ICSE Class X Math 2014 Solved Paper

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Question : 37
Total: 52
Let A=[
21
0−2
]
,B=[
41
−3−2
]
and C=[
−32
−14
]
Find A2+AC−5B.
Solution:  
Given:
A=[
21
0−2
]
,B‌=[
41
−3−2
]
,
C=[
−32
−14
]

A2=A⋅A‌=[
21
0−2
]
[
21
0−2
]

‌=[
4+02−2
0+00+4
]

‌=[
40
04
]

AC‌=[
21
0−2
]
[
−32
−14
]

‌=[
−6−14+4
0+20−8
]
=[
−78
2−8
]

and 5B‌=5[
41
−3−2
]

‌=[
205
−15−10
]

Now,
‌A2+AC−5B=[
40
04
]
+[
−78
2−8
]
‌‌−[
205
−15−10
]

‌‌=[
4−7−200+8−5
0+2+154−8+10
]

‌‌=[
−233
176
]
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