System of equations x + y + z = 2 , 3x - y + 2z = 6 and 3x + y + z =

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3 and 4 .Determinants and Matrices

medium

The system of equations $x + y + z = 2$,$3x - y + 2z = 6$ and $3x + y + z = - 18$ has

A unique solution

No solutions

An infinite number of solutions

Zero solution as the only solution

Solution

(a) Given system of equation can be written as $\left[ {\,\begin{array}{*{20}{c}}1&1&1\\3&{ – 1}&2\\3&1&1\end{array}\,} \right]\,\left[ \begin{array}{l}x\\y\\z\end{array} \right] = \left[ \begin{array}{l}\,\,\,2\\\,\,\,6\\ – 18\end{array} \right]$

On solving the above system we get the unique solution

$x = -10, y = -4, z = 16.$

Standard 12

Mathematics

Solve system of linear equations, using matrix method. $x-y+z=4$ ; $2 x+y-3 z=0$ ; $x+y+z=2$

medium

Equations $x + y = 2,\,\,2x + 2y = 3$will have

easy

Solve system of linear equations, using matrix method. $x-y+2 z=7$ ; $3 x+4 y-5 z=-5$ ; $2 x-y+3 z=12$

hard

If ${a_1}x + {b_1}y + {c_1}z = 0,{a_2}x + {b_2}y + {c_2}z = 0$ ${a_3}x + {b_3}y + {c_3}z = 0$ and $\left| {\,\begin{array}{*{20}{c}}{{a_1}}&{{b_1}}&{{c_1}}\\{{a_2}}&{{b_2}}&{{c_2}}\\{{a_3}}&{{b_3}}&{{c_3}}\end{array}\,} \right| = 0,$then the given system has

normal

Let $S$ be the set of all integer solutions, $(x, y, z)$, of the system of equations

$x-2 y+5 z=0$

$-2 x+4 y+z=0$

$-7 x+14 y+9 z=0$

such that $15 \leq x^{2}+y^{2}+z^{2} \leq 150 .$ Then, the number of elements in the set $S$ is equal to

hard

(JEE MAIN-2020)