Q

# Solve this problem - Algebra - BITSAT-5

With the help of matrices , the solution of the equations  $3x+y+2z=3,\, \, 2x-3y-z=-3\, \, x+2y+z=4, is\, \, given\, \, by$

• Option 1)

x=1, y=2,z=-1

• Option 2)

x=-1,y=2,z=1

• Option 3)

x=6,y=-2,z=-1

• Option 4)

x=-1,y=-2,z=1

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As learnt in

Cramer's rule for solving system of linear equations -

When $\Delta =0$  and $\Delta _{1}=\Delta _{2}=\Delta _{3}=0$ ,

then  the system of equations has infinite solutions.

- wherein

$a_{1}x+b_{1}y+c_{1}z=d_{1}$

$a_{2}x+b_{2}y+c_{2}z=d_{2}$

$a_{3}x+b_{3}y+c_{3}z=d_{3}$

and

$\Delta =\begin{vmatrix} a_{1} &b_{1} &c_{1} \\ a_{2} & b_{2} &c_{2} \\ a_{3}&b _{3} & c_{3} \end{vmatrix}$

$\Delta _{1},\Delta _{2},\Delta _{3}$ are obtained by replacing column 1,2,3 of $\Delta$ by $\left ( d_{1},d_{2},d_{3} \right )$  column

We should solve such questions analytically by cross-checking options because of the time constraint.

$x=1, y=2,z=-1$ , satisfy all these equations

Option 1)

x=1, y=2,z=-1

This option is correct.

Option 2)

x=-1,y=2,z=1

This option is incorrect.

Option 3)

x=6,y=-2,z=-1

This option is incorrect.

Option 4)

x=-1,y=-2,z=1

This option is incorrect.

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