Question:

The values of $\alpha$ for which the system of equation x + y + z = 1, x + 2y + 4z = $\alpha$, x + 4y + 10z = $\alpha^2$ is consistent are given by

Updated On: Apr 19, 2024
  • 1, -2
  • -1, 2
  • 44563
  • 44562
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The Correct Option is C

Solution and Explanation

We have
A:B $\begin{vmatrix}
1 & 1 & 1 & : & 1 \\
1 & 2 &4&:& \alpha \\
1&4&10&:&\alpha^2 \\ \end{vmatrix}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \sim$ $\begin{vmatrix}
1 & 1 & 1 & : & 1 \\
0 & 1 &3&:& \alpha-1 \\
0&3&9&:&\alpha^2-1 \\ \end{vmatrix}$
$\begin{bmatrix}
applying \ R_2 \ \rightarrow \ R_2 \ - \ R_1 \\
\& \ R_3 \ \rightarrow \ R_3 \ - \ R_1 \\ \end{bmatrix}$
$ \ \ \ \sim$ $\begin{vmatrix}
1 & 1 & 1 & : & 1 \\
0 & 1 &3&:& \alpha-1 \\
0&0&0&:&\alpha^2-3\alpha+2 \\ \end{vmatrix}$
$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ [applying \ \ R_3 \ \rightarrow \ R_3 \ - \ 3R_2]$
But the system is consistent
$\therefore \ \ \alpha^2 - \ 3 \alpha + \ 2 \ =0$
$\Rightarrow \ \ (\alpha \ - \ 2)(\alpha \ - \ 1)=0 \ \Rightarrow \ \alpha \ =2 \ or \ \alpha \ = \ 1$
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Concepts Used:

Matrix Transformation

The numbers or functions that are kept in a matrix are termed the elements or the entries of the matrix.

Transpose Matrix:

The matrix acquired by interchanging the rows and columns of the parent matrix is termed the Transpose matrix. The definition of a transpose matrix goes as follows - “A Matrix which is devised by turning all the rows of a given matrix into columns and vice-versa.”