1
Question
If I=pA3+qA, where I is an identity matrix of order 3 and A=⎡⎢⎣10−2−2−22341⎤⎥⎦, then find the value of p and q ?
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Solution
A=⎡⎢⎣10−2−2−22341⎤⎥⎦
A2=⎡⎢⎣10−2−2−22341⎤⎥⎦×⎡⎢⎣10−2−2−22341⎤⎥⎦
=⎡⎢⎣−5−8−48122−2−43⎤⎥⎦
A3=⎡⎢⎣10−2−2−22341⎤⎥⎦×⎡⎢⎣−5−8−48122−2−43⎤⎥⎦
=⎡⎢⎣−10−10−10−16101520−1⎤⎥⎦
Given, I=pA3+qA
⇒⎡⎢⎣100010001⎤⎥⎦=p⎡⎢⎣−10−10−10−16101520−1⎤⎥⎦+q⎡⎢⎣10−2−2−22341⎤⎥⎦
⇒⎡⎢⎣100010001⎤⎥⎦=⎡⎢⎣−p+q0−10p−2q−10p−2q−16p−2q10p+2q15p+3q20p+4q−p+q⎤⎥⎦
On equating both the sides, we get
1=−p+q ...(1)0=−10p−2q ...(2)
⇒0=5p+q ...(3)
Eqn (3)−(1), gives
6p=−1
⇒p=−16, q=56
A2=⎡⎢⎣10−2−2−22341⎤⎥⎦×⎡⎢⎣10−2−2−22341⎤⎥⎦
=⎡⎢⎣−5−8−48122−2−43⎤⎥⎦
A3=⎡⎢⎣10−2−2−22341⎤⎥⎦×⎡⎢⎣−5−8−48122−2−43⎤⎥⎦
=⎡⎢⎣−10−10−10−16101520−1⎤⎥⎦
Given, I=pA3+qA
⇒⎡⎢⎣100010001⎤⎥⎦=p⎡⎢⎣−10−10−10−16101520−1⎤⎥⎦+q⎡⎢⎣10−2−2−22341⎤⎥⎦
⇒⎡⎢⎣100010001⎤⎥⎦=⎡⎢⎣−p+q0−10p−2q−10p−2q−16p−2q10p+2q15p+3q20p+4q−p+q⎤⎥⎦
On equating both the sides, we get
1=−p+q ...(1)0=−10p−2q ...(2)
⇒0=5p+q ...(3)
Eqn (3)−(1), gives
6p=−1
⇒p=−16, q=56
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