0 votes 0 votes $\begin{bmatrix}2 &1 & -4\\ 4 & 3 & -12\\ 1 & 2 & -8 \end{bmatrix}\cdot \begin{bmatrix} x\\ y\\ z \end{bmatrix}= \begin{bmatrix} \alpha \\5 \\ 7 \end{bmatrix}$ For how many values of $\alpha$ , this system has infinite solutions ? Anand Vijayan asked Dec 2, 2016 Anand Vijayan 193 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
Best answer 1 votes 1 votes 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 santhoshdevulapally answered Dec 2, 2016 selected Dec 2, 2016 by Anand Vijayan santhoshdevulapally comment Share Follow See all 0 reply Please log in or register to add a comment.