If I=(1 0 0 1 ) E=(0 1 0 0) show that (aI+bE)=a power3 I+3a square bE
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Step-by-step explanation:
Given If I = (1 0 0 1 ) E = (0 1 0 0) show that (aI+bE)^3 =a^3I + 3a^2bE
- So consider l. h.s we have
- [aI + bE]^3
- So a [1 0 + b [ 0 1
- 0 1] 0 0 ]
- = [ a 0 + [b 0
- 0 a] 0 0]
- = [ a b
- 0 a]
- Now (aI + bE)^2 = [a b [a b
- 0 a] 0 a]
- = [a^2 ab + ab
- 0 a^2 ]
- = [ a^2 2ab
- 0 a^2 ]
- So (aI + bE)^3 = [a^2 2ab [a b
- 0 a^2] 0 a]
- = [ a^3 a^2b + 2a^2b
- 0 a^3 ]
- = [ a^3 3a^2b
- 0 a^3 ] = l. h.s
- Now r. h.s will be
- a^3 I + 3a^2bE
- a^3 [ 1 0 + 3 a^2b [ 0 1
- 0 1] 0 0]
- a^3 [ a^3 0 + [0 3a^2b
- 0 a^3] 0 0]
- [ a^3 3a^2b
- 0 a^3] = r. h.s
Therefore l. h.s = r. h.s
Reference link will be
https://brainly.in/question/25062745
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