If system of equations 2 x+3 y-z=5 ; x+α y+3 z=-4 ; 3 x-y+β z=7 ha

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3 and 4 .Determinants and Matrices

hard

If the system of equations $2 x+3 y-z=5$ ; $x+\alpha y+3 z=-4$ ; $3 x-y+\beta z=7$ has infinitely many solutions, then $13 \alpha \beta$ is equal to

$1110$

$1120$

$1210$

$1220$

(JEE MAIN-2024)

Solution

Using family of planes

$2 \mathrm{x}+3 \mathrm{y}-\mathrm{z}-5=\mathrm{k}_1(\mathrm{x}+\alpha \mathrm{y}+3 \mathrm{z}+4)+\mathrm{k}_2(3 \mathrm{x}-\mathrm{y}+\beta \mathrm{z}-7)$

$2=\mathrm{k}_1+3 \mathrm{k}_2, 3=\mathrm{k}_1 \alpha-\mathrm{k}_2,-1=3 \mathrm{k}_1+\beta \mathrm{k}_2,-5=4{k}_1-7 \mathrm{k}_2$

On solving we get

$\begin{gathered}k_2=\frac{13}{19}, k_1=\frac{-1}{19}, \alpha=-70, \beta=\frac{-16}{13} \\ 13 \alpha \beta=13(-70)\left(\frac{-16}{13}\right) \\ =1120\end{gathered}$

Standard 12

Mathematics

If ${2^{{a_1}}},{2^{{a_2}}},{2^{{a_3}}},{……2^{{a_n}}}$ are in $G.P.$ then $\left| {\begin{array}{*{20}{c}}
{{a_1}}&{{a_2}}&{{a_3}} \\
{{a_{n + 1}}}&{{a_{n + 2}}}&{{a_{n + 3}}} \\
{{a_{2n + 1}}}&{{a_{2n + 2}}}&{{a_{2n + 3}}}
\end{array}} \right|$ is equal to

normal

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medium

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easy

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

hard

(AIEEE-2005)

If $AB = A$ and $BA = B$, then