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TMC1874 Mathematics For Computing

Tutorial Solution
Chapter 3: Determinants

1
...

(a)

t
2

−3
t−1

(b)

t−1
−2
0

0
t
0

1
−1
t+1

Solution

(b)

−3
= t(t − 1) + 6 = t2 − t + 6
...
Compute the following determinants via reduction to triangular form
...
K
...
K
...

13

a1
3
...

c3

Solution

a1
b1
c1
(3R 2 + R 1 → R 1 )

(−2R 3 + R 1 → R 1 )

W
...
Tiong

a2
b2
c2

a 1 + 3b 1
b1
c1

a3
b3
c3

=5

a 2 + 3b 2
b2
c2

a 1 + 3b 1 − 2c 1
b1
c1

a 3 + 3b 3
b3
c3

a 2 + 3b 2 − 2c 2
b2
c2

=5

a 3 + 3b 3 − 2c 3
b3
c3

=5

Page 3 of 12

TMC1874 Mathematics For Computing

a 1 + 3b 1 − 2c 1
b1
c1

∴

a 2 + 3b 2 − 2c 2
b2
c2

a 3 + 3b 3 − 2c 3
b3
= 5
...
Evaluate following determinants via reduction to triangular form
...
K
...

t+2

(d) One possible solution:
t+1
2

t+1
−4
=
t−3

−4

(−

8+( t−3)( t+1)
t +1
2

2
R1 + R2 → R2)
t+1

= 8 + (t − 3)(t + 1) = t − 2t + 5
...
Let A =
4


2
1
3


−3
4 
...
Let A = 


2
1
−1
0

−1
2
1
2

0
−2
−3
−1

(a) det(M12 )

3
4
−2
5






...
K
...
Apply the theorem on page 21 (refer to lecture notes) to evaluate the given determinants
...
K
...
Let A =  −3
4

2
4
−4




8
1 
...

Solution
4
−4

A 11 =
A 21 =
A 31 =

−

1
−3
= 24, A 12 = −
5
4

1
−3
= 19, A 13 =
5
4

4
= −4,
−4

8
6
= −42, A 22 =
5
4

8
6
= −2, A 23 = −
5
4

2
= 32,
−4

2
−4
2
4

8
6
= −30, A 32 = −
1
−3

8
6
= −30, A 33 =
1
−3

2
= 30
...


W
...
Tiong

Page 7 of 12

TMC1874 Mathematics For Computing

Solution
6
| A | = −3
4

2
4
−4

8
4
1 =6
−4
5

1
−3
−2
5
4

1
−3
+8
5
4

4
−4

= 6(24) − 2(−19) + 8(−4)
= 150

(c) Show that A(adj A) = (adj A)A = det(A)I 3
...
Given
A −1 =

1
1
1
5

1
3
2
9

1
(d)  0
1

1
2
1



0
0  = det(A)I 3
1




0
0  = det(A)I 3
1

1
(adj A)
...
K
...

2

Finding cofactors:
1
2

A 21 =

−

A 31 =

2
2
= −4, A 12 = −
0
3

1
2

2
1
= 4, A 22 =
0
3

1
1

A 11 =

2
2
= 6, A 13 =
0
3

1
= 1,
2

2
1
= −6, A 23 = −
0
3

1
= 1,
2

2
1
= 0, A 32 = −
2
2

2
1
= 2, A 33 =
2
2

1
= −1
...

−1

Finding inverse:

−4
1
1
6
A −1 =
(adj A) =
det(A)
4
1

4
−6
1

 
−1
0
2 = 3
2
1
−1
4

1

0

−3
2

1
2
−1
4

1
4




...
K
...

3

By following the same procedure, we can find the value of the rest of cofactors
accordingly
...


1
2 −1 −2 
15 −6 −6 −3
Hence,

A −1 =



−21
−4
1
15

1
1

(adj A) = − 
det(A)
9




= 


7
3
4
9
1
−9
5
−3

3
1
2
−6

−1
3
−1
9
−2
9

3
4
−1
−6

−1
3
−4
9

−2
3
1
9
2
9
1
3

1
9
2
3

2
3

6
−1
−2
−3











...
K
...


Page 10 of 12

TMC1874 Mathematics For Computing

The matrix is singular or noninvertible since the determinant is 0
...

3

Finding cofactors:
2
1

A 21 =

−

A 31 =

3
0
= 1, A 12 = −
2
1

3
0
= 3, A 13 =
2
1

1
1

1
1
= −1, A 22 =
2
1

1
1
= 1, A 23 = −
2
1

1
= 0,
1

1
2

A 11 =

2
= −2,
1

1
1
= 1, A 32 = −
3
0

1
1
= −3, A 33 =
3
0

1
= 2
...

2

Finding inverse:
1
1
1
−1
3
(adj A) =
A =
det(A)
2
−2

−1
1
0



  1
1
2
−3  =  3
2
2
−1

−1
2
1
2

0

0



−3 
...

1

The matrix is singular or noninvertible since the determinant is 0
...
If possible, solve the following linear system by Cramer’s rule:
3x
x
x

+
+
+

4y
2y
3y

+

6z

−

z

=
=
=

4
4
...
K
...

−1

Next, compute the determinant:
3
|A| = 1
1

4
2
3

6
0 = 4
...
Next, find the solution
4
4
8
3
1
1

y =

z

=

|A|
4
6
4
0
8 −1

3
1
1

x =

4
2
3

6
0
−1

|A|
4 4
2 4
3 8
|A|

=

−16
= −4,
4

=

16
= 4,
4

=

0
= 0
...


11
...

3

Solution
1
The coefficient matrix is A =  1
0
1
...
K
...
So the determinant is | A | = 1
0
1

5
4
3

−2
−3
2

3
2 =
1

1
= − = −1

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