If \(u, v\) and \(w\) (all positive) are the \(p^{\text{th}}, q^{\text{th}}\) and \(r^{\text{th}}\) terms of a GP, then the determinant of the matrix is \(\begin{vmatrix} \ln u & p & 1 \\ \ln v & q & 1 \\ \ln w & r & 1 \end{vmatrix}.\)
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Understanding the Problem: Determinant of a Matrix with GP Terms
The question asks us to find the determinant of a specific 3x3 matrix. The elements of the matrix involve the natural logarithms (\(\ln\)) of the p-th, q-th, and r-th terms of a Geometric Progression (GP), along with their respective term numbers (p, q, r) and the number 1.
We are given that u, v, and w are the p-th, q-th, and r-th terms of a GP, and all are positive. This positivity is important because the natural logarithm is only defined for positive numbers.
Properties of Geometric Progression (GP)
A Geometric Progression is a sequence where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. Let the first term of the GP be denoted by a and the common ratio by R. The n-th term of a GP is given by the formula:
\(\text{n-th term} = a \cdot R^{n-1}\)
Using this formula, we can express u, v, and w in terms of a, R, p, q, and r:
- u (p-th term) = \(a R^{p-1}\)
- v (q-th term) = \(a R^{q-1}\)
- w (r-th term) = \(a R^{r-1}\)
Applying Natural Logarithms
Now, let's take the natural logarithm of u, v, and w. Using the properties of logarithms (\(\ln(xy) = \ln x + \ln y\) and \(\ln(x^k) = k \ln x\)):
- \(\ln u = \ln(a R^{p-1}) = \ln a + \ln(R^{p-1}) = \ln a + (p-1) \ln R\)
- \(\ln v = \ln(a R^{q-1}) = \ln a + \ln(R^{q-1}) = \ln a + (q-1) \ln R\)
- \(\ln w = \ln(a R^{r-1}) = \ln a + \ln(R^{r-1}) = \ln a + (r-1) \ln R\)
Setting up the Matrix with Logarithmic Terms
Substitute these expressions for \(\ln u\), \(\ln v\), and \(\ln w\) into the given matrix:
\(\left( {\begin{array}{*{20}{c}} \ln u&p&1\\ \ln v&q&1\\ \ln w&r&1 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} \ln a + (p-1) \ln R & p & 1\\ \ln a + (q-1) \ln R & q & 1\\ \ln a + (r-1) \ln R & r & 1 \end{array}} \right)\)
Calculating the Determinant
We need to find the determinant of this matrix. Let the columns of the matrix be \(C_1\), \(C_2\), and \(C_3\).
- \(C_1 = \begin{pmatrix} \ln a + (p-1) \ln R \\ \ln a + (q-1) \ln R \\ \ln a + (r-1) \ln R \end{pmatrix}\)
- \(C_2 = \begin{pmatrix} p \\ q \\ r \end{pmatrix}\)
- \(C_3 = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}\)
Let's analyze the first column \(C_1\). We can rewrite its elements:
- \(\ln a + (p-1) \ln R = \ln a - \ln R + p \ln R\)
- \(\ln a + (q-1) \ln R = \ln a - \ln R + q \ln R\)
- \(\ln a + (r-1) \ln R = \ln a - \ln R + r \ln R\)
So, we can see that \(C_1\) can be expressed as a linear combination of \(C_2\) and \(C_3\):
\(C_1 = (\ln a - \ln R) \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} + \ln R \begin{pmatrix} p \\ q \\ r \end{pmatrix}\)
\(C_1 = (\ln a - \ln R) C_3 + (\ln R) C_2\)
A fundamental property of determinants states that if one column (or row) of a matrix is a linear combination of the other columns (or rows), then the determinant of the matrix is 0.
Since \(C_1\) is a linear combination of \(C_2\) and \(C_3\), the determinant of the matrix is 0.
Alternative Calculation using Row Operations
We can also calculate the determinant by performing row operations. Subtract the second row from the first row (\(R_1 \rightarrow R_1 - R_2\)) and subtract the third row from the second row (\(R_2 \rightarrow R_2 - R_3\)).
