જો સમીકરણની સંહતિ (lambda-1) x+(lambda- | vedclass

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Question

$(\lambda-1) x+(\lambda-4) y+\lambda z=5$
$\lambda x+(\lambda-1) y+(\lambda-4) z=7$
$(\lambda+1) x+(\lambda+2) y-(\lambda+2) z=9$
For infinitely ma

Vedclass · Accepted Answer

$12$

Vedclass · Answer

$(\lambda-1) x+(\lambda-4) y+\lambda z=5$
$\lambda x+(\lambda-1) y+(\lambda-4) z=7$
$(\lambda+1) x+(\lambda+2) y-(\lambda+2) z=9$
For infinitely many solutions
$D =\left|\begin{array}{ccc}\lambda-1 & \lambda-4 & \lambda \ \lambda & \lambda-1 & \lambda-4 \ \lambda+1 & \lambda+2 & -(\lambda+2)\end{array}ight|=0$
$(\lambda-3)(2 \lambda+1)=0$
$D _{ x }=\left|\begin{array}{ccc}5 & \lambda-4 & \lambda \ 7 & \lambda-1 & \lambda-4 \ 9 & \lambda+2 & -(\lambda+2)\end{array}ight|=0$
$2(3-\lambda)(23-2 \lambda)=0$
$\lambda=3$
$	herefore \lambda^2+\lambda=9+3=12$

જો સમીકરણની સંહતિ $(\lambda-1) x+(\lambda-4) y+\lambda z=5$, $\lambda x+(\lambda-1) y+(\lambda-4) z=7$, $(\lambda+1) x+(\lambda+2) y-(\lambda+2) z=9$ ને અનંત ઉકેલો હોય તો $\lambda^2+\lambda$ ની કિમંત મેળવો.

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જો $\left| {\,\begin{array}{*{20}{c}}5&3&{ - 1}\\{ - 7}&x&{ - 3}\\9&6&{ - 2}\end{array}\,} \right| = 0$, તો $ x$ મેળવો.

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$\left| {\,\begin{array}{*{20}{c}}1&1&1\\1&{1 - x}&1\\1&1&{1 + y}\end{array}\,} \right|$ = . . .

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