Understanding Determinant Value Transformations
This question asks us to identify which of the given determinants has the same value as the original determinant:
\(D = \begin{vmatrix} a & b & c \\ l & m & n \\ p & q & r \end{vmatrix}\)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:
\(D_{ans} = \begin{vmatrix} p & q & r \\ a & b & c \\ l & m & n \end{vmatrix}\)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:
\(D = \begin{vmatrix} a & b & c \\ l & m & n \\ p & q & r \end{vmatrix}\)After swapping \(R_1\) and \(R_3\):
\(\begin{vmatrix} p & q & r \\ l & m & n \\ a & b & c \end{vmatrix}\)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 (\(D_1\)):
\(D_1 = \begin{vmatrix} p & q & r \\ l & m & n \\ a & b & c \end{vmatrix}\)Target Determinant (\(D_{ans}\)):
\(D_{ans} = \begin{vmatrix} p & q & r \\ a & b & c \\ l & m & n \end{vmatrix}\)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 \(D_1\):
\(-(D_1) = \begin{vmatrix} p & q & r \\ a & b & c \\ l & m & n \end{vmatrix}\)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:
\(D_{ans} = -(-D)\) \(D_{ans} = D\)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 \(-D\).
Similar Questions
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for every $m > n$?If B is a non-singular matrix and A is a square matrix, then the value of det (B -1 AB) is equal to
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Select the correct answer using the code given below.
If A is an invertible matrix of order n and k is any positive real number, then the value of [det(kA)] -1 det A is
Important Questions from Determinants
- An $n \times n$ square matrix $A$ satisfies $A^T = A^{-1}$. The determinant of this matrix may take which of the following value(s)?
The determinant of the matrix $\begin{pmatrix} 1 & p & q + r \\ 1 & q & r + p \\ 1 & r & p + q \end{pmatrix}$ is ______.
Let $f(x) = \begin{vmatrix} x^3 & \sin x & \cos x \\ 6 & -1 & 0 \\ p & p^2 & p^3 \end{vmatrix}$
where $p$ is a constant.
The value of $\frac{d^3}{dx^3}f(x)$ at $x = 0$ is
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If A is a square matrix of order 3 and det A = 5, then what is det [(2A) -1 ] equal to ?
