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The value of $\lambda$ such that the system of equation

$$\begin{array}{} 2x & – & y & + & 2z & = & 2 \\ x & – & 2y & + & z & = & -4 \\ x & + & y & + & \lambda z & = & 4 \end{array}$$ has no solution is

  1. $3$
  2. $1$
  3. $0$
  4. $-3$
in Linear Algebra
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2

@ankitgupta.1729

I am getting 1 as the answer. 

Can you please check?

1
I am getting the same.
2
when $\lambda=1$ the 3rd equation becomes : $x+y+z=4$

Now, subtract 2nd equation from 1st, $x+y+z=6$..So, no solution for $\lambda=1$
1
Exactly.

Thanks.

1 Answer

2 votes

Given that $$\begin{array}{} 2x & – & y & + & 2z & = & 2 \\ x & – & 2y & + & z & = & -4 \\ x & + & y & + & \lambda z & = & 4 \end{array}$$

Non-homogeneous eqautions is the form of $AX = B$ and augmented matrix $[A:B]$

Case$1:$ If $\text{rank}(A) \neq \text{rank}([A:B]),$ then no solution. 

Case$2:$ If $\text{rank}(A) = \text{rank}([A:B]),$ then

  1.  If  $\text{rank}(A) = \text{rank}([A:B]) = \text{number of variables(unknown)},$ then unique solution.
  2.  If  $\text{rank}(A) = \text{rank}([A:B]) < \text{number of variables(unknown)},$ then infinite solution.

Now, we can write the above equations into an augmented form

$[A: B]=\begin{bmatrix} 2&-1 &2 &:2 \\ 1&-2 &1 &:-4 \\ 1& 1& \lambda&:4 \end{bmatrix}$

Operation $R_{3}\rightarrow R_{3} + R_{2}$

$[A: B]=\begin{bmatrix} 2&-1 &2 &:2 \\ 1&-2 &1 &:-4 \\ 2& -1& \lambda + 1&:0 \end{bmatrix}$

Operation $R_{3}\rightarrow R_{3} - R_{1}$

 $[A: B]=\begin{bmatrix} 2&-1 &2 &:2 \\ 1&-2 &1 &:-4 \\ 0& 0& \lambda - 1  &:-2 \end{bmatrix}$

No solution means  $\text{rank}(A) \neq \text{rank}([A:B])$

It is only possible when $\lambda - 1 = 0 \implies \lambda = 1$

Here $,\text{rank}(A)  = 2$ and $\text{rank}([A:B]) = 3,$ this is satisfied the no solution condition.

$\therefore \lambda = 1$

So, the correct answer is $(B).$


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