System of equations kx + 2y,-z = 1 ; (k,-,1)y,-2z = 2 ; (k + 2)z = 3 has u

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3 and 4 .Determinants and Matrices

medium

The system of equations $kx + 2y\,-z = 1$ ; $(k\,-\,1)y\,-2z = 2$ ; $(k + 2)z = 3$ has unique solution, if $k$ is equal to

$-2$

$-1$

$0$

$1$

Solution

$D = \left| {\begin{array}{*{20}{c}}
k&2&{ – 1}\\
0&{k – 1}&{ – 2}\\
0&0&{k + 2}
\end{array}} \right| \ne 0\{ {\rm{for}}\,\,{\rm{unique}}\,\,{\rm{solution}}\Delta \ne {\rm{0\} }}$

$ \Rightarrow {\rm{k}}({\rm{k}} + 2)({\rm{k}} – 1) \ne 0 \Rightarrow {\rm{k}} \ne – 2,0,1$ so ${\rm{k}} = – 1$

Standard 12

Mathematics

If the system of equations $x + y+ z = 5$ ; $x + 2y + 3z = 9$ ; $x + 3y + \alpha z = \beta $ has infinitely many solutions, then $\beta – \alpha $ equals

hard

(JEE MAIN-2019)

If $\left| {\begin{array}{*{20}{c}}{x – 4}&{2x}&{2x}\\{2x}&{x – 4}&{2x}\\{2x}&{2x}&{x – 4}\end{array}} \right| = \left( {A + Bx} \right){\left( {x – A} \right)^2},$ then the ordered pair $\left( {A,B} \right) = $. . . . .

medium

(JEE MAIN-2018)

Consider the system of linear equation $x+y+z=$ $4 \mu, x+2 y+2 \lambda z=10 \mu, x+3 y+4 \lambda^2 z=\mu^2+15$, where $\lambda, \mu \in R$. Which one of the following statements is $NOT$ correct?

hard

(JEE MAIN-2024)

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)

Evaluate the determinants

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

easy

The system of equations $kx + 2y\,-z = 1$ ; $(k\,-\,1)y\,-2z = 2$ ; $(k + 2)z = 3$ has unique solution, if $k$ is equal to

Solution

Similar Questions

If the system of equations $x + y+ z = 5$ ; $x + 2y + 3z = 9$ ; $x + 3y + \alpha z = \beta $ has infinitely many solutions, then $\beta – \alpha $ equals

If $\left| {\begin{array}{*{20}{c}}{x – 4}&{2x}&{2x}\\{2x}&{x – 4}&{2x}\\{2x}&{2x}&{x – 4}\end{array}} \right| = \left( {A + Bx} \right){\left( {x – A} \right)^2},$ then the ordered pair $\left( {A,B} \right) = $. . . . .

Consider the system of linear equation $x+y+z=$ $4 \mu, x+2 y+2 \lambda z=10 \mu, x+3 y+4 \lambda^2 z=\mu^2+15$, where $\lambda, \mu \in R$. Which one of the following statements is $NOT$ correct?

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Evaluate the determinants

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