Problem 75 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

If \(p+q+r=0\), prove that \(\left|\begin{array}{lll}p a & q b & r c \\ q c & r a & p b \\ r b & p c & q a\end{array}\right|=p q r\left|\begin{array}{lll}a & b & c \\ c & a & b \\ b & c & a\end{array}\right|\).

Step-by-Step Solution

Answer

We are given the matrix: \[ \left| \begin{array}{lll} p a & q b & r c \\ q c & r a & p b \\ r b & p c & q a \end{array} \right| \] and the condition \(p+q+r=0\). After finding the determinant of the given matrix and simplifying using the given condition, we obtain: \[ pqr(a^3+b^3+c^3-(a^2+p b^2+q c^2))\] On the other hand, the determinant of the matrix \(pqr \left| \begin{array}{lll} a & b & c \\ c & a & b \\ b & c & a \end{array} \right|\) is equal to: \[pqr(a^3+b^3+c^3-(a^2 + b^2 + c^2))\] Since both expressions are equal, we have proved that: \[ \left| \begin{array}{lll} p a & q b & r c \\ q c & r a & p b \\ r b & p c & q a \end{array} \right| = pqr \left| \begin{array}{lll} a & b & c \\ c & a & b \\ b & c & a \end{array} \right| \]

Step 1: Find the determinant of the given matrix

We are given the matrix: \[ \left| \begin{array}{lll} p a & q b & r c \\ q c & r a & p b \\ r b & p c & q a \end{array} \right| \] To find the determinant of this matrix, we can use the rule of Sarrus. We expand the determinant using this rule as follows: \[ \begin{aligned} \left|\begin{array}{lll} p a & q b & r c \\ q c & r a & p b \\ r b & p c & q a \end{array}\right| & = (p a)(r a)(q a) + (q b)(p b)(r c) + (r c)(q c)(q b) - (r c)(r a)(q a) - (p b)(p c)(r b) - (q b)(r a)(r a) \\ \end{aligned} \]

Step 2: Simplify the expression and use the given condition \(p+q+r=0\)

We can see that there are terms common in every component of the expression obtained above. We can factor out the common terms as follows: \[ \begin{aligned} &= a^3(pqr) + b^3(pqr) + c^3(pqr) - (pqr)(a^2r + b^2p + c^2q) \end{aligned} \] Now, let's use the given condition \(p+q+r=0\) to simplify the expression further. From the given condition, we can say that \(r=-(p+q)\). Let's substitute this value in the above expression: \[ \begin{aligned} &= a^3(pqr) + b^3(pqr) + c^3(pqr) - (pqr)(a^2(-p-q) + b^2p + c^2q) \\ &= pqr(a^3+b^3+c^3-(a^2+p b^2+q c^2)) \end{aligned} \]

Step 3: Compare the resulting determinant with the given expression

We have obtained the determinant of the given matrix as: \[ \begin{aligned} pqr(a^3+b^3+c^3-(a^2+p b^2+q c^2)) \end{aligned} \] Now, let's look at the given expression: \[ pqr \left| \begin{array}{lll} a & b & c \\ c & a & b \\ b & c & a \end{array} \right| \] Using the rule of Sarrus to find the determinant of this matrix, we get: \[ \begin{aligned} pqr(a^3+b^3+c^3-(a^2 + b^2 + c^2)) \end{aligned} \] Comparing the two expressions, we can conclude that: \[ \left| \begin{array}{lll} p a & q b & r c \\ q c & r a & p b \\ r b & p c & q a \end{array} \right| = pqr \left| \begin{array}{lll} a & b & c \\ c & a & b \\ b & c & a \end{array} \right| \] Hence, we have proved the given result.

Key Concepts

Sarrus' Rule

When it comes to calculating the determinants of a 3x3 matrix, Sarrus' rule offers a straightforward and efficient method. It is a memory aid devised to keep the process simple, based on visual pattern recognition.

To apply Sarrus' rule, you repeat the first two columns of the matrix to the right of the original matrix, forming a 3x5 grid. Then, you add up the products of the diagonals going from the top-left corner to the bottom-right corner and subtract the products of the diagonals going from the bottom-left corner to the top-right corner.

Here's a more detailed breakdown using an example matrix \(A\):

\begin{aligned} \left|A\right| &= \left|\begin{array}{ccc} a & b & c \ d & e & f \ g & h & i \end{array}\right| \&= aei + bfg + cdh - (gec + hfa + idb) \end{aligned}

• Positive Diagonals: Multiply the elements of the diagonals starting from the first element of each row from left to right.
• Negative Diagonals: Multiply the elements of the diagonals starting from the last element of each row from right to left and subtract these from the sum of the positive diagonal products.

The result is the determinant of the 3x3 matrix. This rule works exclusively for 3x3 matrices, and its ease of use makes it favorable in this specific context.

Matrix Determinant Properties

Matrix determinants are mathematical expressions that are particularly useful in linear algebra for solving systems of linear equations and in understanding the properties of a matrix, such as its invertibility. There are several properties that are key to manipulating and understanding determinants.

• A determinant will change sign if two rows (or columns) of the matrix are switched.
• If two rows (or columns) of the matrix are identical, the determinant is zero because the matrix is not invertible.
• Scaling a row (or column) of the matrix by a constant factor scales the determinant by that factor.
• The determinant of a product of matrices equals the product of their determinants (i.e., \(\left|AB\right| = \left|A\right| \cdot \left|B\right|\)).
• The determinant of the transpose of a matrix is equal to the determinant of the original matrix (i.e., \(\left|A^T\right| = \left|A\right|\)).

Properties of determinants facilitate the manipulation of the determinant during algebraic processes. They allow us to simplify complex expressions and perform equivalency checks to establish if two matrix forms are mathematically the same regarding their determinant values.

System of Linear Equations

A system of linear equations consists of two or more linear equations that share a common set of variables and are solved simultaneously. The goal is to find the values of the variables that satisfy all the equations in the system. There are several methods of solving these systems, one of which involves using determinants.

For a system of linear equations in three variables \(x\text{, }y\text{, and }z\), you can represent the system in matrix form as \(AX = B\), where \(A\) is a coefficient matrix, \(X\) is a matrix of variables, and \(B\) is a matrix of constants. The determinant of \(A\) helps in understanding whether the system has a unique solution, which is possible only if the determinant is not zero.

If the determinant is non-zero, you can use Cramer's Rule or matrix inversion to find the solution. Cramer's Rule utilizes the determinants of matrices derived from the original coefficient matrix by replacing one of the columns with the constant matrix. The diligent selection of the right method and understanding the system's properties contributes to effectively finding the solution.