Let =45; , , R.

Question

Let $\alpha \beta \gamma =45;\alpha ,\beta ,\gamma \in R$. If $x(\alpha ,1,2)+y(1,\beta ,2)+z(2,3,\gamma )=(0,0,0)$ for some $x,y,z\in R,xyz
e 0$, then $6\alpha +4\beta +\gamma $ is equal to \_\_\_\_\_\_\_\_\_.

Answer: 55

Accepted Answer

Given that $\alpha \beta \gamma =45$ and $\alpha ,\beta ,\gamma \in R$, consider the equation $x(\alpha ,1,2)+y(1,\beta ,2)+z(2,3,\gamma )=(0,0,0)$ for some $x,y,z\in R$ where $xyz
e 0$. To find the value of $6\alpha +4\beta +\gamma $, follow these steps:

Express the given equation in matrix form:

$\begin{gathered} \alpha x+y+2z=0 \ x+\beta y+3z=0 \ 2x+2y+\gamma z=0 \end{gathered}$

Since $x,y,z
e 0$, the determinant of the coefficients matrix must be zero:

$\begin{vmatrix} \alpha  \quad  1 \quad  2 \ 1 \quad  \beta  \quad  3 \ 2 \quad  2 \quad  \gamma \end{vmatrix}=0$

Calculate the determinant of the matrix:

$\alpha \beta \gamma -6\alpha -4\beta -\gamma +10=0$

Given $\alpha \beta \gamma =45$, substitute this value into the equation:

$45-6\alpha -4\beta -\gamma +10=0$

Simplify the equation:

$45+10=6\alpha +4\beta +\gamma $

Thus,

$6\alpha +4\beta +\gamma =55$

So, the value of $6\alpha +4\beta +\gamma $ is 55.

Solutions · 1