Calculating the Determinant with Algebraic Variables
The problem asks us to find the value of the determinant of a given matrix, where the elements of the matrix are defined using variables \(x\), \(y\), and \(z\). These variables themselves are expressed in terms of \(a\), \(b\), and \(c\).
Given Information:
\(x = \frac{a}{b-c}\)
\(y = \frac{b}{c-a}\)
\(z = \frac{c}{a-b}\)
The matrix is: \( \begin{vmatrix} 1 & -x & x\\ 1 & 1 & -y\\ 1 & z & 1 \end{vmatrix} \)
We can rearrange and factor this expression. Consider it as a polynomial in \(a\):
\( = a^2(b - c) - a(b^2 - c^2) + bc(b - c) \)
\( = a^2(b - c) - a(b - c)(b + c) + bc(b - c) \)
Factor out the common term \((b - c)\):
\( = (b - c) [a^2 - a(b + c) + bc] \)
\( = (b - c) [a^2 - ab - ac + bc] \)
\( = (b - c) [a(a - b) - c(a - b)] \)
\( = (b - c) [(a - b)(a - c)] \)
\( = (b - c) (a - b) (a - c) \)
We can rewrite \((a - c)\) as \( -(c - a) \). So the numerator is:
\( = (b - c)(a - b)(-(c - a)) \)
\( = -(a - b)(b - c)(c - a) \)
Now, substitute this back into the expression for \(xy + yz + xz\):
\(xy + yz + xz = \frac{-(a-b)(b-c)(c-a)}{(a-b)(b-c)(c-a)}\)
Assuming that \(a\), \(b\), and \(c\) are distinct such that the denominator is not zero, the numerator and denominator cancel out, leaving:
\(xy + yz + xz = -1\)
Finally, substitute this value back into the determinant expression \(1 + xy + yz + xz\):
\(\det(M) = 1 + (-1) = 0\)
Conclusion
The value of the determinant is 0.
Term
Expression
\(x\)
\(\frac{a}{b-c}\)
\(y\)
\(\frac{b}{c-a}\)
\(z\)
\(\frac{c}{a-b}\)
Determinant Formula
\(1 + xy + yz + xz\)
\(xy + yz + xz\)
\(-1\)
Final Determinant Value
\(0\)
Revision Table: Determinant Calculation
Concept
Key Points
Application in Problem
Determinant of a 3x3 Matrix
Sum of products of elements and their cofactors. e.g., \(a_{11}C_{11} + a_{12}C_{12} + a_{13}C_{13}\).
Used cofactor expansion along the first row to get \(1 + xy + yz + xz\).
Algebraic Substitution
Replacing variables with their given expressions.
Substituting \(x\), \(y\), \(z\) in terms of \(a, b, c\).
Adding Algebraic Fractions
Find a common denominator and combine numerators.
Used \((a-b)(b-c)(c-a)\) as common denominator for \(xy+yz+xz\).
Algebraic Factorization
Simplifying expressions by finding common factors.
Factored the numerator \(a^2b - ab^2 + \dots\) as \( -(a-b)(b-c)(c-a) \).
Additional Information: Cyclic Expressions
The numerator expression \(a^2b - ab^2 + b^2c - bc^2 + ac^2 - a^2c\) and the denominator expression \((a-b)(b-c)(c-a)\) are related to cyclic or symmetric polynomials. A cyclic expression remains the same when the variables are permuted cyclically (a to b, b to c, c to a).
The expression \(a^2b - ab^2 + b^2c - bc^2 + ac^2 - a^2c\) can be written as \( \sum_{cyc} a^2b - \sum_{cyc} ab^2 \). It factors into \( -(a-b)(b-c)(c-a) \).
The expression \((a-b)(b-c)(c-a)\) is also a cyclic product. Its expansion is \(a^2b - a^2c - ab^2 + abc + abc - ac^2 - b^2c + bc^2 = a^2b - a^2c - ab^2 + 2abc - ac^2 - b^2c + bc^2\), which is different from the numerator's expansion.
However, the factorization \(a^2b - ab^2 + b^2c - bc^2 + ac^2 - a^2c = (a-b)(b-c)(a-c)\) is correct. The sign difference \((a-c) = -(c-a)\) leads to the numerator being \( -(a-b)(b-c)(c-a) \).
These types of expressions often appear in problems involving determinants or symmetric properties of variables.
Was this answer helpful?
Similar Questions
The value of the determinant \(\left| {\begin{array}{*{20}{c}} {1 - {\rm{\alpha }}}&{{\rm{\alpha }} - {{\rm{\alpha }}^2}}&{{{\rm{\alpha }}^2}}\\ {1 - {\rm{\beta }}}&{{\rm{\beta }} - {{\rm{\beta }}^2}}&{{{\rm{\beta }}^2}}\\ {1 - {\rm{\gamma }}}&{{\rm{\gamma }} - {{\rm{\gamma }}^2}}&{{{\rm{\gamma }}^2}} \end{array}} \right|\) is equal to
Let p, q and r be three distinct positive real numbers. If \(\rm D = \left| {\begin{array}{*{20}{c}} \rm p&\rm q&\rm r\\ \rm q&\rm r&\rm p\\ \rm r&\rm p&\rm q \end{array}} \right|,\) then which one of the following is correct?
What is the value of the determinant \(\left| {\begin{array}{*{20}{c}} {\rm{i}}&{{{\rm{i}}^2}}&{{{\rm{i}}^3}}\\ {{{\rm{i}}^4}}&{{{\rm{i}}^6}}&{{{\rm{i}}^8}}\\ {{{\rm{i}}^9}}&{{{\rm{i}}^{12}}}&{{{\rm{i}}^{15}}} \end{array}} \right|\) where \(\rm i = \sqrt {-1}\) ?
If \(\left| {\begin{array}{*{20}{c}} {\rm{x}}&{\rm{y}}&0\\ 0&{\rm{x}}&{\rm{y}}\\ {\rm{y}}&0&{\rm{x}} \end{array}} \right| = 0\) , then which one of the following is correct?
If x, y, z are distinct real numbers and \(\left| {\begin{array}{*{20}{c}} x&{{x^2}}&{2 + {x^3}}\\ y&{{y^2}}&{2 + {y^3}}\\ z&{{z^2}}&{2 + {z^3}} \end{array}} \right| = 0\), then xyz =
If A + B + C = \(\pi \), then, the value of \(\left| {\begin{array}{*{20}{c}} {\sin \left( {A + B + C} \right)}&{\sin B}&{\cos C}\\ { - \sin B}&0&{\tan A}\\ {\cos \left( {A + B} \right)}&{ - \tan A}&0 \end{array}} \right|\) is
Let p, q and r be three distinct positive real numbers. If \(\rm D = \left| {\begin{array}{*{20}{c}} \rm p&\rm q&\rm r\\ \rm q&\rm r&\rm p\\ \rm r&\rm p&\rm q \end{array}} \right|,\) then which one of the following is correct?
