Practice Set

Question

The system of equation x + y + z = 1, 2x + 3y + 4z = 1 and 4x + 5y + 6z = 2 has

(a) unique solution (b) infinite solution
(c) no solution (d) none of these

Comments

srestha Equation will not be same.after Multiply eq I by 2 and add it to equation ii  the equation will be

4x+5y+6z=3

and equation III is 

4x+5y+6z=2

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if by some simple mathematics , we can directly reach answer , it's better like in this question.

x+y+z=1           ---eq1

2x+3y+4z=1    ---eq2

4x+5y+6z=2     ---eq3

add eq1 and eq2 and subtract from eq3 , got x+y+z=0 , while it's given x+y+z=1 , both things can't happen simultaneously ====>>> No solution , directly we can say

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yes thanks

Answer 1

$A=\begin{bmatrix} 1 &1 &1 \\ 2 & 3 & 4\\ 4 & 5 &6 \end{bmatrix}$

Augmented matrix

$\left [ A \vdots B \right ]=\begin{bmatrix} 1 &1 &1 \ \ \vdots &1\\ 2 & 3 & 4 \ \ \vdots &1\\ 4 & 5 &6 \ \ \vdots &2 \end{bmatrix}$

$\left [ A \vdots B \right ]=\begin{bmatrix} 1 &1 &1 \ \ \vdots &1\\ 2 & 3 & 4 \ \ \vdots &1\\ 2 & 2 &2\ \ \vdots &1 \end{bmatrix} \ \ \ \ \ \ \ \ R_{3} \rightarrow R_{3}-R_{2}$

$\left [ A \vdots B \right ]=\begin{bmatrix} 1 &1 &1 \ \ \vdots &1\\ 2 & 3 & 4 \ \ \vdots &1\\ 0 & 0 &0\ \ \vdots &-1 \end{bmatrix} \ \ \ \ \ \ \ \ R_{3} \rightarrow R_{3}- 2 \times R_{1}$

Here rank of matrix $A$ and rank of augmented matrix $ \left [ A \vdots B \right ] $ is not equal.

So, The system of equation has No Solution.

Hence, Option(C)No Solution.