Understanding Determinant Value Transformations

This question asks us to identify which of the given determinants has the same value as the original determinant:

To solve this, we need to understand the fundamental properties of determinants related to row and column operations.

Key Properties of Determinants

Several properties govern how the value of a determinant changes when its rows or columns are manipulated:

• Row Swapping: If two rows (or two columns) of a determinant are interchanged, the sign of the determinant is reversed (multiplied by -1).
• Transpose: The determinant of a matrix is equal to the determinant of its transpose. That is, \(\det(A) = \det(A^T)\).
• Multiplying a Row/Column by a Scalar: If a single row or column is multiplied by a scalar \(k\), the determinant is multiplied by \(k\).
• Adding a Multiple of One Row/Column to Another: Adding a multiple of one row to another row (or a multiple of one column to another column) does not change the value of the determinant.

Analyzing the Correct Answer Determinant

The determinant provided as the correct answer is:

Let's see how we can obtain this determinant from the original determinant \(D\) using the properties mentioned above. We will perform a sequence of row operations.

Step 1: First Row Swap

Compare the original determinant \(D\) with \(D_{ans}\). The first row of \(D\) is \((a, b, c)\), and the third row is \((p, q, r)\). The first row of \(D_{ans}\) is \((p, q, r)\), and the second row is \((a, b, c)\).

First, let's swap the first row (\(R_1\)) and the third row (\(R_3\)) of the original determinant \(D\). According to the properties, this changes the sign of the determinant.

Original Determinant:

After swapping \(R_1\) and \(R_3\):

Let's call this intermediate determinant \(D_1\). We know that \(D_1 = -D\).

Step 2: Second Row Swap

Now, let's look at the intermediate determinant \(D_1\) and compare it to the target determinant \(D_{ans}\).

Intermediate Determinant (\(D_1\)):

Target Determinant (\(D_{ans}\)):

To get from \(D_1\) to \(D_{ans}\), we need to swap the second row (\(R_2\)) and the third row (\(R_3\)) of \(D_1\). Again, this row swap will multiply the determinant's value by -1.

After swapping \(R_2\) and \(R_3\) of \(D_1\):

This resulting determinant is exactly \(D_{ans}\).

Step 3: Final Value Calculation

We found that \(D_{ans} = -(D_1)\). Since we previously established that \(D_1 = -D\), we can substitute this back:

This shows that the determinant in the correct answer option has the same value as the original determinant.

Analyzing Other Options

Let's briefly analyze the values of the other options relative to the original determinant \(D\).

• Option 1: \(\begin{vmatrix} a & b & c \\ p & q & r \\ l & m & n \end{vmatrix}\) This determinant is obtained by swapping the second (\(R_2\)) and third (\(R_3\)) rows of the original determinant \(D\). Swapping two rows negates the determinant. Value = \(-D\).
• Option 2: \(\begin{vmatrix} l & m & n \\ a & b & c \\ p & q & r \end{vmatrix}\) This determinant is obtained by swapping the first (\(R_1\)) and second (\(R_2\)) rows of the original determinant \(D\). This also negates the determinant. Value = \(-D\).
• Option 4: \(\begin{vmatrix} a & p & l \\ b & q & m \\ c & r & n \end{vmatrix}\) Let the original matrix be \(A\). The matrix for Option 4, let's call it \(M_4\), has columns \((a, b, c)^T\), \((p, q, r)^T\), and \((l, m, n)^T\). These correspond to the first, third, and second rows of the original matrix \(A\), respectively, arranged as columns. This transformation is equivalent to taking the transpose of \(A\) (whose determinant is also \(D\)) and then swapping its second and third columns. Swapping two columns negates the determinant. Value = \(-\det(A^T) = -D\).

Conclusion

By applying the properties of determinants, specifically the effect of row swaps, we've shown that the determinant in the correct answer \(\begin{vmatrix} p & q & r \\ a & b & c \\ l & m & n \end{vmatrix}\) is equal to the original determinant \(D\). This is because it requires two row swaps to transform \(D\) into the target determinant, and \((-1) \times (-1) = 1\). Options 1, 2, and 4 result in determinants with the value \(-D\).