Find the solution set of the system x +...

MathematicsSolution of System of Linear Equations (Matrix Method and Cramer's Rule)JEE Advanced 1980Easy

Find the solution set of the system $x + 2 y + z = 1; 2 x - 3 y - w = 2; x \geq 0; y \geq 0; z \geq 0; w \geq 0$ .

Answer:

x = 1, y = 0, z = 0, w = 0

Visualized Solution (Hindi)

Understanding the Constrained System

Given system: $x + 2 y + z = 1$ and $2 x - 3 y - w = 2$
Constraints: $x \geq 0, y \geq 0, z \geq 0, w \geq 0$
Goal: Find the unique set of values for $(x, y, z, w)$

Analyzing Equation 1 for an Upper Bound

From equation (1): $x + 2 y + z = 1$
Since $y \geq 0$ and $z \geq 0$ , the term $(2 y + z)$ is non-negative.
Isolating $x$ : $x = 1 - (2 y + z)$
Therefore, $x \leq 1$

Analyzing Equation 2 for a Lower Bound

From equation (2): $2 x - 3 y - w = 2$
Rearrange to isolate $x$ terms: $2 x = 2 + 3 y + w$
Since $y \geq 0$ and $w \geq 0$ , then $3 y + w \geq 0$
This implies $2 x \geq 2$ , so $x \geq 1$

Finding the Unique Value of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x

Upper Bound: $x \leq 1$
Lower Bound: $x \geq 1$
The only value satisfying both is $x = 1$

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } z

Substitute $x = 1$ into $x + 2 y + z = 1$ :
$1 + 2 y + z = 1 \Rightarrow 2 y + z = 0$
Since $y \geq 0$ and $z \geq 0$ , the only solution is $y = 0$ and $z = 0$

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } w

Substitute $x = 1, y = 0$ into $2 x - 3 y - w = 2$ :
$2 (1) - 3 (0) - w = 2$
$2 - w = 2 \Rightarrow w = 0$

Final Solution and Takeaway

Final Solution Set: $(x, y, z, w) = (1, 0, 0, 0)$
Key Takeaway: Non-negativity constraints ( $x, y, z, w \geq 0$ ) can restrict an underdetermined system to a single unique point.
Next Challenge: What happens if the constants on the right side are changed? Does a solution always exist?

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