System of linear equation x + y + z = 2, 2x + 3y + 2z = 5 , 2x + 3y + (a^2 -1),z = a + 1 then

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3 and 4 .Determinants and Matrices

hard

The system of linear equation $x + y + z = 2, 2x + 3y + 2z = 5$, $2x + 3y + (a^2 -1)\,z = a + 1$ then

is inconsistent when $a = 4$

has a unique solution for $\left| a \right| = \sqrt 3 $

has infinitely many solutions for $a = 4$

inconsistent when $\left| a \right| = \sqrt 3 $

(JEE MAIN-2019)

Solution

By applying Crammer's Rule

$D = \left| {\begin{array}{*{20}{c}}
1&1&1\\
2&3&2\\
2&3&{{a^2} – 1}
\end{array}} \right|$

$ = 3\left( {{a^2} – 1} \right) – 6 – 2\left( {{a^2} – 1} \right) + 4$

$ = {a^2} – 1 – 2 = {a^2} – 3$

If $\left| a \right| \ne \pm \sqrt 3 \Rightarrow $system has unique solution

If $\left| a \right| = \sqrt 3 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left. \begin{array}{l}
x + y + z = 1\\
2x + 3y + 2z = 1\\
2x + 3y + 2z = \pm \sqrt 3 + 1
\end{array} \right\}$

Hence system is inconsistent for $\left| a \right| = \sqrt 3 $

Standard 12

Mathematics

If the sum of squares of all real values of $\alpha$, for which the lines $2 x-y+3=0,6 x+3 y+1=0$ and $\alpha x+2 y-2=0$ do not form a triangle is $p$, then the greatest integer less than or equal to $\mathrm{p}$ is $………$

medium

(JEE MAIN-2024)

An ordered pair $(\alpha , \beta )$ for which the system of linear equations

$\left( {1 + \alpha } \right)x + \beta y + z = 2$ ; $\alpha x + \left( {1 + \beta } \right)y + z = 3$ ; $\alpha x + \beta y + 2z = 2$ has a unique solution, is

hard

(JEE MAIN-2019)

The determinant $\left| {\begin{array}{*{20}{c}}{\cos \,\,(\theta \, + \,\phi )}&{ – \,\sin \,\,(\theta \, + \,\phi )}&{\cos \,2\phi }\\{\sin \,\theta }&{\cos \,\theta }&{\sin \,\phi }\\{ – \,\cos \,\theta }&{\sin \,\theta }&{\cos \,\phi }\end{array}} \right|$ is :

normal

If area of triangle is $35$ $\mathrm{sq}$ $\mathrm{units}$ with vertices $(2,-6),(5,4)$ and $(\mathrm{k}, 4) .$ Then $\mathrm{k}$ is

medium

The values of the determinant $\left| {\,\begin{array}{*{20}{c}}1&{\cos (\alpha – \beta )}&{\cos \alpha }\\{\cos (\alpha – \beta )}&1&{\cos \beta }\\{\cos \alpha }&{\cos \beta }&1\end{array}\,} \right|$ is

hard

The system of linear equation $x + y + z = 2, 2x + 3y + 2z = 5$, $2x + 3y + (a^2 -1)\,z = a + 1$ then

Solution

Similar Questions

If the sum of squares of all real values of $\alpha$, for which the lines $2 x-y+3=0,6 x+3 y+1=0$ and $\alpha x+2 y-2=0$ do not form a triangle is $p$, then the greatest integer less than or equal to $\mathrm{p}$ is $………$

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An ordered pair $(\alpha , \beta )$ for which the system of linear equations

$\left( {1 + \alpha } \right)x + \beta y + z = 2$ ; $\alpha x + \left( {1 + \beta } \right)y + z = 3$ ; $\alpha x + \beta y + 2z = 2$ has a unique solution, is