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MPSC AE CE Mains 2018 Official (Paper 1)

Option 2 : upper triangular matrix

ST 1: Building Construction and Materials

1095

20 Questions
40 Marks
24 Mins

__Explanation:__

**Gauss elimination method:**

- It is a method to solve the
**linear system**of the form Ax = b, by bringing an**augmented matrix into the upper triangle matrix.**

**For example:** Solve, y + Z = 2, 2X + 3Z = 5, X + Y + Z = 3

The** augmented matrix** of the given equations is

= \(\begin{bmatrix} 0 & 1 & 1 &2 \\ 2 & 0 & 3 &5 \\ 1 & 1 & 1& 3 \end{bmatrix}\)

= \(\begin{bmatrix} 2 & 0 & 3 &5 \\ 0 & 1 & 1 &2 \\ 1 & 1 & 1& 3 \end{bmatrix}\) Interchanging R1 and R2

= \(\begin{bmatrix} 1 & 0 & \frac{3}{2} &\frac{5}{2} \\ 0 & 1 & 1 &2 \\ 1 & 1 & 1& 3 \end{bmatrix}\) R1 -----> R1/2

= \(\begin{bmatrix} 1 & 0 & \frac{3}{2} &\frac{5}{2} \\ 0 & 1 & 1 &2 \\ 0 & 1 & \frac{-1}{2}& \frac{1}{2} \end{bmatrix}\) R_{3} -------> -R_{1 }+R_{3}

= \(\begin{bmatrix} 1 & 0 & \frac{3}{2} &\frac{5}{2} \\ 0 & 1 & 1 &2 \\ 0 & 0 & \frac{-3}{2}& \frac{-3}{2} \end{bmatrix}\) R_{3} -------> -R_{2} + R_{3}

= \(\begin{bmatrix} 1 & 0 & \frac{3}{2} &\frac{5}{2} \\ 0 & 1 & 1 &2 \\ 0 & 0 & 1& 1 \end{bmatrix}\) R_{3} --------> -2R_{2}/3

This is the **upper triangular matrix** after various operations.

So, By comparing we get

Z = 1, y +Z = 2, \(X + \frac{3}{2}Z = \frac{5}{2}\)

**So, The values are x = 1, y = 1, z = 1**