63. Let A = , show that (aI + bA)n = anI + nan – 1 bA, where I is the identity matrix of order 2 and n N.
63. Let A = , show that (aI + bA)n = anI + nan – 1 bA, where I is the identity matrix of order 2 and n N.
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1 Answer
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We shall prove the result by using principal of mathematical induction
we have,
P (n) :- If A = , then (a I + b A)n = an I + nan - 1b A where I is identity matrix of order 2, n∈N
P (1): (a I + b A)1 = a1I + 1 ´a1 - 1b A
= a I + a0bA
= a1I + b A {Qx0 = 1}
So the result is true for n = 1.
Let the result be true for n = k. So,
P (k): (a I + b A)u = auI + u. au-1b A. _____ (1)
Now, we prove that the result holds for n = k + 1,
P (k + 1): (a I + b A)k + 1 = (a I + b A). (a I + b A)k
= (a I + b A) (auI + k au 1b A){using eqn (1)}
= a.akI2 + k. a au - 1b IA + akb.AI + a zzk 1b2k A2
= ak + 1I2 + k au - 1+1b IA + ak b AI + ak - 1b2 k A2
...more
Similar Questions for you
Let
Given ...(1)
∴ x1 + z1 = 2 … (2)
x2 + z2 = 0 … (3)
x3 + z3 = 0 … (4)
Given
⇒ – x1 + z1 = −4 … (5)
–x2 + z2 = 0 &nbs
g (x) = px + q
Compare 8 = ap2 …………… (i)
-2 = a (2pq) + bp
0 = aq2 + bq + c
=>4x2 + 6x + 1 = apx2 + bpx + cp + q
=> Andhra Pradesh = 4 ……………. (ii)
6 = bp
1 = cp + q
From (i) & (ii), p = 2, q = -1
=> b = 3, c = 1, a = 2
f (x) = 2x2 + 3x + 1
f (2) = 8 + 6 + 1 = 15
g (x) = 2x – 1
g (2) = 3
Kindly consider the following figure
B = (I – adjA)5
Kindly consider the following figure
B = (I – adjA)5
System of equation is
R1 – 2 R2, R3 – R2
System of equation will have no solution for 
= -7.
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