Consider the following for the next two (02) items that follow : Let Δ(a, b, c, α) = \(\left|\begin{array}{ccc} a & b & a \alpha+b \\ b & c & b \alpha+c \\ a \alpha+b & b \alpha+c & 0 \end{array}\right|\)
If Δ(a, b, c, α) = 0 for every α > 0, then which one of the following is correct ?
a, b, c are in GP
Analyzing the Determinant Problem
The problem provides a determinant \(\Delta(a, b, c, \alpha)\) and states that its value is 0 for every value of \(\alpha > 0\). We need to determine the relationship between the variables \(a\), \(b\), and \(c\).
The given determinant is:
\[ \Delta(a, b, c, \alpha) = \left|\begin{array}{ccc} a & b & a \alpha+b \\ b & c & b \alpha+c \\ a \alpha+b & b \alpha+c & 0 \end{array}\right| \]Expanding the Determinant
To find the relationship between \(a\), \(b\), and \(c\), we first need to calculate the value of the determinant. We can expand the determinant along the third row (\(R_3\)) as it contains a zero element, which simplifies the calculation.
The expansion along \(R_3\) is given by:
\[ \Delta = (a \alpha+b) \times C_{31} + (b \alpha+c) \times C_{32} + 0 \times C_{33} \]where \(C_{ij}\) is the cofactor of the element in the i-th row and j-th column. The cofactors are calculated as \(C_{ij} = (-1)^{i+j} M_{ij}\), where \(M_{ij}\) is the minor (determinant of the submatrix obtained by removing the i-th row and j-th column).
Calculating the required minors:
- \(M_{31} = \begin{vmatrix} b & a \alpha+b \\ c & b \alpha+c \end{vmatrix} = b(b \alpha+c) - c(a \alpha+b) = b^2 \alpha + bc - ac \alpha - bc = (b^2 - ac) \alpha\)
- \(M_{32} = \begin{vmatrix} a & a \alpha+b \\ b & b \alpha+c \end{vmatrix} = a(b \alpha+c) - b(a \alpha+b) = ab \alpha + ac - ab \alpha - b^2 = ac - b^2\)
Calculating the cofactors:
- \(C_{31} = (-1)^{3+1} M_{31} = +1 \times (b^2 - ac) \alpha = (b^2 - ac) \alpha\)
- \(C_{32} = (-1)^{3+2} M_{32} = -1 \times (ac - b^2) = -(ac - b^2) = b^2 - ac\)
Now, substitute these cofactors back into the expansion along \(R_3\):
\[ \Delta = (a \alpha+b) [(b^2 - ac) \alpha] + (b \alpha+c) [b^2 - ac] + 0 \] \[ \Delta = \alpha (a \alpha+b) (b^2 - ac) + (b \alpha+c) (b^2 - ac) \]We can factor out the term \((b^2 - ac)\) from both terms:
\[ \Delta = (b^2 - ac) [\alpha (a \alpha+b) + (b \alpha+c)] \]Let's simplify the expression inside the square brackets:
\[ \alpha (a \alpha+b) + (b \alpha+c) = a \alpha^2 + b \alpha + b \alpha + c = a \alpha^2 + 2b \alpha + c \]So, the determinant simplifies to:
\[ \Delta = (b^2 - ac) (a \alpha^2 + 2b \alpha + c) \]Condition for the Determinant Being Zero
The problem states that \(\Delta(a, b, c, \alpha) = 0\) for every \(\alpha > 0\).
Thus, we have the equation:
\[ (b^2 - ac) (a \alpha^2 + 2b \alpha + c) = 0 \text{ for all } \alpha > 0 \]This equation represents the product of two factors being zero for all positive values of \(\alpha\). For this to be true, at least one of the factors must cause the product to be zero regardless of \(\alpha\).
Consider the second factor, \(a \alpha^2 + 2b \alpha + c\). This is a quadratic expression in \(\alpha\). A quadratic expression can only be equal to zero for at most two distinct values of \(\alpha\) (its roots), unless the expression is identically zero.
If \(a \alpha^2 + 2b \alpha + c = 0\) for every \(\alpha > 0\), it implies that the quadratic expression must be identically zero for all positive \(\alpha\). For a polynomial to be identically zero, all its coefficients must be zero. Therefore, we would need \(a=0\), \(2b=0\) (which means \(b=0\)), and \(c=0\).
If \(a=b=c=0\), let's check the first factor: \(b^2 - ac = 0^2 - 0 \times 0 = 0\). In this specific case, the first factor is zero. So, if \(a=b=c=0\), the equation \((b^2 - ac) (a \alpha^2 + 2b \alpha + c) = 0\) becomes \(0 \times (0 \cdot \alpha^2 + 2 \cdot 0 \cdot \alpha + 0) = 0\), which is \(0 = 0\), and this holds for all \(\alpha > 0\).
