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$\begin{bmatrix}2 &1 & -4\\ 4 & 3 & -12\\ 1 & 2 & -8 \end{bmatrix}\cdot \begin{bmatrix} x\\ y\\ z \end{bmatrix}= \begin{bmatrix} \alpha \\5 \\ 7 \end{bmatrix}$

For how many values of $\alpha$ , this system has infinite solutions ?

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consider the augmented matrix $\begin{bmatrix} 2 &1 & -4 &\alpha \\ 4 & 3 &-12 &5 \\ 1 & 2& -8 &7 \end{bmatrix}$.

interchange R3<->R1.

$\begin{bmatrix} 1 &2 & -8 &7 \\ 4 & 3 &-12 &5 \\ 2 & 1& -4 &\alpha \end{bmatrix}$

to find Rank of a matrix perform required operations.

R2<-R2-4R1.

R3<-R3-2R1.

$\begin{bmatrix} 1 &2 & -8 &7 \\ 0 & -5 &20 &-23\\ 0& -3& 12 &\alpha-14 \end{bmatrix}$

one more operation is R3<-5R3-3R1.

$\begin{bmatrix} 1 &2 & -8 &7 \\ 0 & -5 &20 &-23\\ 0& 0& 0 &5\alpha-1 \end{bmatrix}$.

to get infinite solutions RANK OF MATRIX A=RANK OF AUGMENETED MATRIX <NO OF VARIABLES.

RANK(A)=RANK(AB) //

5$\alpha$-1 =0.AND $\alpha$=0.2
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