Question

The system of equations \[3x - 2y\; + z = 0\], \[\lambda x - 14y + 15z = 0\;\] and \[x + 2y - 3z = 0\] have non-zero solution, then $\lambda = $
A $1$
B $3$
C $5$
D $0$

Answer 1

Hint: Given, the system of equations \[3x - 2y\; + z = 0\], \[\lambda x - 14y + 15z = 0\;\] and \[x + 2y - 3z = 0\] have non-zero solution. We have to find the value of $\lambda $. First, we will find the determinant of the given equation then put that determinant equals zero to find the value of $\lambda $.

Formula Used: System have non zero solution then
$\left| {\begin{array}{*{20}{c}}
  {{a_1}}&{{b_1}}&{{c_1}} \\
  {{a_2}}&{{b_2}}&{{c_2}} \\
  {{a_3}}&{{b_3}}&{{c_3}}
\end{array}} \right| = 0$

Complete step by step solution: Given, \[3x - 2y\; + z = 0\]
\[\lambda x - 14y + 15z = 0\;\]
\[x + 2y - 3z = 0\]
We know that if ${a_1}x + {b_1}y + {c_1}z = 0$
${a_2}x + {b_2}y + {c_2}z = 0$
${a_3}x + {b_3}y + {c_3}z = 0$
Have non zero solution then
$\left| {\begin{array}{*{20}{c}}
  {{a_1}}&{{b_1}}&{{c_1}} \\
  {{a_2}}&{{b_2}}&{{c_2}} \\
  {{a_3}}&{{b_3}}&{{c_3}}
\end{array}} \right| = 0$
A unique number known only for square matrices is the determinant of a matrix (matrices that have the same number of rows and columns). In calculus and other algebraic matrices, a determinant is frequently used to describe the matrix in terms of a real integer, which is then used to solve systems of linear equations and determine the inverse of matrices.
Now we will find the determinant and put that equal to zero for a non-zero solution.
$\left| {\begin{array}{*{20}{c}}
  3&{ - 2}&1 \\
  \lambda &{ - 14}&{15} \\
  1&2&{ - 3}
\end{array}} \right| = 0$
Expanding along ${R_1}$
$3\left( {\left( { - 14} \right) \times \left( { - 3} \right) - 2 \times 15} \right) - \left( { - 2} \right)\left( {\lambda \times \left( { - 3} \right) - 1 \times 15} \right) + 1\left( {2\lambda - \left( { - 14} \right)} \right) = 0$
After simplifying the above equation
$3\left( {42 - 30} \right) + 2\left( { - 3\lambda - 15} \right) + \left( {2\lambda + 14} \right) = 0$
After solving the above expression
$36 - 6\lambda - 30 + 2\lambda + 14 = 0$
Writing simplified equation for above
$ - 4\lambda + 20 = 0$
Shifting 20 to other side
$4\lambda = 20$
Dividing both sides with 5
$\lambda = 5$
Hence, option C is correct.

Note: Students should do calculations carefully to avoid any mistakes. They should use concepts correctly to get the required solution. They should know how to find the determinant of a square matrix. To get the correct value of $\lambda $ they should do calculations step by step.