If the system of linear equations : gathered x+y+z=6 x+2y+5z=10 2x+3y+ z= gathered has inf…

Question

If the system of linear equations :

$\begin{gathered} x+y+z=6 \ x+2y+5z=10 \ 2x+3y+\lambda z=\mu \end{gathered}$

has infinitely many solutions, then the value of $\lambda +\mu $ equals:

Accepted Answer

To find the condition for infinitely many solutions, we first calculate the determinant $D$ of the coefficients of $x,y,z$:

$D=\begin{vmatrix} 1 \quad  1 \quad  1 \ 1 \quad  2 \quad  5 \ 2 \quad  3 \quad  \lambda \end{vmatrix}=\lambda -6$

For infinitely many solutions, $D=0$.

So, $\lambda -6=0\Rightarrow \lambda =6.$

Now, let us find ${D}_{1}$, the determinant obtained by replacing the first column with the constants:

${D}_{1}=\begin{vmatrix} 6 \quad  1 \quad  1 \ 10 \quad  2 \quad  5 \ \mu  \quad  3 \quad  \lambda \end{vmatrix}=2\lambda +3\mu -60$

Next, the determinant ${D}_{2}$ is obtained by replacing the second column with the constants:

${D}_{2}=\begin{vmatrix} 1 \quad  6 \quad  1 \ 1 \quad  10 \quad  5 \ 2 \quad  \mu  \quad  \lambda \end{vmatrix}=4(\lambda -\mu +10)$

Finally, ${D}_{3}$ is obtained by replacing the third column with the constants:

${D}_{3}=\begin{vmatrix} 1 \quad  1 \quad  6 \ 1 \quad  2 \quad  10 \ 2 \quad  3 \quad  \mu \end{vmatrix}=\mu -16$

For the system to have infinitely many solutions, all the determinants must be zero:

$D={D}_{1}={D}_{2}={D}_{3}=0$

Substitute $\lambda =6$:

From ${D}_{3}=0$,

$\mu -16=0\Rightarrow \mu =16.$

Thus,

$\lambda +\mu =6+16=22$

Solutions · 1