Problem 54 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If \(\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right] \cdot\left[\begin{array}{ll}1 & 2 \\ 0 & 1\end{array}\right] \cdot\left[\begin{array}{ll}1 & 3 \\ 0 & 1\end{array}\right] \ldots \ldots \ldots . .\left[\begin{array}{cc}1 & \mathrm{n}-1 \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}1 & 78 \\ 0 & 1\end{array}\right]\), then the inverse of \(\left[\begin{array}{ll}1 & n \\ 0 & 1\end{array}\right]\) is: \(\quad\) [April 09, \(\left.\mathbf{2 0 1 9}(\mathbf{I I})\right]\) (a) \(\left[\begin{array}{cc}1 & 0 \\ 12 & 1\end{array}\right]\) (b) \(\left[\begin{array}{cc}1 & -13 \\ 0 & 1\end{array}\right]\) (c) \(\left[\begin{array}{cc}1 & -12 \\ 0 & 1\end{array}\right]\) (d) \(\left[\begin{array}{cc}1 & 0 \\ 13 & 1\end{array}\right]\)

Step-by-Step Solution

Answer

The inverse is \( \left[ \begin{array}{cc}1 & -13 \ 0 & 1 \end{array} \right] \).

Step 1: Analyzing the Matrix Multiplication

The given problem involves multiplying several matrices of the form \( \left[\begin{array}{cc}1 & k \ 0 & 1\end{array}\right] \) for values of \( k \) ranging from 1 to \( n-1 \). Each of these matrices is an upper triangular matrix, and when multiplied, they combine the upper right element linearly.

Step 2: Simplifying Matrix Product

The result of multiplying these matrices is a single matrix with the form \( \left[\begin{array}{cc}1 & 1 + 2 + \ldots + (n-1) \ 0 & 1\end{array}\right] \). The sequence \( 1 + 2 + \ldots + (n-1) \) is an arithmetic series.

Step 3: Using Arithmetic Series Formula

The sum of the first \( n-1 \) natural numbers is given by the formula \( \frac{(n-1)n}{2} \). We set this equal to 78, as given in the problem: \( \frac{(n-1)n}{2} = 78 \).

Step 4: Solving for n

Solving \( \frac{(n-1)n}{2} = 78 \) gives \( (n-1)n = 156 \). This can be rearranged to \( n^2 - n - 156 = 0 \). Solve this quadratic equation using the quadratic formula: \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = -1, c = -156 \).

Step 5: Calculating the Discriminant and Solutions

The discriminant \( b^2 - 4ac = 1 + 624 = 625 \). The solutions to the quadratic equation are \( n = \frac{1 \pm 25}{2} \). So, \( n = 13 \) or a negative solution, which we discard because \( n \) must be positive.

Step 6: Finding the Inverse Matrix

The inverse of a matrix \( \left[ \begin{array}{cc}1 & n \ 0 & 1 \end{array} \right] \) is \( \left[ \begin{array}{cc}1 & -n \ 0 & 1 \end{array} \right] \). Substituting \( n = 13 \), we get the inverse \( \left[ \begin{array}{cc}1 & -13 \ 0 & 1 \end{array} \right] \).

Key Concepts

Inverse Matrices

Inverse matrices are a fundamental concept in linear algebra. An inverse of a matrix is like the reciprocal of a number. It is a matrix that, when multiplied with the original matrix, yields an identity matrix. The identity matrix acts like the number one in matrix multiplication, leaving a matrix unchanged when multiplied by. Consider a matrix \( A \). The inverse of \( A \), denoted as \( A^{-1} \), satisfies the equation:

• \( A \cdot A^{-1} = I \)
• where \( I \) is the identity matrix

For a matrix to have an inverse, it must be square (same number of rows and columns) and invertible, which means it has a non-zero determinant.In the given exercise, we work with the matrix \( \begin{bmatrix}1 & n \ 0 & 1\end{bmatrix} \). This is a special type of matrix called an upper triangular matrix with ones along the diagonal. Its inverse was found to be \( \begin{bmatrix}1 & -n \ 0 & 1\end{bmatrix} \), which directly follows from the property of these triangular matrices being easy to invert.

Quadratic Equation

Quadratic equations are polynomial equations of degree two that can be written in the standard form:

• \( ax^2 + bx + c = 0 \)

To solve quadratic equations, one can use the quadratic formula:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

This formula provides solutions for the variable \( x \), based on the coefficients \( a \), \( b \), and \( c \) from the equation. The expression under the square root, \( b^2 - 4ac \), is known as the discriminant. The discriminant determines the nature of the roots:

• If positive, there are two distinct real roots.
• If zero, there is exactly one real root.
• If negative, the equation has no real roots, only complex ones.

In the problem, we used the quadratic equation \( n^2 - n - 156 = 0 \). Solving it involves calculating the discriminant, finding it to be 625 (a perfect square), and then using the quadratic formula to discover \( n = 13 \), which provided the necessary value for the inverse matrix solution.

Arithmetic Series

An arithmetic series is a sequence of numbers in which the difference between consecutive terms is constant. This difference is called the common difference. When we talk about the sum of terms in an arithmetic sequence, we refer to it as an arithmetic series.The formula to find the sum of the first \( n \) terms \( S_n \) of an arithmetic series is:

• \( S_n = \frac{n}{2} (a + l) \)

Here, \( a \) is the first term, \( l \) is the last term, and \( n \) is the number of terms. Alternatively, if the common difference \( d \) is known, we replace \( l \) with \( a + (n-1)d \).In our exercise, we needed to calculate the sum \( 1 + 2 + \ldots + (n-1) \), which forms an arithmetic series with a common difference of 1. The sum formula for the first \( n-1 \) natural numbers simplifies to \( \frac{(n-1)n}{2} \). Given that this sum equals 78, we used it to form and solve the quadratic equation, ultimately finding \( n = 13 \). Understanding arithmetic series is crucial for solving problems involving sequential additions or gaps, as often seen in algebra.