>
CONFIDENTIAL 201904/BSC124/Midterm
Question 1 (25 marks)
2 3 2 3 2 3
2 1 3 1 4
(a) Let A = 4 5, B = 4 5 and u = 4 5. Compute the following:
1 3 0 2 3
(i) Au [2]
(ii) A + B [2]
(iii) 3A [2]
(iv) B T [2]
(v) AB [3]
(vi) B T AT [3]
2 32 3 2 3
he
2 1 4 5
(i) Au = 4 54 5=4 5 [2]
1 3 3 13
2 3 2 3 2 3
2 1 3 1 5 2
(ii) A + B = 4 5+4 5=4 5 [2]
1 3 0 2 1 5
2 3 2 3
2 1 6 3
(iii) 3A = 3 4 5=4 5 [2]
1 3 3 9
2 3T 2 3
3 1 3 0
(iv) B T = 4 5 =4 5 [2]
0 2 1 2
2 32 3 2 3
2 1 3 1 6 4
(v) AB = 4 54 5=4 5 [3]
1 3 0 2 3 5
2 3
6 3
(vi) B T AT = (AB)T = 4 5 [3]
4 5
·
-1pt for each minor mistake.
(b) Solve the following linear system and write the solution in parametric vector form.
x1 + 3x2 + 2x3 = 1
3x1 7x2 8x3 = 3
2x1 + 2x2 + 8x3 = 10
Page 1 of 8
, ·
·
CONFIDENTIAL 201904/BSC124/Midterm
[5]
We transform the augmented matrix into row echelon form.
2 3
1 3 2 1
t
6 7
6 7
6 3 7 8 37 [1pt]
4 5
2 2 8 10
2 3
Rat3Ri t Ra 1 3 2 1
6 7
6 7
>
R2 + 3R1 , R3 2R1 60 2
4
2 67
5
[+1pt]
0 4 4 12
2 3
1 3 2 1
6 7
t R3
6 7
R3 + 2R2 60 2 2 67 [+1pt]
> 4 5
0 0 0 0
This gives
I 0518
&
x3 = t
x2 = t 3
R
>
-
, ER2TRI
L86(3)
·
x1 = 5t + 8 [+1pt]
2 3 2 3
8 5
6 7 6 71/
6 7 6 7
or in Isparametric vector form x = 6 37 + t 6 1 7. [+1pt]
- 4 5 4 5
0 1
(c) For which h is the following linear system consistent?
3x + hy = 1
[ R [Con)
2x 4y = 2
[3]
Transforming the augmented matrix in row echelon form or
se
solving the linear system leads to the equation h 6= 6 [2pt].
Hence the system is consistent if and only if h 6= 6. ·[+1pt]
h = d then OxitOx= 2 *(
2
1 0
3
& if
(d) Let A = 4 5. Is there a matrix B such that AB = 2BA? Justify your answer. [3]
0 2
Page 2 of 8
CONFIDENTIAL 201904/BSC124/Midterm
Question 1 (25 marks)
2 3 2 3 2 3
2 1 3 1 4
(a) Let A = 4 5, B = 4 5 and u = 4 5. Compute the following:
1 3 0 2 3
(i) Au [2]
(ii) A + B [2]
(iii) 3A [2]
(iv) B T [2]
(v) AB [3]
(vi) B T AT [3]
2 32 3 2 3
he
2 1 4 5
(i) Au = 4 54 5=4 5 [2]
1 3 3 13
2 3 2 3 2 3
2 1 3 1 5 2
(ii) A + B = 4 5+4 5=4 5 [2]
1 3 0 2 1 5
2 3 2 3
2 1 6 3
(iii) 3A = 3 4 5=4 5 [2]
1 3 3 9
2 3T 2 3
3 1 3 0
(iv) B T = 4 5 =4 5 [2]
0 2 1 2
2 32 3 2 3
2 1 3 1 6 4
(v) AB = 4 54 5=4 5 [3]
1 3 0 2 3 5
2 3
6 3
(vi) B T AT = (AB)T = 4 5 [3]
4 5
·
-1pt for each minor mistake.
(b) Solve the following linear system and write the solution in parametric vector form.
x1 + 3x2 + 2x3 = 1
3x1 7x2 8x3 = 3
2x1 + 2x2 + 8x3 = 10
Page 1 of 8
, ·
·
CONFIDENTIAL 201904/BSC124/Midterm
[5]
We transform the augmented matrix into row echelon form.
2 3
1 3 2 1
t
6 7
6 7
6 3 7 8 37 [1pt]
4 5
2 2 8 10
2 3
Rat3Ri t Ra 1 3 2 1
6 7
6 7
>
R2 + 3R1 , R3 2R1 60 2
4
2 67
5
[+1pt]
0 4 4 12
2 3
1 3 2 1
6 7
t R3
6 7
R3 + 2R2 60 2 2 67 [+1pt]
> 4 5
0 0 0 0
This gives
I 0518
&
x3 = t
x2 = t 3
R
>
-
, ER2TRI
L86(3)
·
x1 = 5t + 8 [+1pt]
2 3 2 3
8 5
6 7 6 71/
6 7 6 7
or in Isparametric vector form x = 6 37 + t 6 1 7. [+1pt]
- 4 5 4 5
0 1
(c) For which h is the following linear system consistent?
3x + hy = 1
[ R [Con)
2x 4y = 2
[3]
Transforming the augmented matrix in row echelon form or
se
solving the linear system leads to the equation h 6= 6 [2pt].
Hence the system is consistent if and only if h 6= 6. ·[+1pt]
h = d then OxitOx= 2 *(
2
1 0
3
& if
(d) Let A = 4 5. Is there a matrix B such that AB = 2BA? Justify your answer. [3]
0 2
Page 2 of 8
