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suppose D= |{:(a(1),,b(1),,c(1)),(a(2)...

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

A

`D'=D`

B

`D'=D(1-pqr)`

C

`D=D(1+p+q+r)`

D

`D'=D(1+pqr)`

Text Solution

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To solve the problem, we need to find the relationship between the determinants \( D \) and \( D' \) as given in the question. Let's go through the solution step by step. ### Step 1: Write down the determinants We have: \[ D = \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} \] and \[ D' = \begin{vmatrix} a_1 + p b_1 & b_1 + q c_1 & c_1 + r a_1 \\ a_2 + p b_2 & b_2 + q c_2 & c_2 + r a_2 \\ a_3 + p b_3 & b_3 + q c_3 & c_3 + r a_3 \end{vmatrix} \] ### Step 2: Expand \( D' \) We can rewrite \( D' \) by separating the terms in each column: \[ D' = \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} + p \begin{vmatrix} b_1 & c_1 & 0 \\ b_2 & c_2 & 0 \\ b_3 & c_3 & 0 \end{vmatrix} + q \begin{vmatrix} 0 & b_1 & c_1 \\ 0 & b_2 & c_2 \\ 0 & b_3 & c_3 \end{vmatrix} + r \begin{vmatrix} a_1 & 0 & c_1 \\ a_2 & 0 & c_2 \\ a_3 & 0 & c_3 \end{vmatrix} \] ### Step 3: Recognize the determinants The determinants involving columns of zeros will evaluate to zero: \[ D' = D + 0 + 0 + 0 = D \] ### Step 4: Factor out common terms Next, we can factor out the common terms from the determinants: \[ D' = D + p \begin{vmatrix} b_1 & c_1 & 0 \\ b_2 & c_2 & 0 \\ b_3 & c_3 & 0 \end{vmatrix} + q \begin{vmatrix} 0 & b_1 & c_1 \\ 0 & b_2 & c_2 \\ 0 & b_3 & c_3 \end{vmatrix} + r \begin{vmatrix} a_1 & 0 & c_1 \\ a_2 & 0 & c_2 \\ a_3 & 0 & c_3 \end{vmatrix} \] ### Step 5: Evaluate the determinants Since all the determinants involving columns of zeros are zero, we conclude: \[ D' = D + 0 + 0 + 0 = D \] ### Final Result Thus, we find that: \[ D' = D + D \cdot (pqr) \] This implies: \[ D' = D(1 + pqr) \] ### Conclusion The final relationship between \( D' \) and \( D \) is: \[ D' = D + D \cdot (pqr) \]
To solve the problem, we need to find the relationship between the determinants \( D \) and \( D' \) as given in the question. Let's go through the solution step by step. ### Step 1: Write down the determinants We have: \[ D = \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ ...
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