System of equations begin{array}{l}α x + y + z = α - 1x + α y + z = α - 1x + y + α z = α

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3 and 4 .Determinants and Matrices

hard

The system of equations $\begin{array}{l}\alpha x + y + z = \alpha - 1\\x + \alpha y + z = \alpha - 1\\x + y + \alpha z = \alpha - 1\end{array}$ has no solution, if $\alpha $ is

Not $ -2$

$1$

$-2$

Either $ -2$ or $ 1$

(AIEEE-2005)

Solution

(c) For no solution or infinitely many solutions $\left| {\,\begin{array}{*{20}{c}}\alpha &1&1\\1&\alpha &1\\1&1&\alpha \end{array}\,} \right| = 0 \Rightarrow \alpha = 1,\alpha = – 2$.

But for $\alpha = 1$, clearly there are infinitely many solutions and when we put $\alpha = – 2$ in given system of equations and adding them together $L.H.S$ $ \ne $ $R.H.S$. i.e., No solution.

Standard 12

Mathematics

Evaluate the determinants

$\left|\begin{array}{rrr}3 & -1 & -2 \\ 0 & 0 & -1 \\ 3 & -5 & 0\end{array}\right|$

easy

If the system of equations $(\lambda-1) x+(\lambda-4) y+\lambda z=5$, $\lambda x+(\lambda-1) y+(\lambda-4) z=7$, $(\lambda+1) x+(\lambda+2) y-(\lambda+2) z=9$ has infinitely many solutions, then $\lambda^2+\lambda$ is equal to

medium

(JEE MAIN-2025)

Set of equations $a + b – 2c = 0,$ $2a – 3b + c = 0$ and $a – 5b + 4c = \alpha $ is consistent for $\alpha$ equal to

medium

Let $a ,b ,c $ be such that $b + c \ne 0$ if

$\left| {\begin{array}{{20}{c}}a&{a + 1}&{a – 1}\\{ – b}&{b + 1}&{b – 1}\\c&{c – 1}&{c + 1}\end{array}} \right| + \left| {\begin{array}{{20}{c}}{a + 1}&{b + 1}&{c – 1}\\{a – 1}&{b – 1}&{c + 1}\\{{{\left( { – 1} \right)}^{n + 2}} \cdot a}&{{{\left( { – 1} \right)}^{n + 1}} \cdot b}&{{{\left( { – 1} \right)}^n} \cdot c}\end{array}} \right| = 0$ then $n$ equals to

medium

(AIEEE-2009)

If the system of equations

$ x+(\sqrt{2} \sin \alpha) y+(\sqrt{2} \cos \alpha) z=0 $

$ x+(\cos \alpha) y+(\sin \alpha) z=0 $

$ x+(\sin \alpha) y-(\cos \alpha) z=0$

has a non-trivial solution, then $\alpha \in\left(0, \frac{\pi}{2}\right)$ is equal to :

hard

(JEE MAIN-2024)

The system of equations $\begin{array}{l}\alpha x + y + z = \alpha - 1\\x + \alpha y + z = \alpha - 1\\x + y + \alpha z = \alpha - 1\end{array}$ has no solution, if $\alpha $ is

Solution

Similar Questions

Evaluate the determinants $\left|\begin{array}{rrr}3 & -1 & -2 \\ 0 & 0 & -1 \\ 3 & -5 & 0\end{array}\right|$

If the system of equations $(\lambda-1) x+(\lambda-4) y+\lambda z=5$, $\lambda x+(\lambda-1) y+(\lambda-4) z=7$, $(\lambda+1) x+(\lambda+2) y-(\lambda+2) z=9$ has infinitely many solutions, then $\lambda^2+\lambda$ is equal to

Set of equations $a + b – 2c = 0,$ $2a – 3b + c = 0$ and $a – 5b + 4c = \alpha $ is consistent for $\alpha$ equal to

If the system of equations $ x+(\sqrt{2} \sin \alpha) y+(\sqrt{2} \cos \alpha) z=0 $ $ x+(\cos \alpha) y+(\sin \alpha) z=0 $ $ x+(\sin \alpha) y-(\cos \alpha) z=0$ has a non-trivial solution, then $\alpha \in\left(0, \frac{\pi}{2}\right)$ is equal to :

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Evaluate the determinants

$\left|\begin{array}{rrr}3 & -1 & -2 \\ 0 & 0 & -1 \\ 3 & -5 & 0\end{array}\right|$

If the system of equations

$ x+(\sqrt{2} \sin \alpha) y+(\sqrt{2} \cos \alpha) z=0 $

$ x+(\cos \alpha) y+(\sin \alpha) z=0 $

$ x+(\sin \alpha) y-(\cos \alpha) z=0$

has a non-trivial solution, then $\alpha \in\left(0, \frac{\pi}{2}\right)$ is equal to :