Let \alpha \beta \gamma=45 ; \alpha, \beta, \gamma \in \mathbbR . If x(\alpha, 1,2)+y(1, \beta, 2)+z(2,3, \gamma)=(0,0,0) for some x, y, z \in \mathbbR, x y z eq 0 , then 6 \alpha+4 \beta+\gamma is equal to \underline\hspace2cm .

Matrices and Determinants PYQ — JEE Main 2024 · 6 Apr · Shift 1 · Q54 with Solution

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Correct answer: 55

$x(\alpha,1,2)+y(1,\beta,2)+z(2,3,\gamma)=(0,0,0).$

This represents a homogeneous system in $x,y,z$ :

\alpha x + y + 2z = 0,\\ x + \beta y + 3z = 0,\\ 2x + 2y + \gamma z = 0. \end{cases}$$ Since $x,y,z \neq 0$ and there exists a non-trivial solution, the coefficient matrix must be singular. 2. So the determinant of $$A=\begin{pmatrix} \alpha & 1 & 2\\ 1 & \beta & 3\\ 2 & 2 & \gamma \end{pmatrix}$$ must be zero: $$\det A=0.$$ Now compute the determinant: $$\det A= \alpha\begin{vmatrix}\beta & 3\\ 2 & \gamma\end{vmatrix} -1\begin{vmatrix}1 & 3\\ 2 & \gamma\end{vmatrix} +2\begin{vmatrix}1 & \beta\\ 2 & 2\end{vmatrix}.$$ So, $$\det A=\alpha(\beta\gamma-6)-(\gamma-6)+2(2-2\beta).$$ Hence, $$\det A=\alpha\beta\gamma-6\alpha-\gamma+6+4-4\beta.$$ Thus, $$\det A=\alpha\beta\gamma-6\alpha-4\beta-\gamma+10.$$ Since $\det A=0$, $$\alpha\beta\gamma-6\alpha-4\beta-\gamma+10=0.$$ 3. Given $$\alpha\beta\gamma=45,$$ substitute this into the determinant condition: $$45-6\alpha-4\beta-\gamma+10=0.$$ So, $$55-(6\alpha+4\beta+\gamma)=0.$$ Therefore, $$6\alpha+4\beta+\gamma=55.$$ 4. Hence the required integer is $$\boxed{55}.$$

View written solutionFree

Correct answer: 55

$x(\alpha,1,2)+y(1,\beta,2)+z(2,3,\gamma)=(0,0,0).$

This represents a homogeneous system in $x,y,z$ :

\alpha x + y + 2z = 0,\\ x + \beta y + 3z = 0,\\ 2x + 2y + \gamma z = 0. \end{cases}$$ Since $x,y,z \neq 0$ and there exists a non-trivial solution, the coefficient matrix must be singular. 2. So the determinant of $$A=\begin{pmatrix} \alpha & 1 & 2\\ 1 & \beta & 3\\ 2 & 2 & \gamma \end{pmatrix}$$ must be zero: $$\det A=0.$$ Now compute the determinant: $$\det A= \alpha\begin{vmatrix}\beta & 3\\ 2 & \gamma\end{vmatrix} -1\begin{vmatrix}1 & 3\\ 2 & \gamma\end{vmatrix} +2\begin{vmatrix}1 & \beta\\ 2 & 2\end{vmatrix}.$$ So, $$\det A=\alpha(\beta\gamma-6)-(\gamma-6)+2(2-2\beta).$$ Hence, $$\det A=\alpha\beta\gamma-6\alpha-\gamma+6+4-4\beta.$$ Thus, $$\det A=\alpha\beta\gamma-6\alpha-4\beta-\gamma+10.$$ Since $\det A=0$, $$\alpha\beta\gamma-6\alpha-4\beta-\gamma+10=0.$$ 3. Given $$\alpha\beta\gamma=45,$$ substitute this into the determinant condition: $$45-6\alpha-4\beta-\gamma+10=0.$$ So, $$55-(6\alpha+4\beta+\gamma)=0.$$ Therefore, $$6\alpha+4\beta+\gamma=55.$$ 4. Hence the required integer is $$\boxed{55}.$$