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Summary matrix and linear de with solution

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a study guide of matrix and linear de with solution

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lOMoARcPSD|16248954




EUM114- Advanced Engineering Calculus


Tutorial 4 (Matrix and Linear Differential Equations)

1. Let
a b c
A =  d e f  if det( A) = −7 .

g h i 
a g d
Compute (i) det (2 A )  (ii) b
−1

 
h e.
c i f
[Ans: i) -1/56 ii) 7 ]

Solution:

(i)
det (2 A )−1  = 1
= 3
1
= 3
1
=
−1
  det(2 A) 2 det( A) 2  (−7) 56
(ii)
a g d a b c
b h e = g h i property (a)
cifde f
a b c
= −1 d e f interchanging the 2nd and 3rd rows
g h i
= − det( A) = 7



2. Solve




by using Cramer's rule. [ans:x1=1 , x2=2 , x3=3 ]

Solution First we calculate :




Now substitute


SK/EUM114/MATRIX & LINEAR_DE/TUT1/2020
1

, lOMoARcPSD|16248954




EUM114- Advanced Engineering Calculus




for column 1 and calculate




Similarly,




3. Solve the system of linear equations by reducing the augmented matrix to its row echelon
form.
x1 – 2x2 + x3 = –5
2x1 + x2 – x3 = –
2 x1 + 3x2 + x3 =

1 −2 01 −5
[Ans: 0 1 0  1  ]
0 0 1 −1

Solution
The augmented matrix is



SK/EUM114/MATRIX & LINEAR_DE/TUT1/2020
2

, lOMoARcPSD|16248954




EUM114- Advanced Engineering Calculus


1 −2 1 −5 1 −2 1 −5 1 −2 1 −5
2 1 −1 −2  ⎯R⎯ ⎯⎯⎯
2−2 R1 R 2
→ 0 5 −3 8 ⎯ ⎯ ⎯ ⎯→ 0 5 −3 8 
  R 3− R 2 R 3
  R3−R1R3    
1 3 1 0  0 5 0 5 
 0 0 3 −3

1 −2 1 −5 1 1 −2 1 −5
⎯ ⎯ ⎯ ⎯⎯ 
→ 0 5 0 5 ⎯⎯⎯⎯
R 2+ R 3 R 2

 5
1
→ 0 1 0 1 

R 3 R 3
 3
R 2 R 2
0 0 1 −1
0 0 3 −3
1 1 1 

4.Compute A −1 , given A =  2 0 −1 .
0 1 1 

1 0 −1
[Ans: A −1
=  −2 1 3  ]

 2 −1 −2




Solution

First, we form the augmented matrix A I  and perform row operations on A I  so that the

augmented matrix becomes the form of I B where B is the inverse of A.

Thus we have,

1 1 1 1 0 0 1 1 1 1 0 0
2   
 0 −1 0 1 0  → 0 −2 −3 −2 1 0 
0 1 1 0 0 1 0 1 1 0 0 1

1 1 1 1 0 0 1 1 1 1 0 0

→ 0 −2 −3 −2 1 0 → 0 −2 −3 −2
 1 0
0 0 −1 −2 1 2 0 0 1 2 −1 −2

1 1 1 1 0 0  1 1 1 1 0 0
→ 0 −2 0 4 −2 −6 → 0

1 0 −2 1 3
0 0 1 2 −1 −2 0 0 1 2 −1
−2


1 0 1 3 −1 −3 1 0 0 1 0 −1

→ 0 1 0 −2 1 3  → 0 1 0 −2 1 3 
0 0 1 2 −1 −2 0 0 1 2 −1 −2

SK/EUM114/MATRIX & LINEAR_DE/TUT1/2020

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