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The determinant |(b(1)+c(1),c(1)+a(1),...

The determinant
`|(b_(1)+c_(1),c_(1)+a_(1),a_(1)+b_(1)),(b_(2)+c_(2),c_(2)+a_(2),a_(2)+b_(2)),(b_(3)+c_(3),c_(3)+a_(3),a_(3)+b_(3))|`

A

`|(a_(1),b_(1),c_(1)),(a_(2),b_(2),c_(2)),(a_(3),b_(3),c_(3))|`

B

`2|(a_(1),b_(1),c_(1)),(a_(2),b_(2),c_(2)),(a_(3),b_(3),c_(3))|`

C

`3|(a_(1),b_(1),c_(1)),(a_(2),b_(2),c_(2)),(a_(3),b_(3),c_(3))|`

D

None of these

Text Solution

AI Generated Solution

The correct Answer is:
To solve the determinant \[ D = \begin{vmatrix} b_1 + c_1 & c_1 + a_1 & a_1 + b_1 \\ b_2 + c_2 & c_2 + a_2 & a_2 + b_2 \\ b_3 + c_3 & c_3 + a_3 & a_3 + b_3 \end{vmatrix} \] we will perform a series of row operations to simplify it without expanding the determinant directly. ### Step 1: Row Operations We will first perform some row operations to make the determinant easier to compute. Let's subtract the second column from the first column and the third column from the second column. \[ D = \begin{vmatrix} (b_1 + c_1) - (c_1 + a_1) & (c_1 + a_1) & (a_1 + b_1) \\ (b_2 + c_2) - (c_2 + a_2) & (c_2 + a_2) & (a_2 + b_2) \\ (b_3 + c_3) - (c_3 + a_3) & (c_3 + a_3) & (a_3 + b_3) \end{vmatrix} \] This simplifies to: \[ D = \begin{vmatrix} b_1 - a_1 & c_1 + a_1 & a_1 + b_1 \\ b_2 - a_2 & c_2 + a_2 & a_2 + b_2 \\ b_3 - a_3 & c_3 + a_3 & a_3 + b_3 \end{vmatrix} \] ### Step 2: Further Simplification Next, we will perform another operation by subtracting the first column from the second column and the first column from the third column: \[ D = \begin{vmatrix} b_1 - a_1 & (c_1 + a_1) - (b_1 - a_1) & (a_1 + b_1) - (b_1 - a_1) \\ b_2 - a_2 & (c_2 + a_2) - (b_2 - a_2) & (a_2 + b_2) - (b_2 - a_2) \\ b_3 - a_3 & (c_3 + a_3) - (b_3 - a_3) & (a_3 + b_3) - (b_3 - a_3) \end{vmatrix} \] This simplifies to: \[ D = \begin{vmatrix} b_1 - a_1 & c_1 + 2a_1 - b_1 & 2a_1 \\ b_2 - a_2 & c_2 + 2a_2 - b_2 & 2a_2 \\ b_3 - a_3 & c_3 + 2a_3 - b_3 & 2a_3 \end{vmatrix} \] ### Step 3: Factor Out Common Terms Now we can factor out 2 from the third column: \[ D = 2 \begin{vmatrix} b_1 - a_1 & c_1 + 2a_1 - b_1 & a_1 \\ b_2 - a_2 & c_2 + 2a_2 - b_2 & a_2 \\ b_3 - a_3 & c_3 + 2a_3 - b_3 & a_3 \end{vmatrix} \] ### Step 4: Final Form Now, we can see that we have a determinant that has been simplified significantly. The determinant can be expressed in terms of the variables \(b_i\), \(c_i\), and \(a_i\). ### Conclusion The final expression for the determinant can be represented as: \[ D = 2 \begin{vmatrix} b_1 - a_1 & c_1 + 2a_1 - b_1 & a_1 \\ b_2 - a_2 & c_2 + 2a_2 - b_2 & a_2 \\ b_3 - a_3 & c_3 + 2a_3 - b_3 & a_3 \end{vmatrix} \]
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suppose D= |{:(a_(1),,b_(1),,c_(1)),(a_(2),,b_(2),,c_(2)),(a_(3),,b_(3),,c_(3)):}| and Dprime= |{:(a_(1)+pb_(1),,b_(1)+qc_(1),,c_(1)+ra_(1)),(a_(2)+pb_(2),,b_(2)+qc_(2),,c_(2)+ra_(2)),(a_(3)+pb_(3),,b_(3)+qc_(3),,c_(3)+ra_(3)):}| . Then

Delta_(1)=det[[a_(1)+pb_(1),b_(1)+qc_(1),c_(1)+ra_(1)a_(2)+pb_(2),b_(1)+qc_(2),c_(2)+ra_(2)a_(3)+pb_(3),b_(3)+qc_(3),c_(3)+ra_(3)]]

The value of |(a_(1) x_(1) + b_(1) y_(1),a_(1) x_(2) + b_(1) y_(2),a_(1) x_(3) + b_(1) y_(3)),(a_(2) x_(1) +b_(2) y_(1),a_(2) x_(2) + b_(2) y_(2),a_(2) x_(3) + b_(2) y_(3)),(a_(3) x_(1) + b_(3) y_(1),a_(3) x_(2) + b_(3) y_(2),a_(3) x_(3) + b_(3) y_(3))| , is

Let = |(2a_(1)b_(1),a_(1)b_(2)+a_(2)b_(1),a_(1)b_(3)+a_(3)b_(1)),(a_(1)b_(2)+a_(2)b_(1),2a_(2)b_(2),a_(2)b_(3)+a_(3)b_(2)),(a_(1)b_(3)+a_(3)b_(1),a_(3)b_(2)+a_(2)b_(3),2a_(3)b_(3))| Express the determinant D as a product of two determinants. Hence or otherwise show that D = 0.