If \(a, b, c\) are the sides of triangle \(ABC\), then what is
\(\begin{vmatrix}a^2 & b\sin A & c\sin A \\b\sin A & 1 & \cos A \\c\sin A & \cos A & 1\end{vmatrix} equal to ?\)
Solving the Determinant for Triangle Sides and Angles
This problem asks us to find the value of a specific determinant involving the sides (\(a, b, c\)) and one angle (\(A\)) of a triangle \(ABC\).
Understanding the Triangle Determinant Problem
We are given a \(3 \times 3\) determinant: \(D = \begin{vmatrix} a^2 & b\sin A & c\sin A \\ b\sin A & 1 & \cos A \\ c\sin A & \cos A & 1 \end{vmatrix}\) Our goal is to simplify this expression using the properties of triangles and basic algebra.
Key Triangle Formulas
To solve this, we need to recall some fundamental formulas related to triangles:
- Sine Rule: For any triangle \(ABC\) with sides \(a, b, c\) opposite to angles \(A, B, C\) respectively, the rule states: \(\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R\) where \(R\) is the circumradius of the triangle.
- Cosine Rule: This rule relates the sides of a triangle to one of its angles: \(a^2 = b^2 + c^2 - 2bc\cos A\) This formula can be rearranged as: \(b^2 + c^2 - 2bc\cos A = a^2\) This rearranged form will be particularly useful.
Step-by-Step Determinant Calculation
Let's calculate the determinant \(D\) using the cofactor expansion method. We will expand along the first row:
\(D = a^2 \begin{vmatrix} 1 & \cos A \\ \cos A & 1 \end{vmatrix} - b\sin A \begin{vmatrix} b\sin A & \cos A \\ c\sin A & 1 \end{vmatrix} + c\sin A \begin{vmatrix} b\sin A & 1 \\ c\sin A & \cos A \end{vmatrix}\)
Now, let's compute the \(2 \times 2\) determinants:
- \(\begin{vmatrix} 1 & \cos A \\ \cos A & 1 \end{vmatrix} = (1 \cdot 1) - (\cos A \cdot \cos A) = 1 - \cos^2 A\)
- \(\begin{vmatrix} b\sin A & \cos A \\ c\sin A & 1 \end{vmatrix} = (b\sin A \cdot 1) - (\cos A \cdot c\sin A) = b\sin A - c\sin A \cos A\)
- \(\begin{vmatrix} b\sin A & 1 \\ c\sin A & \cos A \end{vmatrix} = (b\sin A \cdot \cos A) - (1 \cdot c\sin A) = b\sin A \cos A - c\sin A\)
Substitute these back into the expression for \(D\):
\(D = a^2(1 - \cos^2 A) - b\sin A (b\sin A - c\sin A \cos A) + c\sin A (b\sin A \cos A - c\sin A)\)
Now, distribute and simplify:
\(D = a^2\sin^2 A - (b^2\sin^2 A - bc\sin^2 A \cos A) + (bc\sin^2 A \cos A - c^2\sin^2 A)\)
\(D = a^2\sin^2 A - b^2\sin^2 A + bc\sin^2 A \cos A + bc\sin^2 A \cos A - c^2\sin^2 A\)
Combine like terms:
\(D = a^2\sin^2 A - b^2\sin^2 A - c^2\sin^2 A + 2bc\sin^2 A \cos A\)
Factor out \(\sin^2 A\):
\(D = \sin^2 A (a^2 - b^2 - c^2 + 2bc \cos A)\)
Now, let's use the rearranged Cosine Rule: \(b^2 + c^2 - 2bc \cos A = a^2\).
Substitute this into the expression:
\(D = \sin^2 A (a^2 - (b^2 + c^2 - 2bc \cos A))\)
\(D = \sin^2 A (a^2 - a^2)\)
\(D = \sin^2 A (0)\)
\(D = 0\)
Conclusion
By applying the cofactor expansion and utilizing the Cosine Rule for triangles, we find that the value of the given determinant simplifies to zero.
Similar Questions
If Δ(a, b, c, α) = 0 for every α > 0, then which one of the following is correct ?
What are the values of x that satisfy the equation \(\left| {\begin{array}{*{20}{c}} x&0&2\\ {2x}&2&1\\ 1&1&1 \end{array}} \right| + \left| {\begin{array}{*{20}{c}} {3x}&0&2\\ {{x^2}}&2&1\\ 0&1&1 \end{array}} \right| = 0\;?\)
If x + a + b + c = 0, then what is the value of \(\left| {\begin{array}{*{20}{c}} {x + a}&b&c\\ a&{x + b}&c\\ a&b&{x + c} \end{array}} \right|?\)
Which one of the following factors does the expansion of the determinant
\(\left| {\begin{array}{c} x&y&3\\ {{x^2}}&{5{y^3}}&9\\ {{x^3}}&{10{y^3}}&{27} \end{array}} \right|\) Contain?
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 \(\begin{vmatrix} \ln u & p & 1 \\ \ln v & q & 1 \\ \ln w & r & 1 \end{vmatrix}.\)
Let matrix B be the adjoint of a square matrix A, l be the identify matrix of same order as A. If k (≠ 0) is the determinate of the matrix A, then what is AB equal to?
What is the determinant of the matrix?
| x y y+z |
| z x z+x |
| y z x+y |
If B is a non-singular matrix and A is a square matrix, then the value of det (B -1 AB) is equal to
Which of the following determinants have value zero?
1. \(\left| {\begin{array}{*{20}{c}} {41}&1&5\\ {79}&7&9\\ {29}&5&3 \end{array}} \right|\)
2. \(\left| {\begin{array}{*{20}{c}} 1&a&{b + c}\\ 1&b&{c + a}\\ 1&c&{a + b} \end{array}} \right|\)
3. \(\left| {\begin{array}{*{20}{c}} 0&c&b\\ { - c}&0&a\\ { - b}&{ - a}&0 \end{array}} \right|\)
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
Let $A$ and $B$ be two invertible matrices of order $3 \times 3$. If $\det(A^2 B (A^T)^3) = 16$ and $\det(A^3 B^{-2}) = 32$, then $\det(B^2 A^{-1} (B^T)^3)$ is equal to:
The value of determinant \(\left| {\begin{array}{*{20}{c}} {a - b - c}&{2a}&{2a}\\ {2b}&{b - c - a}&{2b}\\ {2c}&{2c}&{c - a - b} \end{array}} \right|\) is:
If \(\left| {\begin{array}{*{20}{c}} 5&a\\ a&2 \end{array}} \right| = \left| {\begin{array}{*{20}{c}} 2&1\\ 3&2 \end{array}} \right|\), then the values of a are:
The value of the determinant \(\left| {\begin{array}{*{20}{c}} {\sqrt {13} + \sqrt 3 }&{2\sqrt 5 }&{\sqrt 5 }\\ {\sqrt {15} + \sqrt {26} }&5&{\sqrt {10} }\\ {3 + \sqrt {65} }&{\sqrt {15} }&5 \end{array}} \right|\) is
The determinant \(\left| {\begin{array}{*{20}{c}} {xp + y}&x&y\\ {yp + z}&y&z\\ 0&{xp + y}&{yp + z} \end{array}} \right| = 0,\) if
