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If $a$ and $b$ are any two real numbers, then
$\left|\begin{array}{ccc}
2 a-2 b-4 & 4 a & 4 a \\
4 & 2-b-a & 4 \\
2 b & 2 b & b-a-2
\end{array}\right|=$
Options:
$\left|\begin{array}{ccc}
2 a-2 b-4 & 4 a & 4 a \\
4 & 2-b-a & 4 \\
2 b & 2 b & b-a-2
\end{array}\right|=$
Solution:
1699 Upvotes
Verified Answer
The correct answer is:
$2\left[(a+b)^3+6(a+b)^2+12(a+b)+8\right]$
We have,
$\begin{aligned}
& \left|\begin{array}{ccc}
2 a-2 b-4 & 4 a & 4 a \\
4 & 2-b-a & 4 \\
2 b & 2 b & b-a-2
\end{array}\right| \\
& =2\left|\begin{array}{ccc}
a-b-2 & 2 a & 2 a \\
4 & 2-b-a & 4 \\
2 b & 2 b & b-a-2
\end{array}\right|
\end{aligned}$
$\begin{aligned} & =2\left|\begin{array}{ccc}-(a+b+2) & 0 & 2 a \\ a+b+2 & -(a+b+2) & 4 \\ 0 & a+b+2 & b-a-2\end{array}\right| \\ & =2(a+b+2)^2\left|\begin{array}{ccc}-1 & 0 & 2 a \\ 1 & -1 & 4 \\ 0 & 1 & b-a-2\end{array}\right| \\ & =2(a+b+2)^2(-1(-b+a+2-4)+2 a(1) \\ & =2(a+b+2)^2(a+b+2)=2(a+b+2)^3 \\ & =2\left[(a+b)^3+6(a+b)^2+12(a+b)+8\right]\end{aligned}$
$\begin{aligned}
& \left|\begin{array}{ccc}
2 a-2 b-4 & 4 a & 4 a \\
4 & 2-b-a & 4 \\
2 b & 2 b & b-a-2
\end{array}\right| \\
& =2\left|\begin{array}{ccc}
a-b-2 & 2 a & 2 a \\
4 & 2-b-a & 4 \\
2 b & 2 b & b-a-2
\end{array}\right|
\end{aligned}$
$\begin{aligned} & =2\left|\begin{array}{ccc}-(a+b+2) & 0 & 2 a \\ a+b+2 & -(a+b+2) & 4 \\ 0 & a+b+2 & b-a-2\end{array}\right| \\ & =2(a+b+2)^2\left|\begin{array}{ccc}-1 & 0 & 2 a \\ 1 & -1 & 4 \\ 0 & 1 & b-a-2\end{array}\right| \\ & =2(a+b+2)^2(-1(-b+a+2-4)+2 a(1) \\ & =2(a+b+2)^2(a+b+2)=2(a+b+2)^3 \\ & =2\left[(a+b)^3+6(a+b)^2+12(a+b)+8\right]\end{aligned}$
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