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−R1R3
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
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−R1R3
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