Value of 2a(a^3+7a-5a+4) for a =(-1) is :
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Class 12
>>Maths
>>Determinants
>>Inverse of a Matrix Using Adjoint
>>If A = 3 1 | - 1 2 , show that A^2 -
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If A=[
3
−1
1
2
], show that A
2
−5A+7I=0. Hence, find A
−1
.
Medium
Solution
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We have, [
3
−1
1
2
]
∴A
2
=AA= [
3
−1
1
2
][
3
−1
1
2
]=[
8
−5
5
3
]
So, A
2
−5A+7I= [
8
−5
5
3
]−5[
3
−1
1
2
]+7[
1
0
0
1
]=[
8−15+7
−5+5+0
5−5+0
3−10+7
]=[
0
0
0
0
]=O
Now, A
2
−5A+7I=O
⇒A
−1
(A
2
−5A+7I)=A
−1
=O [Multiplying throughout by A
−1
]
⇒A
−1
A
2
−5A
−1
A+7A
−1
I=O
⇒A−5I+7A
−1
=O
⇒7A
−1
=5I−A
7A
−1
= [
5
0
0
5
]−[
3
−1
1
2
]=[
2
1
−1
3
]
⇒A
−1
=
7
1
[
2
1
−1
3
]
Answer dentify the terms, their coefficients for each of the following expressions:
(i) 5xyz2 – 3zy (ii) 1 + x + x2 (iii) 4x2y2 – 4x2y2z2 + z2 (iv) 3 – pq + qr – rp (v) (vi) 0.3 a – 0.6 ab + 0.5 b
Sol. (i) 5xyz2 – 3zy
Step-by-step explanation: