Find the value of k if the following equations are consistent: (k+1)x+(k-1)y+(k-1)=0,(k-1)x+(k+1)y+(k-1)=0,(k-

Miscellaneous Exercise 4(A) · Q68

(k+1)x+(k-1)y+(k-1)=0,\ (k-1)x+(k+1)y+(k-1)=0,\ (k-1)x+(k-1)y+(k+1)=0

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$D=\begin{vmatrix}k+1&k-1&k-1\\k-1&k+1&k-1\\k-1&k-1&k+1\end{vmatrix}$ . Replace $R_1\to R_1+R_2+R_3=(3k-1,3k-1,3k-1)=(3k-1)(1,1,1)$ , so $D=(3k-1)\begin{vmatrix}1&1&1\\k-1&k+1&k-1\\k-1&k-1&k+1\end{vmatrix}$ .

Apply $C_2\to C_2-C_1,\ C_3\to C_3-C_1$ on the inner determinant: it becomes $\begin{vmatrix}1&0&0\\k-1&2&0\\k-1&0&2\end{vmatrix}=1(4-0)=4$ . …

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Q.Find the value of k if the following equations are consistent: $(k+1)x+(k-1)y+(k-1)=0,\ (k-1)x+(k+1)y+(k-1)=0,\ (k-1)x+(k-1)y+(k+1)=0$

Concept understanding — Consistency of a System of Linear Equations

Q.Find the value of k if the following equations are consistent: (k+1)x+(k−1)y+(k−1)=0, (k−1)x+(k+1)y+(k−1)=0, (k−1)x+(k−1)y+(k+1)=0(k+1)x+(k-1)y+(k-1)=0,\ (k-1)x+(k+1)y+(k-1)=0,\ (k-1)x+(k-1)y+(k+1)=0(k+1)x+(k−1)y+(k−1)=0, (k−1)x+(k+1)y+(k−1)=0, (k−1)x+(k−1)y+(k+1)=0

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Q.Find the value of k if the following equations are consistent: $(k+1)x+(k-1)y+(k-1)=0,\ (k-1)x+(k+1)y+(k-1)=0,\ (k-1)x+(k-1)y+(k+1)=0$

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