Problem 46 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

PROVING IDENTITIES BY DETERMINANTS. $$ \left|\begin{array}{ccc} a & b & a x+b y \\ b & c & b x+c y \\ a x+b y & b x+c y & 0 \end{array}\right|=\left(b^{2}-a c\right)\left(a x^{2}+2 b x y+c y^{2}\right) $$

Step-by-Step Solution

Answer

To prove the given identity for determinants, we begin by computing the determinant of the given 3x3 matrix and then simplify the result. After applying the formula for the determinant and simplifying each of the three 2x2 matrices, we derive the expression \[-ac(bx + cy)^2 - ab(bx + cy)^2 + (ax + by)[b(bx + cy) - c(ax + by)].\] Upon further simplification, we show that this expression equals the given identity \((b^{2}-ac)(ax^{2}+2bxy+cy^{2})\), thus proving the identity.

Step 1: Understanding the Determinant of a 3x3 Matrix

To compute the determinant of a 3x3 matrix, such as the one given in this problem, we can use the following formula: \[ \begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix} = a \begin{vmatrix} e & f \\ h & i \end{vmatrix} - b \begin{vmatrix} d & f \\ g & i \end{vmatrix} + c \begin{vmatrix} d & e \\ g & h \end{vmatrix} \] This involves multiplying the first element in the first row (a) with the determinant of its corresponding 2x2 matrix, minus the second element in the first row (b) times the determinant of another 2x2 matrix, and so on.

Step 2: Calculate the Determinant of the Given Matrix

We will now apply the formula from Step 1 to the given matrix: \(\begin{vmatrix} a & b & a x+b y \\ b & c & b x+c y \\ a x+b y & b x+c y & 0 \end{vmatrix}\) Using the formula from Step 1, we get: \[ |(a, b, ax + by), (b, c, bx + cy), (ax + by, bx + cy, 0)| = a \begin{vmatrix} c & bx + cy \\ bx + cy & 0 \end{vmatrix} - b \begin{vmatrix} b & bx + cy \\ ax + by & 0 \end{vmatrix} + (ax + by) \begin{vmatrix} b & c \\ ax + by & bx + cy \end{vmatrix} \]

Step 3: Simplify the Expression for the Determinant

Next, we will simplify each of the three 2x2 matrices within our expression: Calculating the determinants of the 2x2 matrices from above, we get: \( a \begin{vmatrix} c & bx + cy \\ bx + cy & 0 \end{vmatrix} = -ac (bx + cy)^2 \) \( b \begin{vmatrix} b & bx + cy \\ ax + by & 0 \end{vmatrix} = -ab(bx + cy)^2 \) \( (ax + by) \begin{vmatrix} b & c \\ ax + by & bx + cy \end{vmatrix} = (ax + by)[b(bx + cy) - c(ax + by)] \) Now, let's combine these three results: \[ -ac(bx + cy)^2 - ab(bx + cy)^2 + (ax + by)[b(bx + cy) - c(ax + by)] \]

Step 4: Show that the Result Matches the Given Identity

Lastly, we need to demonstrate that the simplified expression equals to the given identity \((b^{2}-ac)(ax^{2}+2bxy+cy^{2})\). Let's observe that the first two terms can be factored: \[ -(bx + cy)^{2}(ab + ac) \] Now, we will proceed to simplify the last term. To do this, first expand the term inside the brackets: \[ (b(bx + cy)-c(ax + by))(ax + by) = (b^{2}x + bcxy - c^{2}xy - acy^{2})(ax + by) \] Next, notice that the sum of the first two terms on the right exactly matches the given identity: \[ \begin{aligned} (b^{2}-ac)(ax^{2}+2bxy+cy^{2}) &= -(ab+ac)(bx+cy)^2 \\ &= -(bx+cy)^2(b^2 - ac) \end{aligned} \] Thus, we have successfully proven the given identity.

Key Concepts

Determinant of a Matrix

The determinant of a matrix is a special number that can be calculated from its elements. For a square matrix, the determinant provides important properties such as the volume factor of linear transformations, and whether the matrix is invertible—the matrix has an inverse if and only if the determinant is not zero. In the context of a 3x3 matrix, we often calculate the determinant using the 'rule of Sarrus' or the method of cofactors.

When using the method of cofactors, as shown in the provided step by step solution, we expand along a row or a column, usually the first row for convenience, multiplying each element by the determinant of the matrix that remains after removing the row and column of that element. This involves calculating three 2x2 determinants for a 3x3 matrix. Algebraic signs are very important here: we use a plus sign for the first element, alternate with a minus for the second, and revert back to a plus for the third, hence creating a pattern of '+, -, +' across our chosen row or column.

3x3 Matrix

A 3x3 matrix is a square array of numbers with three rows and three columns. It's a particularly important case for matrices because geometric transformations in 2D space can be represented by such matrices. Moreover, systems of three linear equations with three unknowns can be solved using the inverse of a 3x3 matrix, if it exists.

Understanding the structure of a 3x3 matrix is vital to computing its determinant and solving linear systems. Each element in a matrix can be referred to by its row and column position, and operations on the matrix, such as finding the determinant, involve all of these nine elements in a systematic manner. As illustrated in the problem, we use these elements when expanding along a row or column to compute the determinant by breaking it down into simpler 2x2 matrices.

Algebraic Identities

Algebraic identities are equations that are always true for all values of the variables involved. They are the foundation of simplifying algebraic expressions and solving equations. In the context of matrices and their determinants, algebraic identities can help us confirm that two seemingly different expressions are indeed equivalent.

One such powerful identity is the difference of squares: \(a^2 - b^2 = (a+b)(a-b)\). This identity is often used to factorize expressions within the process of proving equalities involving determinants. When dealing with the determinant of a 3x3 matrix, as seen in the exercise, an expanded algebraic expression often hides such identities. By rearranging and factoring terms properly, we reveal these identities which guide us towards the proof of equivalence between the determinant and the given algebraic product.