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Determine the value of b that would guarantee that the below linear system is consistent.

$$\eqalign{x_1 − 2x_2 − 6x_3 &= -4\cr 5x_1 − 4x_2 − 2x_3 &= -7\cr −11x_1 + 4x_2 − 18x_3 &= b\cr}$$

I did the row reductions and ended up with ( $0x_1 + 0x_2 + 0x_3 = -2+b$ ) for the last row. So the answer I got was $b=2$, but that answer was incorrect?


[ 1 -2 -6 | -4 ]

[ 5 -4 -2 | -7 ]

[ -11 4 -18 | b ]


[ 1 -2 -6 | -4 ]

[ 0 -6 -28 | -14 ]

[ -11 4 -18 | b ]


[ 1 -2 -6 | -4 ]

[ 0 -6 -28 | -14 ]

[ 0 -18 -84 | -44+b]


[ 1 -2 -6 | -4 ]

[ 0 -6 -28 | -14 ]

[ 0 0 0 | -2+b ]

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    $\begingroup$ Check your row operations for arithmetic mistakes. Post them, if you can't find a mistake. $\endgroup$ – Christopher Carl Heckman May 14 '16 at 0:15
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By basic passages you can transform this linear system to $$ \begin{equation} \begin{bmatrix} 1 &-2 &-6 \\ 0 &6 &28 \\ 0 &-6 &-28 \\ \end{bmatrix} \begin{bmatrix} x_1\\ x_2 \\ x_3 \\ \end{bmatrix}= \begin{bmatrix} -4\\ 13\\ b-18\\ \end{bmatrix} \end{equation} $$

$$ \begin{equation} \begin{bmatrix} 1 &-2 &-6 \\ 0 &6 &28 \\ 0 &0 &0 \\ \end{bmatrix} \begin{bmatrix} x_1\\ x_2 \\ x_3 \\ \end{bmatrix}= \begin{bmatrix} -4\\ 13\\ b-5\\ \end{bmatrix} \end{equation} $$

As you see the coefficient matrix is singular, with rank 2.

So to ensure consistency we need $b=5$.

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