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If the system of equations
$\left[\begin{array}{ccc}\alpha & -1 & -1 \\ 1 & -\alpha & -1 \\ 1 & -1 & -\alpha\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{c}\alpha-1 \\ \alpha-1 \\ \alpha-1\end{array}\right]$ is inconsistent, then $\alpha=$
Options:
$\left[\begin{array}{ccc}\alpha & -1 & -1 \\ 1 & -\alpha & -1 \\ 1 & -1 & -\alpha\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{c}\alpha-1 \\ \alpha-1 \\ \alpha-1\end{array}\right]$ is inconsistent, then $\alpha=$
Solution:
2703 Upvotes
Verified Answer
The correct answer is:
$-2$
For $\alpha=1$, the system reduces to a
homogeneous system which is always consistent. So, $\alpha \neq 1$
For $\alpha \neq 1$, we have
$\begin{aligned} D= & \left|\begin{array}{ccc}\alpha & -1 & -1 \\ 1 & -\alpha & -1 \\ 1 & -1 & -\alpha\end{array}\right|=\left|\begin{array}{ccc}\alpha & 1 & 1 \\ 1 & \alpha & 1 \\ 1 & 1 & \alpha\end{array}\right| \\ = & \left|\begin{array}{ccc}\alpha+2 & \alpha+2 & \alpha+2 \\ 1 & \alpha & 1 \\ 1 & 1 & \alpha\end{array}\right| \\ \Rightarrow \quad D & (\alpha+2)\left|\begin{array}{ccc}1 & 1 & 1 \\ 1 & \alpha & 1 \\ 1 & 1 & \alpha\end{array}\right| \\ = & (\alpha+2)\left|\begin{array}{ccc}1 & 0 & 0 \\ 1 & \alpha-1 & 0 \\ 1 & 0 & \alpha-1\end{array}\right| \\ \Rightarrow \quad & \left(\text { Applying } C_2 \rightarrow C_2-C_1, C_3 \rightarrow C_3-C_1\right)\end{aligned}$
$\begin{aligned} & \Rightarrow \quad D=(\alpha+2)(\alpha-1)^2 \\ & \text { and } D_1=\left|\begin{array}{lll}\alpha-1 & 1 & 1 \\ \alpha-1 & \alpha & 1 \\ \alpha-1 & 1 & \alpha\end{array}\right|=(\alpha-1)\left|\begin{array}{ccc}1 & 1 & 1 \\ 1 & \alpha & 1 \\ 1 & 1 & \alpha\end{array}\right| \\ & =(\alpha-1)\left|\begin{array}{ccc}1 & 0 & 0 \\ 1 & \alpha-1 & 0 \\ 1 & 0 & \alpha-1\end{array}\right| \\ & \text { (Applying } C_2 \rightarrow C_2-C_1, C_3 \rightarrow C_3-C_1 \text { ) } \\ & \Rightarrow \quad D_1=(\alpha-1)^3 \\ & \text { Clearly } D=0 \text { for } \alpha=-2 \text {, but } D_1 \neq 0 \\ & \text { So, the system is inconsistent for } \alpha=-2 \\ & \end{aligned}$
homogeneous system which is always consistent. So, $\alpha \neq 1$
For $\alpha \neq 1$, we have
$\begin{aligned} D= & \left|\begin{array}{ccc}\alpha & -1 & -1 \\ 1 & -\alpha & -1 \\ 1 & -1 & -\alpha\end{array}\right|=\left|\begin{array}{ccc}\alpha & 1 & 1 \\ 1 & \alpha & 1 \\ 1 & 1 & \alpha\end{array}\right| \\ = & \left|\begin{array}{ccc}\alpha+2 & \alpha+2 & \alpha+2 \\ 1 & \alpha & 1 \\ 1 & 1 & \alpha\end{array}\right| \\ \Rightarrow \quad D & (\alpha+2)\left|\begin{array}{ccc}1 & 1 & 1 \\ 1 & \alpha & 1 \\ 1 & 1 & \alpha\end{array}\right| \\ = & (\alpha+2)\left|\begin{array}{ccc}1 & 0 & 0 \\ 1 & \alpha-1 & 0 \\ 1 & 0 & \alpha-1\end{array}\right| \\ \Rightarrow \quad & \left(\text { Applying } C_2 \rightarrow C_2-C_1, C_3 \rightarrow C_3-C_1\right)\end{aligned}$
$\begin{aligned} & \Rightarrow \quad D=(\alpha+2)(\alpha-1)^2 \\ & \text { and } D_1=\left|\begin{array}{lll}\alpha-1 & 1 & 1 \\ \alpha-1 & \alpha & 1 \\ \alpha-1 & 1 & \alpha\end{array}\right|=(\alpha-1)\left|\begin{array}{ccc}1 & 1 & 1 \\ 1 & \alpha & 1 \\ 1 & 1 & \alpha\end{array}\right| \\ & =(\alpha-1)\left|\begin{array}{ccc}1 & 0 & 0 \\ 1 & \alpha-1 & 0 \\ 1 & 0 & \alpha-1\end{array}\right| \\ & \text { (Applying } C_2 \rightarrow C_2-C_1, C_3 \rightarrow C_3-C_1 \text { ) } \\ & \Rightarrow \quad D_1=(\alpha-1)^3 \\ & \text { Clearly } D=0 \text { for } \alpha=-2 \text {, but } D_1 \neq 0 \\ & \text { So, the system is inconsistent for } \alpha=-2 \\ & \end{aligned}$
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