Problem 68 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Let \(A=\left(\begin{array}{lll}1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1\end{array}\right)\). If \(u_{1}\) and \(u_{2}\) are column matrices such that \(A u_{1}=\left(\begin{array}{l}1 \\ 0 \\ 0\end{array}\right)\) and \(A u_{2}=\left(\begin{array}{l}0 \\ 1 \\ 0\end{array}\right)\), then \(u_{1}+u_{2}\) is equal to: \([2012]\) (a) \(\left(\begin{array}{c}-1 \\ 1 \\ 0\end{array}\right)\) (b) \(\left(\begin{array}{c}-1 \\ 1 \\ -1\end{array}\right)\) (c) \(\left(\begin{array}{c}-1 \\ -1 \\ 0\end{array}\right)\) (d) \(\left(\begin{array}{c}1 \\ -1 \\ -1\end{array}\right)\)

Step-by-Step Solution

Answer

The answer is option (d) \(\begin{pmatrix} 1 \\ -1 \\ -1 \end{pmatrix}\).

Step 1: Express System Using Matrix Equation

We have the matrix \(A\) and are given two equations: \(A\mathbf{u}_1 = \begin{pmatrix} 1 \ 0 \ 0 \end{pmatrix}\) and \(A\mathbf{u}_2 = \begin{pmatrix} 0 \ 1 \ 0 \end{pmatrix}\). We need to solve these equations for \(\mathbf{u}_1\) and \(\mathbf{u}_2\).

Step 2: Solve for \(\mathbf{u}_1\)

To find \(\mathbf{u}_1\), we consider the equation \(A\mathbf{u}_1 = \begin{pmatrix} 1 \ 0 \ 0 \end{pmatrix}\). The structure of matrix \(A\) is such that it is an upper triangular matrix, making it relatively straightforward to solve using back substitution.

Step 3: Compute the Back Substitution for \(\mathbf{u}_1\)

Since \(A\) is an upper triangular matrix, consider its form when performing back substitution. By inspection (or multiplication), we find that \(\mathbf{u}_1 = \begin{pmatrix} 1 \ -2 \ 1 \end{pmatrix}\) satisfies the equation, as shown by checking each row.

Step 4: Solve for \(\mathbf{u}_2\)

Next, solve \(A\mathbf{u}_2 = \begin{pmatrix} 0 \ 1 \ 0 \end{pmatrix}\). Again, use back substitution and recognize that \(A\) being upper triangular facilitates this solving.

Step 5: Compute the Back Substitution for \(\mathbf{u}_2\)

Using similar steps as before, we find \(\mathbf{u}_2 = \begin{pmatrix} 0 \ 1 \ -2 \end{pmatrix}\). Check by multiplying to ensure correctness with each row of \(A\).

Step 6: Add \(\mathbf{u}_1\) and \(\mathbf{u}_2\)

Compute \(\mathbf{u}_1 + \mathbf{u}_2 = \begin{pmatrix} 1 \ -2 \ 1 \end{pmatrix} + \begin{pmatrix} 0 \ 1 \ -2 \end{pmatrix} = \begin{pmatrix} 1+0 \ -2+1 \ 1-2 \end{pmatrix} = \begin{pmatrix} 1 \ -1 \ -1 \end{pmatrix}\).

Step 7: Match with Given Options

Compare the result \(\begin{pmatrix} 1 \ -1 \ -1 \end{pmatrix}\) to the provided multiple-choice options. The correct choice is (d) \(\begin{pmatrix} 1 \ -1 \ -1 \end{pmatrix}\).

Key Concepts

Upper Triangular Matrix

In the world of linear algebra, an upper triangular matrix is a special kind of square matrix. Why is it special? Well, all the elements below its main diagonal are zero. Simple enough, right? Now let's break this down a little further.
When you look at the main diagonal of a matrix, you will see a line of elements from the top left to the bottom right. In an upper triangular matrix, any element located below this line is, quite literally, a zero.
Here's a quick visual: if you have a 3x3 matrix, and when you glance below its diagonal you see zeros, congratulations—it's upper triangular! This makes solving matrices, especially in solving equations, much easier through methods like back substitution.
An example of an upper triangular matrix is:
- \[\begin{pmatrix}1 & 2 & 3 \0 & 1 & 4 \0 & 0 & 1\end{pmatrix}\]
This structure simplifies computational processes considerably because we can build straightforward methods like back substitution into our solutions.

System of Linear Equations

A system of linear equations consists of two or more equations made up of two or more variables. The entire goal is to find values for these variables that satisfy all the equations simultaneously. But how do we do this?
The solution process involves considering both the coefficients of the variables and the constants from each linear equation. It is like having a puzzle where each piece is integral to seeing the complete solution.
Solving these systems can be achieved in various methods like substitution, elimination, or with matrices. Matrices, in particular, provide a powerful way to handle complex systems efficiently.
With matrix multiplication and notation, you can express these systems compactly. Take for example a system represented as \( A\mathbf{x} = \mathbf{b} \), where \( A \) is a matrix of coefficients, \( \mathbf{x} \) is a column matrix of variables, and \( \mathbf{b} \) is a column matrix of solutions.
This matrix equation setup budgets for tasks like row equivalences or transformations which streamline the process of finding solutions to variable-heavy equations.

Back Substitution

Back substitution is a method specifically suited for solving equations that have been simplified into an upper triangular form. It's like working backward—solving from bottom to top.
Imagine you have already transformed your system of equations into an upper triangular matrix form. You begin solving from the last row, where you often find that only a single variable exists since others are zeroed out beneath the matrix's diagonal.
This helps in simplifying the calculations tremendously:
- First solve the bottom-most equation for its one variable.- Then substitute that solution back into the equations above it.- Continue substitution until you reach the top row.
This quick unraveling process is efficient and systematic, and is especially handy when employing matrices in tandem with methods like matrix reduction. For example, if you have an equation like \( 1x + 2y + 3z = 5 \) where you've zeroed out other rows below, you will solve for \( z \) in the last row.Back substitution is not only powerful but vital when solving systems with upper triangular matrices. It handles potentially complex systems simply and effectively, allowing for accurate solution computation.