Problem 157 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

Find \(x, y, z\) if \(\left[\begin{array}{cc}x+y & y-z \\ z-2 x & y-x\end{array}\right]=\left[\begin{array}{cc}3 & -1 \\ 1 & 1\end{array}\right]\)

Step-by-Step Solution

Answer

The solutions for \(x, y, z\) are \(1, 2, 3\) respectively.

Step 1: Create the system of equations

First thing to do is to create equations by equating corresponding elements from both sides of the given equation. This results in four equations: 1) \(x + y = 3\), 2) \(y - z = -1\), 3) \(z - 2x = 1\), 4) \(y - x = 1\).

Step 2: Solve the system of equations

Next, the system of equations can be solved. It can be simplified by eliminating one variable at a time. Substituting equation 1) into equation 4), we get \(2x = 2\) which gives \(x = 1\). Substituting \(x = 1\) into equation 1), we find \(y = 2\). Finally, substituting \(y = 2\) into equation 2), gives \(z = 3\).

Step 3: Verify the solution

The final step is to validate our answers for the variables \(x, y, z\) by substituting them back into the original equation. It can be noticed that the equality holds true, thus validating our solution.

Key Concepts

Matrix Equations

Matrix equations are a compact and organized way of representing systems of linear equations. They bring a visual and structural clarity that can be especially helpful in solving multiple equations at once. In the given exercise, we're presented with two matrices: the matrix on the left is a 2x2 matrix containing expressions in terms of the variables \(x, y,\) and \(z\), and the matrix on the right holds constant numbers. By equating the corresponding elements of these matrices, we can derive a set of linear equations. This process simplifies dealing with many equations all at once:

• Matrix position \( (1,1) \) leads to the equation: \(x + y = 3\)
• Matrix position \( (1,2) \) becomes: \(y - z = -1\)
• Matrix position \( (2,1) \) results in: \(z - 2x = 1\)
• Matrix position \( (2,2) \) simplifies to: \(y - x = 1\)

Equating these elements allows us to tackle the problem step-by-step, by turning the matrix into a more familiar set of algebraic equations.

Substitution Method

The substitution method is a handy algebraic technique for solving systems of equations. This method involves solving one of the equations for one variable and then using that expression to replace the variable in another equation. This method systematically reduces the number of variables until each can be solved. In our exercise, we used substitution several times:
1. From the equation \(x + y = 3\), solve for \(y\): \(y = 3 - x\).
- Substituting \(x = 1\) into this expression, we found \(y = 2\).
2. Next, substitute \(y = 2\) into \(y - z = -1\):
- This returns us \(2 - z = -1\), leading to \(z = 3\).
By repeating these steps for other variables, substitution method efficiently isolates each variable, leading us straight to the solution.

Equation Solving

Equation solving is the foundational skill we're applying in this exercise when working out each individual variable from the system of equations. This involves identifying and performing operations that maintain equality to move variables and numbers around. Here are the steps we followed:

• From \(x + y = 3\), isolate \(y\) by subtracting \(x\) from both sides, leading to \(y = 3 - x\).
• Use \(y - x = 1\) to find \(x\). Substituting \(x = 1\) allows finding \(y=2\).
• Find \(z\) with \(y - z = -1\), substitute \(y = 2\) to obtain \(z = 3\).

Each equation gently nudges us closer to resolving the values. Simplifying step-by-step preserves accuracy and ensures all variables are accounted for.

Variable Elimination

Variable elimination is often used alongside substitution to simplify and solve systems of equations. It deals with strategically removing variables to simplify calculations and solve the system. Through elimination, we pair up equations with a common variable which can be removed by addition or subtraction:
1. Combine the two equations \( x + y = 3 \) and \( y - x = 1 \) by adding both equations, leading to \(2y = 4\), hence \(y = 2\).

2. Now substitute \(y = 2\) into \(y - z = -1\) to find \(z = 3\).
3. Finally, from \(z - 2x = 1\), substitute \(z = 3\) to solve for \(x\).
This method quickly pointed us to solutions, one small calculation at a time. It complements substitution by acting in brief, calculated bursts to reduce complexity, making it one of the essential strategies in controlling cumbersome algebraic problems.