If the system of linear equations2x+y−z=7,x−3y+2z=1,x+4y+δz=k, where δ, k∈R has infinitely many solutions, then δ+k is equal to

1−32:121−1:714δ:k~1−32:107−5:507δ−2:k−1~1−32:107−5:500δ+3:k−6Infinite solutions ⇒8+3=0,k−6=0⇒δ=−3,k=6⇒δ+k=3

If the system of linear equations2x+y−z=7,x−3y+2z=1,x+4y+δz=k, where δ, k∈R has infinitely many solutions, then δ+k is equal to

1−32:121−1:714δ:k~1−32:107−5:507δ−2:k−1~1−32:107−5:500δ+3:k−6Infinite solutions ⇒8+3=0,k−6=0⇒δ=−3,k=6⇒δ+k=3

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If the system of linear equations
$2 x + y - z = 7$ ,
$x - 3 y + 2 z = 1$ ,
$x + 4 y + δz = k,$ where $δ, k \in R$ has infinitely many solutions, then $δ + k$ is equal to

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-3

answer is B.

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Detailed Solution

$[\begin{array}{ccc} 1 & - 3 & 2 : 1 \\ 2 & 1 & - 1 : 7 \\ 1 & 4 & δ : k \end{array}] ~ [\begin{array}{crc} 1 & - 3 & 2 : 1 \\ 0 & 7 & - 5 : 5 \\ 0 & 7 & δ - 2 : k - 1 \end{array}] ~ [\begin{array}{ccc} 1 & - 3 & 2 : 1 \\ 0 & 7 & - 5 : 5 \\ 0 & 0 & δ + 3 : k - 6 \end{array}]$
Infinite solutions $\Rightarrow 8 + 3 = 0, k - 6 = 0$
$\begin{array}{l} \Rightarrow δ = - 3, k = 6 \\ \Rightarrow δ + k = 3 \end{array}$