. If = −1
(
+
−
), then show that
2
2
2 + 2
2
+
2
2
2 = 0.
Answers
Answer:
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MATHS
If A=
⎣
⎢
⎢
⎡
1
2
1
3
0
2
2
−1
3
⎦
⎥
⎥
⎤
, then show that A satisfies the equation A
3
−4A
2
−3A+11I=O
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ANSWER
A
2
=A×A=
⎣
⎢
⎢
⎡
1
2
1
3
0
2
2
−1
3
⎦
⎥
⎥
⎤
×
⎣
⎢
⎢
⎡
1
2
1
3
0
2
2
−1
3
⎦
⎥
⎥
⎤
⎣
⎢
⎢
⎡
1+6+2
2+0−1
1+4+3
1+0+4
6+0−2
3+0+6
2−3+6
4+0−3
2−2+9
⎦
⎥
⎥
⎤
=
⎣
⎢
⎢
⎡
9
1
8
7
4
9
5
1
9
⎦
⎥
⎥
⎤
And A
3
=A
2
×A=
⎣
⎢
⎢
⎡
9
1
8
7
4
9
5
1
9
⎦
⎥
⎥
⎤
×
⎣
⎢
⎢
⎡
1
2
1
3
0
2
2
−1
3
⎦
⎥
⎥
⎤
=
⎣
⎢
⎢
⎡
9+14+5
1+8+1
8+18+9
28+0+10
3+0+2
27+0+18
18−7+15
2−4+3
16−9+27
⎦
⎥
⎥
⎤
=
⎣
⎢
⎢
⎡
28
10
35
37
5
42
26
1
34
⎦
⎥
⎥
⎤
Now A
3
−4A
2
−3A+11(I)
=
⎣
⎢
⎢
⎡
28
10
35
37
5
42
26
1
34
⎦
⎥
⎥
⎤
−4
⎣
⎢
⎢
⎡
9
1
8
7
4
9
5
1
9
⎦
⎥
⎥
⎤
−3
⎣
⎢
⎢
⎡
1
2
1
3
0
2
2
−1
3
⎦
⎥
⎥
⎤
+11
⎣
⎢
⎢
⎡
1
0
0
0
1
0
0
0
1
⎦
⎥
⎥
⎤
=
⎣
⎢
⎢
⎡
28−36−3+11
10−4−6+0
35−32−3+0
37−28−9+0
5−16+0+11
42−36−6+0
26−20−6+0
1−4+3+0
34−36−9+11
⎦
⎥
⎥
⎤
=
⎣
⎢
⎢
⎡
0
0
0
0
0
0
0
0
0
⎦
⎥
⎥
⎤
=0
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