Discussion Forum

1)

The number of values of k, for which the system of equations (k+1)x+8y=4k

    kx+(k+3)y=3k-1   has no solution , is 

Answer:

Option B

Explanation:

Given equations can be written in matrix form

 AX=B

 Where, A= $\begin{bmatrix}k+1 & 8 \\k& k+3 \end{bmatrix}=\begin{bmatrix}x   \\y  \end{bmatrix} and \begin{bmatrix}4k  \\3k-1  \end{bmatrix}$

 For no solution , |A|= 0  and (adj A)B≠ 0

nOW, |A|=   $\begin{bmatrix}k+1 & 8 \\k& k+3 \end{bmatrix}=0$

  $\Rightarrow$     (k+1)(k+3)-8K=0

$\Rightarrow$     k²+4k+3-8k=0

$\Rightarrow$    k²-4k+3=0

$\Rightarrow$     (k-1) (k-3)=0

$\Rightarrow$    k=1,k=3

   Now, adj A=   $\begin{bmatrix}k+3 & -8 \\-k & k+1 \end{bmatrix}$

  Now, (adj A)B=  $\begin{bmatrix}k+3 & -8 \\-k & k+1 \end{bmatrix}$   $\begin{bmatrix}4k  \\3k-1  \end{bmatrix}$

     =   $\begin{bmatrix}(k+3)(4k)-8(3k-1)  \\-4k^{2}+(k+1)(3k-1)  \end{bmatrix}$

   =  $\begin{bmatrix}4k^{2}-12k+8 \\-k^{2}+2k-1  \end{bmatrix}$

  Put k=1

(adj A)B=   $\begin{bmatrix}4-12-8\\-1+2-1 \end{bmatrix}=\begin{bmatrix}0  \\0  \end{bmatrix}$   not true

 Put k=3

 (adj A) B=   $\begin{bmatrix}36-36+8  \\-9+6-1  \end{bmatrix}=\begin{bmatrix}8  \\-4  \end{bmatrix}\neq0$   true

 Hence , required value of k is 3

    only onne value of k exist