𝑓𝑜𝑟 𝐵 𝑡𝑜 𝑏𝑒 𝑖𝑛𝑣𝑒𝑟𝑡𝑖𝑏𝑙𝑒.
10 15




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, 2 1 𝑥 𝑦
(c) 𝐴=[ ] and B = [𝑦 𝑧 ] 𝑖𝑓 𝑥 − 𝑧 = 7𝑦 𝑡ℎ𝑒𝑛 𝑥 = 7𝑦 + 𝑧
1 −5
2 1 7𝑦 + 𝑧 𝑦 (2)(7𝑦 + 𝑧) + (1)(𝑦) (2)(𝑦) + (1). (𝑧)
⟹ AB = [ ][ ]=[ ]
1 −5 𝑦 𝑧 (1)(7𝑦 + 𝑧) + (−5)(𝑦) (1)(𝑦 ) + (−5)(𝑧)
15𝑦 + 2𝑧 2𝑦 + 𝑧
=[ ]
2𝑦 + 𝑧 𝑦 − 5𝑧
2 1 𝑥 𝑦
𝐴=[ ] and B = [𝑦 𝑧 ] 𝑖𝑓 𝑥 − 𝑧 = 7𝑦 𝑡ℎ𝑒𝑛 𝑥 = 7𝑦 + 𝑧
1 −5
7𝑦 + 𝑧 𝑦 2 1 (7𝑦 + 𝑧)(2) + (𝑦)(1) (7𝑦 + 𝑧)(1) + (𝑦)(−5)
⟹ BA = [ ][ ]=[ ]
𝑦 𝑧 1 −5 (𝑦)(2) + (𝑧)(1) (𝑦 )(1) + (𝑧)(−5)
15𝑦 + 2𝑧 2𝑦 + 𝑧
=[ ]
2𝑦 + 𝑧 𝑦 − 5𝑧
∴ 𝐴𝐵 = 𝐵𝐴 𝑖𝑓 𝑥 − 𝑧 = 7𝑦∎




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