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
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
