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MAT1503 Assignment 3 2021
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UNISA MAT1503 Linear Algebra Assignment THREE 2021 solutions. The necessary working is shown when each problem is solved. Topics covered are: Systems of linear equations. Identifying the number of solutions to a system of linear equations. Gaussian elimination. The row echelon form of a matrix....

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  • May 16, 2021
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MAT1503 ASSIGNMENT 3 2021



Question 1


1 −1 2 1
[ 3 −1 5 | −2 ]
−4 2 𝑥 2 − 8 𝑥 + 2
𝑅2 𝑐ℎ𝑎𝑛𝑔𝑒𝑠 𝑡𝑜 𝑅2 − 3𝑅1
𝑅3 𝑐ℎ𝑎𝑛𝑔𝑒𝑠 𝑡𝑜 𝑅3 + 4𝑅1
1 −1 2 1
[ 3 − 3(1) −1 − 3(−1) 5 − 3(2) | −2 − 3(1) ]
−4 + 4(1) 2 + 4(−1) 𝑥 2 − 8 + 4(2) 𝑥 + 2 + 4(1)


1 −1 2 1
[0 2 −1| −5 ]
0 −2 𝑥 2 𝑥 + 6
𝑅3 𝑐ℎ𝑎𝑛𝑔𝑒𝑠 𝑡𝑜 𝑅3 + 𝑅2
1 −1 2 1
[ 0 2 −1 | −5 ]
0 + 0 −2 + 2 𝑥 2 + (−1) 𝑥 + 6 + (−5)


1 −1 2 1
[0 2 −1 | −5 ]
0 0 𝑥2 − 1 𝑥 + 1


(i)

𝑁𝑜 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛

𝑥2 − 1 = 0 𝑎𝑛𝑑 𝑥 + 1 ≠ 0
(𝑥 + 1)(𝑥 − 1) = 0 𝑎𝑛𝑑 𝑥 ≠ −1
[𝑥 + 1 = 0 𝑜𝑟 𝑥 − 1 = 0] 𝑎𝑛𝑑 𝑥 ≠ −1
[𝑥 = −1 𝑜𝑟 𝑥 = 1] 𝑎𝑛𝑑 𝑥 ≠ −1
𝑥=1
𝑇ℎ𝑒 𝑠𝑦𝑠𝑡𝑒𝑚 𝑜𝑓 𝑙𝑖𝑛𝑒𝑎𝑟 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛𝑠 𝑤𝑖𝑙𝑙 ℎ𝑎𝑣𝑒 𝑛𝑜 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑖𝑓 𝑥 = 1.

,(ii)

𝐸𝑥𝑎𝑐𝑡𝑙𝑦 𝑜𝑛𝑒 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛

𝑥 2 − 1 ≠ 0 𝑎𝑛𝑑 𝑥 + 1 𝑚𝑎𝑦 𝑏𝑒 𝑎𝑛𝑦 𝑟𝑒𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟.
(𝑥 + 1)(𝑥 − 1) ≠ 0
𝑥 + 1 ≠ 0 𝑎𝑛𝑑 𝑥 − 1 ≠ 0
𝑥 ≠ −1 𝑎𝑛𝑑 𝑥 ≠ 1



𝑇ℎ𝑒 𝑠𝑦𝑠𝑡𝑒𝑚 𝑜𝑓 𝑙𝑖𝑛𝑒𝑎𝑟 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛𝑠 𝑤𝑖𝑙𝑙 ℎ𝑎𝑣𝑒 𝑒𝑥𝑎𝑐𝑡𝑙𝑦 𝑜𝑛𝑒 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑖𝑓 𝑥 ≠ −1 𝑎𝑛𝑑 𝑥 ≠ 1.
𝑇ℎ𝑒 𝑠𝑦𝑠𝑡𝑒𝑚 𝑜𝑓 𝑙𝑖𝑛𝑒𝑎𝑟 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛𝑠 𝑤𝑖𝑙𝑙 ℎ𝑎𝑣𝑒 𝑒𝑥𝑎𝑐𝑡𝑙𝑦 𝑜𝑛𝑒 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑖𝑓 𝑥 ∈ ℝ \ {−1, 1}.



