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The values of $\eta$ for which the following system of equations

$$\begin{array} {} x & + & y & + & z & = & 1 \\ x & + & 2y & + & 4z & = & \eta \\ x & + & 4y & + & 10z & = & \eta ^2 \end{array}$$ has a solution are

  1. $\eta = 1, -2$
  2. $\eta = -1, -2$
  3. $\eta = 3, -3$
  4. $\eta = 1, 2$
in Linear Algebra by Veteran (431k points)
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1 Answer

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From the given set of equations,augmented matrix=

$\begin{bmatrix} 1&1 &1 &1 \\ 1& 2 & 4 & a\\ 1& 4& 10& a^{2} \end{bmatrix}$

$R_2$$\rightarrow$$R_2$-$R_1$

$R_3$$\rightarrow$$R_3$-$R_1$

$\begin{bmatrix} 1 & 1 & 1& 1\\ 0& 1 & 3 & a-1\\ 0& 3&9 & a^{2}-1 \end{bmatrix}$

$R_3$$\rightarrow$$R_3$-$3R_2$

$\begin{bmatrix} 1 & 1 & 1& 1\\ 0& 1 & 3 & a-1\\ 0& 0&0 & a^{2}-3a+2 \end{bmatrix}$

For the system of equations to have solution,Rank of augmented matrix must be equal to rank of coefficient matrix

Rank of coefficient matrix=2

so for the rank of augmented matrix to be equal to 2, $a^{2}-3a+2$ must be equal to 0.

Therefore,a=1,2
by Active (4.3k points)

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