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The value of $\lambda$ for which the system of linear equations $2x-y-z=12$, $x-2y+z=-4$ and $x+y+\lambda z=4$ has no solution is

  1. $2$
  2. $-2$
  3. $3$
  4. $-3$
in Numerical Ability by Boss (42.5k points) | 89 views

1 Answer

+2 votes

Matrix A=$\begin{bmatrix}2&-1&-1\\1&-2&1\\1&1&\lambda \end{bmatrix}$

$|A|=0\implies 2(-2\lambda-1)+(\lambda-1)-3=0$

$\implies -3\lambda -6=0$

$\implies \lambda=-2$

But det(A)=0 is not a sufficient condition for no solution.

Infinitely many solutions $\implies$ |A|=0 and rank(A)=rank(A|B)

No solution $\implies$ |A|=0 and rank(A)$\neq$ rank(A|B)

For $\lambda=-2$,

Augmented matrix (A|B)

$=\begin{bmatrix} 2&-1&-1&12\\1&-2&1&-4\\1&1&-2&4\end{bmatrix}$

Apply $R1\leftarrow R1-(R2+R3)$

$\implies \begin{bmatrix} 0&0&0&12\\1&-2&1&-4\\1&1&-2&4\end{bmatrix}$

Apply $R1\leftrightarrow R3$

$\implies \begin{bmatrix}1&1&-2&4\\1&-2&1&-4\\0&0&0&12\end{bmatrix} $

So rank(A)=2 and rank(A|B)=3

Hence $\lambda=-2\implies \text{no solution}$


Alternative approach--

$2x-y-z=12$         $ \ldots (1)$

$x-2y+z=-4$          $\ldots (2)$

$x+y+\lambda z=4$           $\ldots (3)$

$equation(1)-(2)\implies  x+y+(-2)z=16$

So eqⁿ  3 will be inconsistent with this equation if $\lambda=-2$

by Boss (13.1k points)
edited by
0

@Verma Ashish

|A|=0 may give infinite solution also if rank(A)=rank(A| b). So, It does not ensure system Ax=b has no solution. right ? 

+1
Yes.

But if a non-homogenous system has to have no solution then rank(A) should be less than rank(A|B)..   ?
+1
yes or we can say for no solution in case of non-homogeneous system, rank(A) is not equal to rank(A|b)...I think your alternative approach is correct but in 1st one, I think you have to show rank(A) is not equal to rank(A|b) for $\lambda=-2$
0

Sir i think  no solution implies |A|=0

but  |A|=0 not necessarily implies no solution.

So according to question i applied det(A)=0 for no solution..

Isn't it a correct way ??

+1
@ankitgupta you are right..

(k+1)x+8y=4k

kx+(k+3)y=3k-1

Value of k for which equations become inconsistent??

If we do determinant=0 then we get k=1,3

k=3 ==> no solution

k=1 ==> infinitely many solution.

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