Now consider the first factor, \((b^2 - ac)\). If this factor is zero, then the equation becomes:
\[ 0 \times (a \alpha^2 + 2b \alpha + c) = 0 \] \[ 0 = 0 \]This equation \(0=0\) is true for any values of \(a\), \(b\), \(c\), and any value of \(\alpha\) (including all \(\alpha > 0\)), provided that \(b^2 - ac = 0\).
Since the problem states that \(\Delta = 0\) for every \(\alpha > 0\), the condition \(b^2 - ac = 0\) is sufficient to guarantee this. If \(b^2 - ac \neq 0\), then \(a \alpha^2 + 2b \alpha + c\) would have to be zero for all \(\alpha > 0\), which only happens if \(a=b=c=0\), a case already covered by \(b^2 - ac = 0\).
Therefore, the condition that must hold for \(a\), \(b\), and \(c\) is \(b^2 - ac = 0\).
Relationship between a, b, c
The condition \(b^2 - ac = 0\) can be rewritten as \(b^2 = ac\). This is the defining property of a Geometric Progression (GP).
Three non-zero numbers \(a, b, c\) are in Geometric Progression if the ratio of consecutive terms is constant, i.e., \(b/a = c/b\), which leads to \(b^2 = ac\). If one or more terms are zero, the most general condition that covers all cases for \(a, b, c\) to be in GP is \(b^2 = ac\).
Thus, the condition \(\Delta(a, b, c, \alpha) = 0\) for every \(\alpha > 0\) implies that \(a, b, c\) are in GP.
Checking the Options
Let's evaluate the given options based on our finding that \(a, b, c\) are in GP (\(b^2 = ac\)).
- Option 1: a, b, c are in AP. This means \(2b = a+c\). This is not equivalent to \(b^2 = ac\).
- Option 2: a, b, c are in GP. This means \(b^2 = ac\). This matches our derived condition.
- Option 3: a, 2b, c are in AP. This means \(2(2b) = a+c\), or \(4b = a+c\). This is not equivalent to \(b^2 = ac\).
- Option 4: a, 2b, c are in GP. This means \((2b)^2 = ac\), or \(4b^2 = ac\). This is not equivalent to \(b^2 = ac\).
Our analysis shows that the only condition implied by \(\Delta(a, b, c, \alpha) = 0\) for all \(\alpha > 0\) is that \(a, b, c\) are in GP.
Conclusion
Expanding the determinant \(\Delta(a, b, c, \alpha)\) reveals that it simplifies to \((b^2 - ac) (a \alpha^2 + 2b \alpha + c)\). For this expression to be zero for every \(\alpha > 0\), the factor \((b^2 - ac)\) must be zero. This condition, \(b^2 = ac\), is the definition of \(a, b, c\) being in Geometric Progression.
Revision Table: Determinant Properties
| Property | Description |
|---|---|
| Determinant Expansion | The determinant of a matrix can be computed by expanding along any row or column, using the elements and their corresponding cofactors. |
| Determinant as a Polynomial | If the entries of a determinant involve a variable, the determinant's value can be expressed as a polynomial in that variable. |
| Polynomial Identity | A polynomial in a variable is identically zero over an infinite set of values if and only if all its coefficients are zero. |
| Zero Determinant Condition | A determinant is zero if its rows (or columns) are linearly dependent. In this problem, the condition \(b^2=ac\) leads to linear dependence between rows/columns for specific values. |
Additional Information: Sequences (AP and GP)
Understanding the definitions and properties of Arithmetic Progressions (AP) and Geometric Progressions (GP) is fundamental in sequence and series problems.
- Arithmetic Progression (AP): A sequence where the difference between consecutive terms is constant. If \(a, b, c\) are three consecutive terms in an AP, then \(b - a = c - b\), which simplifies to \(2b = a + c\). The middle term is the arithmetic mean of the other two.
- Geometric Progression (GP): A sequence where the ratio between consecutive terms is constant. If \(a, b, c\) are three consecutive terms in a GP (with \(a, c \neq 0\)), then \(b/a = c/b\), which simplifies to \(b^2 = ac\). The middle term (if non-zero) is the geometric mean of the other two. If \(a=0\), then for \(b^2=ac\) to hold, \(b\) must be \(0\). \(c\) can be any value. However, standard GP definition usually implies a common ratio. The condition \(b^2=ac\) is the most general form for three numbers \(a,b,c\) to be in GP, covering cases including zeros.
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