(iii)

𝐼𝑛𝑓𝑖𝑛𝑖𝑡𝑒𝑙𝑦 𝑚𝑎𝑛𝑦 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛𝑠

𝑥2 − 1 = 0 𝑎𝑛𝑑 𝑥 + 1 = 0

𝑥2 − 1 = 0 𝑎𝑛𝑑 𝑥 + 1 = 0
(𝑥 + 1)(𝑥 − 1) = 0 𝑎𝑛𝑑 𝑥 = −1
[𝑥 + 1 = 0 𝑜𝑟 𝑥 − 1 = 0] 𝑎𝑛𝑑 𝑥 = −1
[𝑥 = −1 𝑜𝑟 𝑥 = 1] 𝑎𝑛𝑑 𝑥 = −1
[𝑥 = −1 ] 𝑎𝑛𝑑 𝑥 = −1
𝑥 = −1



𝑇ℎ𝑒 𝑠𝑦𝑠𝑡𝑒𝑚 𝑜𝑓 𝑙𝑖𝑛𝑒𝑎𝑟 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛𝑠 𝑤𝑖𝑙𝑙 ℎ𝑎𝑣𝑒 𝑖𝑛𝑓𝑖𝑛𝑖𝑡𝑒𝑙𝑦 𝑚𝑎𝑛𝑦 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛𝑠 𝑖𝑓 𝑥 = −1.



Question 2



Method 1 (Attempt to find the inverse by using the adjoint)

𝐿𝑒𝑡 𝑇 𝑏𝑒 𝑎𝑛 𝑛 × 𝑛 𝑚𝑎𝑡𝑟𝑖𝑥.
𝑁𝑜𝑡𝑒 𝑡ℎ𝑎𝑡 𝑤𝑒 𝑚𝑎𝑦 𝑓𝑖𝑛𝑑 𝑡ℎ𝑒 𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑛𝑡 𝑜𝑓 𝑎 𝑚𝑎𝑡𝑟𝑖𝑥 𝑏𝑦 𝑎𝑑𝑑𝑖𝑛𝑔 𝑎𝑙𝑜𝑛𝑔 𝑎𝑛𝑦 𝑟𝑜𝑤.
𝑇ℎ𝑒 𝑟𝑒𝑠𝑢𝑙𝑡 𝑤𝑖𝑙𝑙 𝑏𝑒 𝑡ℎ𝑒 𝑠𝑎𝑚𝑒 𝑛𝑜 𝑚𝑎𝑡𝑡𝑒𝑟 𝑤ℎ𝑖𝑐ℎ 𝑟𝑜𝑤 𝑤𝑒 𝑐ℎ𝑜𝑜𝑠𝑒.

, 𝐼𝑓 𝑡ℎ𝑒 𝑟𝑜𝑤 𝑖𝑠 𝑜𝑛 𝑎𝑛 𝑜𝑑𝑑 𝑝𝑜𝑠𝑖𝑡𝑖𝑜𝑛 (𝑒. 𝑔. 1𝑠𝑡 𝑟𝑜𝑤 𝑜𝑟 5𝑡ℎ 𝑟𝑜𝑤) 𝑡ℎ𝑒𝑛:
det(𝑇) = +0(𝑐𝑜𝑟𝑟𝑒𝑠𝑝𝑜𝑛𝑑𝑖𝑛𝑔 𝑚𝑖𝑛𝑜𝑟) − 0(𝑐𝑜𝑟𝑟𝑒𝑠𝑝𝑜𝑛𝑑𝑖𝑛𝑔 𝑚𝑖𝑛𝑜𝑟) + ⋯ 𝑢𝑛𝑡𝑖𝑙 𝑡ℎ𝑒 𝑙𝑎𝑠𝑡 𝑧𝑒𝑟𝑜(𝑐𝑜𝑟𝑟𝑒𝑠𝑝𝑜𝑛𝑑𝑖𝑛𝑔 𝑚𝑖𝑛𝑜𝑟)
det(𝑇) = +0 − 0 + 0 − 0 + ⋯ ± 0
det(𝑇) = 0



𝐼𝑓 𝑡ℎ𝑒 𝑟𝑜𝑤 𝑖𝑠 𝑜𝑛 𝑎𝑛 𝑒𝑣𝑒𝑛 𝑝𝑜𝑠𝑖𝑡𝑖𝑜𝑛 (𝑒. 𝑔. 2𝑛𝑑 𝑟𝑜𝑤 𝑜𝑟 8𝑡ℎ 𝑟𝑜𝑤) 𝑡ℎ𝑒𝑛:
det(𝑇) = −0(𝑐𝑜𝑟𝑟𝑒𝑠𝑝𝑜𝑛𝑑𝑖𝑛𝑔 𝑚𝑖𝑛𝑜𝑟) + 0(𝑐𝑜𝑟𝑟𝑒𝑠𝑝𝑜𝑛𝑑𝑖𝑛𝑔 𝑚𝑖𝑛𝑜𝑟) − ⋯ 𝑢𝑛𝑡𝑖𝑙 𝑡ℎ𝑒 𝑙𝑎𝑠𝑡 𝑧𝑒𝑟𝑜(𝑐𝑜𝑟𝑟𝑒𝑠𝑝𝑜𝑛𝑑𝑖𝑛𝑔 𝑚𝑖𝑛𝑜𝑟)
det(𝑇) = −0 + 0 − 0 + 0 − ⋯ ± 0
det(𝑇) = 0



