If `A=[[0, 1,2],[1,2,3],[3,a,1]]and A^(-1)[[1//2,-1//2,1//2],[-4,3,b],[5//2,-3//2,1//2]]` thenA. `a = 1, b = -1`B. `a = 2, b = -1/2`C. `a = -1, b=1`D. `a=1/2, b=1/2`

If `A=[[0, 1,2],[1,2,3],[3,a,1]]and A^(-1)[[1//2,-1//2,1//2],[-4,3,b],

1.	If `A=[[0, 1,2],[1,2,3],[3,a,1]]and A^(-1)[[1//2,-1//2,1//2],[-4,3,b],[5//2,-3//2,1//2]]` thenA. `a = 1, b = -1`B. `a = 2, b = -1/2`C. `a = -1, b=1`D. `a=1/2, b=1/2`
Answer» Correct Answer - A We have, `A A^(-1) = I` `rArr [[0,1,2],[1,2,3],[3,a,1]] [[1/2,-1/2,1/2],[-4,3,6],[5/2,-3/2,1/2]]= [[1,0,0],[0,1,0],[0,0,1]]` `rArr [[1,0,b+1],[0,1,2(b+1)],[4(1-a),3(a-1),ab+2]]= [[1,0,0],[0,1,0],[0,0,1]]` On comparing, we get `b+ 1 + 0 , ab + 2 =1, a- 1 = 0` `therefore a = 1, b= -1`

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If `A=[[0, 1,2],[1,2,3],[3,a,1]]and A^(-1)[[1//2,-1//2,1//2],[-4,3,b],[5//2,-3//2,1//2]]` thenA. `a = 1, b = -1`B. `a = 2, b = -1/2`C. `a = -1, b=1`D. `a=1/2, b=1/2`

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