Understanding the Determinant Problem
We are asked to find the value of the determinant:
\(\begin{vmatrix} p+c & a & b \\ c & p+a & b \\ c & a & p+b \end{vmatrix}\)Here, \(a\), \(b\), and \(c\) represent the lengths of the sides of a triangle ABC, and \(p\) is its perimeter. The perimeter \(p\) is defined as the sum of the lengths of its sides: \(p = a+b+c\).
Step-by-Step Determinant Calculation
To solve this, we can use properties of determinants and substitute the definition of the perimeter.
1. Substitute Perimeter Definition
First, let's replace \(p\) with \(a+b+c\) in the matrix elements:
- Top-left element: \(p+c = (a+b+c)+c = a+b+2c\)
- Middle element, first column: \(c\)
- Bottom-left element: \(c\)
- Top-middle element: \(a\)
- Middle element, second column: \(p+a = (a+b+c)+a = 2a+b+c\)
- Bottom-middle element: \(a\)
- Top-right element: \(b\)
- Middle element, third column: \(b\)
- Bottom-right element: \(p+b = (a+b+c)+b = a+b+2c\)
The determinant now looks like this:
\(\begin{vmatrix} a+b+2c & a & b \\ c & 2a+b+c & b \\ c & a & a+b+2c \end{vmatrix}\)2. Apply Column Operations for Simplification
A useful strategy is to simplify the matrix by making elements zero or identical. Let's apply the column operation \(C_1 \rightarrow C_1 + C_2 + C_3\). This means we add the elements of the second and third columns to the first column.
- New \(C_1\) element (Row 1): \((a+b+2c) + a + b = 2a + 2b + 2c = 2(a+b+c) = 2p\)
- New \(C_1\) element (Row 2): \(c + (2a+b+c) + b = 2a + 2b + 2c = 2(a+b+c) = 2p\)
- New \(C_1\) element (Row 3): \(c + a + (a+b+2c) = 2a + 2b + 2c = 2(a+b+c) = 2p\)
The determinant becomes:
\(\begin{vmatrix} 2p & a & b \\ 2p & p+a & b \\ 2p & a & p+b \end{vmatrix}\)3. Factor out Common Term
Notice that the first column has a common factor of \(2p\). We can take this factor out of the determinant:
\(2p \begin{vmatrix} 1 & a & b \\ 1 & p+a & b \\ 1 & a & p+b \end{vmatrix}\)4. Apply Row Operations
Now, let's simplify further using row operations. We'll use \(R_2 \rightarrow R_2 - R_1\) and \(R_3 \rightarrow R_3 - R_1\).
- New Row 2: \((1, p+a, b) - (1, a, b) = (1-1, (p+a)-a, b-b) = (0, p, 0)\)
- New Row 3: \((1, a, p+b) - (1, a, b) = (1-1, a-a, (p+b)-b) = (0, 0, p)\)
The determinant simplifies to:
\(2p \begin{vmatrix} 1 & a & b \\ 0 & p & 0 \\ 0 & 0 & p \end{vmatrix}\)5. Calculate the Final Value
The resulting matrix is an upper triangular matrix. The determinant of an upper triangular matrix is simply the product of its diagonal entries.
Determinant = \(1 \times p \times p = p^2\).
Multiplying by the factor \(2p\) we took out earlier:
\(\text{Final Value} = 2p \times p^2 = 2p^3\)Conclusion
Therefore, the value of the given determinant is \(2p^3\).
Similar Questions
How many matrices of different orders are possible with elements comprising all prime numbers less than 30?
If (a - b) (b - c) (c - a) = 2 and abc = 6, then what is the value of \(\begin{vmatrix}a & b & c \\ a^2 & b^2 & c^2 \\ a^3 & b^3 & c^3 \end{vmatrix}?\)
Under which of the following conditions does the determinant \(\begin{vmatrix}a & b & c \\ b & c & a \\ c & a & b\end{vmatrix}\) vanish?
1. a + b + c = 0
2. a 3+ b 3+ c 3= 3abc
3. a 2+ b 2+ c 2- ab - bc - ca = 0
Select the correct answer using the code given below:
If \(ABC\) is a triangle, then what is the value of the determinant
\(\begin{vmatrix}\cos C & \sin B & 0 \\\tan A & 0 & \sin B \\0 & \tan(B+C) & \cos C\end{vmatrix}?\)
If \(a, b, c\) are the sides of triangle \(ABC\), then what is
\(\begin{vmatrix}a^2 & b\sin A & c\sin A \\b\sin A & 1 & \cos A \\c\sin A & \cos A & 1\end{vmatrix} equal to ?\)
If in a triangle ABC, sin3A + sin3B + sin3C = 3sinAsinBsinc, then what is the value of the determinant \(\begin{vmatrix}a & b & c \\b & c & a \\c & a & b\end{vmatrix}\) where a,b,c are sides of the triangle?
For what value of $n$ is the determinant
$[\begin{vmatrix}C\binom{9}{4} & C\binom{9}{3} & C\binom{10}{n-2} \\C\binom{11}{6} & C\binom{11}{5} & C\binom{12}{n} \\C\binom{m}{7} & C\binom{m}{6} & C\binom{m+1}{n+1} \end{vmatrix} = 0]$
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
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
- 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
How many matrices of different orders are possible with elements comprising all prime numbers less than 30?
If A is a square matrix of order 3 and det A = 5, then what is det [(2A) -1 ] equal to ?
