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Question : 133 of 180
Marks:
+1,
-0
Solution:
We have,
A=[]Now,
‌‌A‌=[][]‌=[| 1−6 | −3−3k |
| 2+2k | −6+k2 |
]=[]Now,
A2−4A+10I=A‌⇒[]−4[]‌+10[]=[]‌⇒[| −5−4+10 | −3−3k+12+0 |
| 2+2k−8+0 | k2−6−4k+10 |
]=[]‌⇒[| −1 | −3k+9 |
| 2k−6 | k2−4k+4 |
]=[]On comparing both sides, we get
‌k2−4k+4=k⇒k2−5k+4=0⇒k2−4k−k+4=0⇒k(k−4)−1(k−4)=0⇒(k−4)(k−1)=0⇒k=4,1
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