Let $\alpha, \beta$ and $\gamma$ be real numbers such that the system of linear equations $\begin{aligned}…

Let $\alpha, \beta$ and $\gamma$ be real numbers such that the system of linear equations $\begin{aligned} x+2 y+3 z&=\alpha \\ 4 x+5 y+6 z&=\beta \\ 7 x+8 y+9 z&=\gamma-1 \end{aligned}$ is consistent. Let $|M|$ represent the determinant of the matrix $ M=\left[\begin{array}{ccc} \alpha & 2 & \gamma \\ \beta & 1 & 0 \\ -1 & 0 & 1 \end{array}\right] $ Let $P$ be the plane containing all those $(\alpha, \beta, \gamma)$ for which the above system of linear equations is consistent, and $D$ be the square of the distance of the point $(0,1,0)$ from the plane $P$. The value of $|M|$ is

Solution

$\begin{aligned} x + 2y + 3z = \alpha \end{aligned}$ $\begin{aligned} 4x + 5y + 6z = \beta \end{aligned}$ $\begin{aligned} 7x + 8y + 9z = \gamma - 1 \end{aligned}$ The system of linear equations is consistent it means it has unique solution or infinite solutions Here, $\Delta = \begin{vmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{vmatrix} = 0$ Hence, there are infinitely many solutions. If equations have infinitely many solutions, then the equations are linearly connected, i.e., $L_1 + \lambda L_2 = L_3$ $\Rightarrow (x + 2y + 3z - \alpha) + \lambda (4x + 5y + 6z - \beta) = 7x + 8y + 9z - \gamma + 1$ $\frac{1 + 4\lambda}{7} = \frac{2 + 5\lambda}{8} = \frac{3 + 6\lambda}{9} = \frac{\alpha + \lambda \beta}{\gamma - 1}$ $\frac{1 + 4\lambda}{7} = \frac{2 + 5\lambda}{8} \Rightarrow \lambda = -2$ Also, $\frac{1 + 4\lambda}{7} = \frac{\alpha + \lambda \beta}{\gamma - 1}$ $\Rightarrow -1 = \frac{\alpha - 2\beta}{\gamma - 1}$ $\Rightarrow \alpha - 2\beta + \gamma = 1$ Now, $|M| = \begin{vmatrix} \alpha & 2 & \gamma \\ \beta & 1 & 0 \\ -1 & 0 & 1 \end{vmatrix}$ $= \alpha + 2(-\beta) + \gamma(1) = \alpha - 2\beta + \gamma = 1$

Asked in: JEE Advanced 2021 (Paper 1)

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