If a and b are any two real numbers, then {vmatrix} 2a-2b-4 & 4a & 4a \\ 4 & 2-b-a & 4 \\ 2b & 2b & b-a-2 {vmatrix}=

Correct Answer is : 2[(a+b)3+6(a+b)2+12(a+b)+8]2[(a+b)³+6(a+b)²+12(a+b)+8]2[(a+b)3+6(a+b)2+12(a+b)+8]

Correct Answer is : 2[(a+b)3+6(a+b)2+12(a+b)+8]2[(a+b)³+6(a+b)²+12(a+b)+8]2[(a+b)3+6(a+b)2+12(a+b)+8]

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TS EAMCET 14-Sep-2020 Shift 2 Solved Paper

Section: Mathematics

Question : 4 of 160

Marks: +1, -0

\begin{vmatrix} 2a-2b-4 & 4a & 4a \\ 4 & 2-b-a & 4 \\ 2b & 2b & b-a-2 \end{vmatrix}=

Solution:

We have,

=2\begin{vmatrix} 2a-2b-4 & 4a & 4a \\ 4 & 2-b-a & 4 \\ 2b & 2b & b-a-2 \end{vmatrix}

=2\begin{vmatrix} a-b-2 & 2a & 2a \\ 4 & 2-b-a & 4 \\ 2b & 2b & b-a-2 \end{vmatrix}

=2\begin{vmatrix} -(a+b+2) & 0 & 2a \\ a+b+2 & -(a+b+2) & 4 \\ 0 & a+b+2 & b-a-2 \end{vmatrix}

C_{1} \rightarrow C_{1}-C_{2}

and

C_{2} \rightarrow C_{2}-C_{3}

= \,\; 2(a+b+2)^{2}\begin{vmatrix} -1 & 0 & 2a \\ 1 & -1 & 4 \\ 0 & 1 & b-a-2 \end{vmatrix}

= \,\; 2(a+b+2)^{2}(-1(-b+a+2-4)+2 a(1)

= \,\; 2(a+b+2)^{2}(a+b+2)=2(a+b+2)^{3}

= \,\; 2[(a+b)^{3}+6(a+b)^{2}+12(a+b)+8]

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