𝑂𝑛 𝑏𝑜𝑡ℎ 𝑜𝑓 𝑡ℎ𝑒 𝑎𝑏𝑜𝑣𝑒 𝑐𝑎𝑠𝑒𝑠 𝑤𝑒 𝑛𝑜𝑡𝑖𝑐𝑒 𝑡ℎ𝑎𝑡 𝑡ℎ𝑒 𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑛𝑡 𝑜𝑓 𝑇 𝑖𝑠 𝑧𝑒𝑟𝑜.
𝑂𝑛𝑒 𝑤𝑎𝑦 𝑜𝑓 𝑓𝑖𝑛𝑑𝑖𝑛𝑔 𝑡ℎ𝑒 𝑖𝑛𝑣𝑒𝑟𝑠𝑒 𝑜𝑓 𝑇 𝑖𝑠 𝑏𝑦 𝑢𝑠𝑖𝑛𝑔 𝑡ℎ𝑒 𝑓𝑜𝑙𝑙𝑜𝑤𝑖𝑛𝑔 𝑓𝑜𝑟𝑚𝑢𝑙𝑎:


1
𝑇 −1 = [𝑎𝑑𝑗𝑜𝑖𝑛𝑡(𝑇)]
det(𝑇)
1
𝑆𝑜, 𝑜𝑛 𝑡ℎ𝑒 𝑎𝑏𝑜𝑣𝑒 𝑓𝑜𝑟𝑚𝑢𝑙𝑎 𝑎𝑙𝑙 𝑡ℎ𝑒 𝑛2 𝑒𝑛𝑡𝑟𝑖𝑒𝑠 𝑖𝑛 𝑡ℎ𝑒 𝑎𝑑𝑗𝑜𝑖𝑛𝑡 𝑚𝑎𝑡𝑟𝑖𝑥 𝑚𝑢𝑠𝑡 𝑏𝑒 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑖𝑒𝑑 𝑏𝑦 .
det(𝑇)
1
𝑁𝑜𝑤, det(𝑇) = 0. 𝑇ℎ𝑖𝑠 𝑖𝑚𝑝𝑙𝑖𝑒𝑠 𝑡ℎ𝑎𝑡 𝑖𝑠 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑.
det(𝑇)
𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑇 −1 𝑖𝑠 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑.
𝑇 −1 𝑑𝑜𝑒𝑠 𝑛𝑜𝑡 𝑒𝑥𝑖𝑠𝑡.
𝑆𝑜, 𝑇 𝑑𝑜𝑒𝑠 𝑛𝑜𝑡 ℎ𝑎𝑣𝑒 𝑎𝑛 𝑖𝑛𝑣𝑒𝑟𝑠𝑒.



Method 2 (Verify if we can multiply T by a certain matrix then get the identity matrix)

𝐿𝑒𝑡 𝑇 𝑏𝑒 𝑎𝑛 𝑛 × 𝑛 𝑚𝑎𝑡𝑟𝑖𝑥.



𝐼𝑓 𝑇 ℎ𝑎𝑠 𝑎𝑛 𝑖𝑛𝑣𝑒𝑟𝑠𝑒, 𝑡ℎ𝑒𝑛 𝑡ℎ𝑒𝑟𝑒 𝑒𝑥𝑖𝑠𝑡𝑠 𝑎𝑛 𝑛 × 𝑛 𝑚𝑎𝑡𝑟𝑖𝑥 𝑋 𝑠𝑢𝑐ℎ 𝑡ℎ𝑎𝑡:
𝑇𝑋 = 𝐼𝑛 𝑎𝑛𝑑 𝑋𝑇 = 𝐼𝑛
1 0 0 0 0 0 1 0 0 0 0 0
0 1 0 … 0 0 0 0 1 0 … 0 0 0
0 0 1 0 0 0 0 0 1 0 0 0
𝑇𝑋 = ⋮ ⋱ ⋮ 𝑎𝑛𝑑 𝑋𝑇 = ⋮ ⋱ ⋮
0 0 0 1 0 0 0 0 0 1 0 0
0 0 0 … 0 1 0 0 0 0 … 0 1 0
[0 0 0 0 0 1] [0 0 0 0 0 1]

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