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If the system of linear equations
x+ay+z=3
x+2y+2z=6
x+5y+3z=b
has no solution, then :

Option 1)
$a=-1, b=9$

Option 2)
$a=-1,b\neq 9$

Option 3)
$a\neq-1,b= 9$

Option 4)
$a=1,b\neq9$

Answers (3)

best_answer

As we learned,

Cramer's rule for solving system of linear equations -

When $\Delta =0$ and atleast one of $\Delta_{1},\Delta _{2} and \Delta _{3}$ is non-zero , system of equations has no solution

- wherein

$a_{1}x+b_{1}y+c_{1}z=d_{1}$

$a_{2}x+b_{2}y+c_{2}z=d_{2}$

$a_{3}x+b_{3}y+c_{3}z=d_{3}$

and

$\Delta =\begin{vmatrix} a_{1} &b_{1} &c_{1} \\ a_{2} & b_{2} &c_{2} \\ a_{3}&b _{3} & c_{3} \end{vmatrix}$

x+ay+z=3
x+2y+2z=6
x+5y+3z=b

$Determinant\, \bigtriangleup \: = \begin{vmatrix} 1 &a &1 \\ 1&2 &2 \\ 1& 5 &3 \end{vmatrix}$

$\Rightarrow \: 1\left ( 6-10 \right )-a\left ( 3-2 \right )+1\left ( 5-2 \right )$

$\Rightarrow \: -4-a+3=0$

$\Rightarrow \: -a-1=0$

$\Rightarrow \: a=-1$

And atleast one of $\bigtriangleup _{1}\bigtriangleup _{2}\bigtriangleup _{3}$ is non zero

Thus $\begin{vmatrix} 1 &-1 &3 \\ 1&2 &6 \\ 1& 5 &b \end{vmatrix}\neq 0$

$1\left ( 2b-30 \right )+1\left ( b-6 \right )+3\left ( 3 \right )\neq 0$

$2b-30+b-6+9\neq 0$

$3b-27\neq 0\: \Rightarrow \: b\neq 9$

Option 1)

$a=-1, b=9$

Option 2)

$a=-1,b\neq 9$

Option 3)

$a\neq-1,b= 9$

Option 4)

$a=1,b\neq9$

Posted by

Himanshu

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