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A = \begin{bmatrix} 2 &1 \\ 1& -3\\ 3 &4 \end{bmatrix}

B = \begin{bmatrix} 3\\ -1\\ k \end{bmatrix}

X = \begin{bmatrix} x\\ y \end{bmatrix}

AX = B

\begin{bmatrix} 2 & 1\\ 1& -3\\ 3&4 \end{bmatrix} \begin{bmatrix} x\\ y \end{bmatrix} = \begin{bmatrix} 5\\ -1\\ k \end{bmatrix}

augmented \ matrix(A/B) = \begin{bmatrix} 2 & 1 & 5\\ 1&-3 &-1 \\ 3& 4 &k \end{bmatrix}

-> R2 → 2R2 -R1 ; R3 →2R3 - 3R1

\begin{bmatrix} 2 & 1 & 5\\ 0&-7 &-7 \\ 0& 5 &2k-15 \end{bmatrix}

-> R3 → 7R3+5R2

\begin{bmatrix} 2 & 1 & 5\\ 0&-7 &-7 \\ 0& 0 &14k-140 \end{bmatrix}

rank of ( A/B ) is depending on the value of k

rank of A is 2

As the system is to be consistent Rank(A) = Rank(A/B)

∴   14k - 140 =0 

     ⇒ k = 140/14

           k  = 10

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