# ISI2014-DCG-64

147 views

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. $-3$

recategorized
2

@ankitgupta.1729

I am getting 1 as the answer.

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.

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 $(B).$

edited

## Related questions

1
160 views
The values of $\eta$ for which the following system of equations $\begin{array} {} x & + & y & + & z & = & 1 \\ x & + & 2y & + & 4z & = & \eta \\ x & + & 4y & + & 10z & = & \eta ^2 \end{array}$ has a solution are $\eta=1, -2$ $\eta=-1, -2$ $\eta=3, -3$ $\eta=1, 2$
1 vote
If $M$ is a $3 \times 3$ matrix such that $\begin{bmatrix} 0 & 1 & 2 \end{bmatrix}M=\begin{bmatrix}1 & 0 & 0 \end{bmatrix}$ and $\begin{bmatrix}3 & 4 & 5 \end{bmatrix} M = \begin{bmatrix}0 & 1 & 0 \end{bmatrix}$ then $\begin{bmatrix}6 & 7 & 8 \end{bmatrix}M$ is ... $\begin{bmatrix}0 & 0 & 1 \end{bmatrix}$ $\begin{bmatrix} -1 & 2 & 0 \end{bmatrix}$ $\begin{bmatrix} 9 & 10 & 8 \end{bmatrix}$
Suppose that $A$ is a $3 \times 3$ real matrix such that for each $u=(u_1, u_2, u_3)’ \in \mathbb{R}^3, \: u’Au=0$ where $u’$ stands for the transpose of $u$. Then which one of the following is true? $A’=-A$ $A’=A$ $AA’=I$ None of these
For the matrices $A = \begin{pmatrix} a & a \\ 0 & a \end{pmatrix}$ and $B = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$, $(B^{-1}AB)^3$ is equal to $\begin{pmatrix} a^3 & a^3 \\ 0 & a^3 \end{pmatrix}$ $\begin{pmatrix} a^3 & 3a^3 \\ 0 & a^3 \end{pmatrix}$ $\begin{pmatrix} a^3 & 0 \\ 3a^3 & a^3 \end{pmatrix}$ $\begin{pmatrix} a^3 & 0 \\ -3a^3 & a^3 \end{pmatrix}$