Problem 3 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Let \(A=\left[\begin{array}{ll}x & 1 \\ 1 & 0\end{array}\right], x \in \mathbf{R}\) and \(A^{4}=\left[a_{i j}\right]\). If \(a_{11}=109\), then \(a_{22}\) is equal to [NA Sep. 03,2020 (I)]

Step-by-Step Solution

Answer

The value of \(a_{22}\) is 10.

Step 1: Compute A^2

First, calculate \(A^2\). We have: \[A^2 = A \cdot A = \left(\begin{array}{cc} x & 1 \ 1 & 0 \end{array}\right) \cdot \left(\begin{array}{cc} x & 1 \ 1 & 0 \end{array}\right) = \left(\begin{array}{cc} x^2 + 1 & x \ x & 1 \end{array}\right)\]

Step 2: Compute A^4 Using A^2

Next, we find \(A^4\) by computing \((A^2)^2\).\[A^4 = A^2 \cdot A^2 = \left(\begin{array}{cc} x^2 + 1 & x \ x & 1 \end{array}\right) \cdot \left(\begin{array}{cc} x^2 + 1 & x \ x & 1 \end{array}\right)\]Carrying out the multiplication yields:\[A^4 = \left(\begin{array}{cc} (x^2 + 1)^2 + x^2 & x(x^2 + 1) + x \ x(x^2 + 1) + x & x^2 + 1 \end{array}\right)\]

Step 3: Set Up Equations from Given Conditions

From the problem, \(a_{11} = 109\) which corresponds to:\[(x^2 + 1)^2 + x^2 = 109\]Simplifying gives us:\[x^4 + 2x^2 + 1 + x^2 = 109\]\[x^4 + 3x^2 + 1 = 109\]\[x^4 + 3x^2 - 108 = 0\]

Step 4: Solve the Polynomial Equation

We solve the equation \(x^4 + 3x^2 - 108 = 0\) by setting \(y = x^2\), which changes the equation to:\[y^2 + 3y - 108 = 0\]To factor, find the roots using the quadratic formula:\[y = \frac{-3 \pm \sqrt{9 + 432}}{2} = \frac{-3 \pm \sqrt{441}}{2} = \frac{-3 \pm 21}{2}\]Thus, the roots are \(y = 9\) and \(y = -12\), but since \(y = x^2\), \(y\) must be \(9\) (since \(x^2 = 9\)). Thus, \(x = \pm 3\).

Step 5: Determine A_{22}

Returning to matrix \(A^4\), we need \(a_{22}\), which is:\[a_{22} = x^2 + 1\]For \(x^2 = 9\), we have:\[a_{22} = 9 + 1 = 10\]

Key Concepts

Matrix Multiplication

Matrix multiplication is a crucial concept in linear algebra, allowing for the combination of two matrices to produce a new matrix. This operation is not like regular number multiplication; instead, it involves a specific set of rules:

• The number of columns in the first matrix must match the number of rows in the second matrix.
• Each element in the resulting matrix is computed by taking the dot product of corresponding rows and columns from the two matrices being multiplied.

For example, if you have two matrices, like in this case with \( A = \begin{pmatrix} x & 1 \ 1 & 0 \end{pmatrix} \), multiplying it by itself involves calculating each element in the new matrix by combining rows from the first \( A \) and columns from the second \( A \).
When we calculated \( A^2 \), each element was derived using this approach, resulting in: \(A^2 = \begin{pmatrix} x^2 + 1 & x \ x & 1 \end{pmatrix} \).
Matrix multiplication preserves certain properties, like the associative property, and does not generally commute, meaning \( AB eq BA \) in most cases. This makes it distinctly different from normal arithmetic operations.

Quadratic Equations

Quadratic equations are polynomial equations of degree two, typically written in the form \( ax^2 + bx + c = 0 \). These equations play a pointed role in algebra because their solutions can be found using simple algebraic methods like factoring, completing the square, or using the quadratic formula:

• The quadratic formula is \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
• Solutions using this formula can result in real or complex numbers, depending on the discriminant \( b^2 - 4ac \).

In our exercise, we transformed a polynomial by setting \( y = x^2 \), which simplified our task to solving \( y^2 + 3y - 108 = 0 \), a cleaner and more direct path to the desired roots.
Here, solving using the quadratic formula produced a discriminant with real solutions, leading us to \( y = 9 \) and \( y = -12 \). Since \( y \) represents \( x^2 \), only \( y = 9 \) is valid because squares of real numbers cannot be negative.

Polynomial Equations

Polynomial equations involve expressions with variables raised to whole number exponents, encompassing a wide range of complexities from linear to quadratic and beyond. A polynomial of degree \( n \) has the general form:

• \( a_nx^n + a_{n-1}x^{n-1} + \, ... \, + a_1x + a_0 = 0 \)
• Where \( n \) is the highest power of the variable, determining the degree.

In more intricate scenarios, like the exercise we're exploring, we dealt with a fourth-degree equation \( x^4 + 3x^2 - 108 = 0 \). To simplify, we substituted \( y = x^2 \), reducing it to a quadratic form, streamlining the solution process.
Solving polynomial equations can often involve multiple steps, such as finding real roots, factoring, or making substitutions to simplify the process, as seen here. These equations provide fundamental insights and solutions fundamental to various fields, from physics to economics.