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CBSE Class 12 Math 2008 Solved Paper

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Question : 13
Total: 29
Let A = |
325
413
067
|. Express A as sum of two matrices such that one is symmetric and the other is skew symmetric.
OR
If A = |
122
212
221
|, verify that A2 - 4A - 5I = 0
Solution:  
A = [
325
413
067
]
A' = [
340
216
537
]
Now, A can be written as:
A =
1
2
 A + A' +
1
2
 A - A'
A + A' = [
325
413
067
]+[
340
216
537
]
= [
3+32+45+0
4+21+13+6
0+56+37+7
]
= [
665
629
5914
]
1
2
 A + A' =
1
2
[
665
629
5914
] = [
33
5
2
31
9
2
5
2
9
2
7
] = P, say
Now, P' = [
33
5
2
31
9
2
5
2
9
2
7
]
Thus, P =
1
2
 A + A' is a symmetric matrix.
Now, A - A' = [
3−32−45−0
4−21−13−6
0−56−37−7
] = [
0−25
20−3
−530
]
1
2
 A - A' = [
0−1
5
2
10−
3
2
−
5
2
3
2
0
] = Q, say
Now, Q' = [
01−
5
2
−10
3
2
5
2
−
3
2
0
] = - [
0−1
5
2
10−
3
2
−
5
2
3
2
0
] = - Q
Thus, Q =
1
2
 A - A' is a skew symmetric matrix.
∴ A = [
325
413
067
] = [
33
5
2
31
9
2
5
2
9
2
7
] + [
0−1
5
2
10−
3
2
−
5
2
3
2
0
]
OR
A2 = [
122
212
221
] [
122
212
221
]
=
[
1×1+2×2+2×22×1+2×1+2×21×2+2×2+2×1
2×1+1×2+2×22×2+1×1+2×22×2+1×2+2×1
2×1+2×2+1×22×2+2×1+1×22×2+2×2+1×1
]

=
[
1+4+42+2+42+4+2
2+2+44+1+44+2+2
2+4+24+2+24+4+1
]
= [
988
898
889
]
4A = 4[
122
212
221
] =
[
1×42×42×4
2×41×42×4
2×42×41×4
]
= [
488
848
884
]
5I = 5 [
100
010
001
] = [
500
050
005
]
A2 - 4A - 5I = [
988
898
889
] - [
488
848
884
] - [
500
050
005
]
=
[
9−4−58−8−08−8−0
8−8−09−4−58−8−0
8−8−08−8−09−4−5
]

= [
000
000
000
] = 0 = R.H.S.
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