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Question : 19 of 90
Marks:
+1,
-0
Solution:
Let
A=[] Given
A[]=[] .......(1)
‌∴[]=[] ‌∴x1+z1=2 ...........(2)
‌x2+z2=0 .............(3)
‌x3+z3=0 ..............(4)
‌‌ Given ‌A[]=[] ‌∴[]=[] ‌⇒−x1+z1=−4 ............(5)
‌−x2+x2=0 .............(6)
‌−x3+z3=4 ................(7)
‌‌ Given ‌A[]=[]‌∴[]=[]‌∴y1=0,y2=2,y3=0‌∴‌ from ‌(2),(3),(4),(5),(6)‌ and ‌(7)‌x1=3x,x2=0,x3=−1‌y1=0,y2=2,y3=0‌z1=−1,z2=0,z3=3‌∴A=[] ‌∴‌ Now ‌(A−31)[] =[] ‌∴[][] =[] ‌[]=[] ‌[z=−1],[y=−2],[x=−3]
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