The new matrix is:
\(\left( {\begin{array}{*{20}{c}} \ln u - \ln v & p - q & 1 - 1\\ \ln v - \ln w & q - r & 1 - 1\\ \ln w & r & 1 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} \ln u - \ln v & p - q & 0\\ \ln v - \ln w & q - r & 0\\ \ln w & r & 1 \end{array}} \right)\)
Now, calculate \(\ln u - \ln v\) and \(\ln v - \ln w\):
- \(\ln u - \ln v = (\ln a + (p-1)\ln R) - (\ln a + (q-1)\ln R) = (p-1-q+1)\ln R = (p-q)\ln R\)
- \(\ln v - \ln w = (\ln a + (q-1)\ln R) - (\ln a + (r-1)\ln R) = (q-1-r+1)\ln R = (q-r)\ln R\)
Substitute these back into the matrix:
\(\left( {\begin{array}{*{20}{c}} (p-q)\ln R & p - q & 0\\ (q-r)\ln R & q - r & 0\\ \ln w & r & 1 \end{array}} \right)\)
Calculate the determinant by expanding along the third column:
\(\text{Determinant} = 0 \cdot \det\left( {\begin{array}{*{20}{c}} (q-r)\ln R & q - r\\ \ln w & r \end{array}} \right) - 0 \cdot \det\left( {\begin{array}{*{20}{c}} (p-q)\ln R & p - q\\ \ln w & r \end{array}} \right) + 1 \cdot \det\left( {\begin{array}{*{20}{c}} (p-q)\ln R & p - q\\ (q-r)\ln R & q - r \end{array}} \right)\)
\(\text{Determinant} = 1 \cdot [((p-q)\ln R)(q-r) - (p-q)((q-r)\ln R)]\)
\(\text{Determinant} = (p-q)(q-r)\ln R - (p-q)(q-r)\ln R\)
\(\text{Determinant} = 0\)
Conclusion
Both methods show that the determinant of the given matrix is 0.
| Concept | Formula / Property | Application Here |
|---|---|---|
| GP n-th term | \(a_n = a R^{n-1}\) | \(u=aR^{p-1}, v=aR^{q-1}, w=aR^{r-1}\) |
| Logarithm Property | \(\ln(xy) = \ln x + \ln y\) | Used to expand \(\ln(aR^{n-1})\) |
| Logarithm Property | \(\ln(x^k) = k \ln x\) | Used to simplify \(\ln(R^{n-1})\) |
| Determinant Property | Linear dependent columns/rows | Determinant is 0 if one column/row is a linear combination of others. |
| Determinant Calculation | Expansion by cofactors | Used in the alternative method to calculate the 2x2 determinant. |
Revision Table: Key Mathematical Concepts
| Topic | Description | Relevance to Problem |
|---|---|---|
| Geometric Progression (GP) | A sequence with a constant ratio between successive terms. | Defines u, v, w and their relationship based on p, q, r. |
| Natural Logarithm (\(\ln\)) | The inverse function of the exponential function \(e^x\). Converts multiplication/division into addition/subtraction, and powers into multiplication. | Transforms the multiplicative structure of GP terms into additive structure suitable for matrix rows. |
| Determinant of a Matrix | A scalar value that can be computed from the elements of a square matrix. Indicates properties of the linear transformation represented by the matrix (e.g., whether it is invertible). | The core value we need to compute for the given matrix. |
| Matrix Column/Row Operations | Operations like adding a multiple of one column/row to another that change the matrix but preserve its determinant (under certain operations). | Used to simplify the matrix before calculating the determinant or to demonstrate linear dependency. |
| Linear Dependency | When one vector in a set can be expressed as a linear combination of the others. If columns (or rows) of a matrix are linearly dependent, the determinant is zero. | Explains why the determinant is 0 in this specific case due to the structure derived from GP properties. |
Additional Information: Determinant Properties and Applications
The determinant of a matrix is a powerful concept in linear algebra with many applications.
- Invertibility: A square matrix is invertible if and only if its determinant is non-zero. If the determinant is 0, the matrix is singular and does not have an inverse.
- Area/Volume Scaling: For a 2x2 matrix, the absolute value of the determinant represents the scaling factor by which the matrix transforms the area of a unit square. For a 3x3 matrix, it represents the scaling factor for the volume of a unit cube.
- Systems of Linear Equations: Determinants can be used to solve systems of linear equations using Cramer's rule, although this method is often inefficient for large systems.
- Eigenvalues: Determinants are used in finding the eigenvalues of a matrix by solving the characteristic equation \(\det(A - \lambda I) = 0\).
- Linear Dependency Test: As seen in this problem, the determinant being zero indicates that the columns (or rows) of the matrix are linearly dependent. This means one column (or row) can be written as a combination of the others.
In this problem, the relationship between the terms of a GP when expressed logarithmically leads directly to a linear dependency among the columns of the resulting matrix, causing the determinant to be zero.
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