Linear Algebra – How to Determine Value of h for Consistent Linear System

linear algebramatrices

Determine the value of h such that the matrix is the augmented matrix of a consistent linear system.
$$
\begin{bmatrix}
1 && h && 4 \\
3 && 6 && 8
\end{bmatrix}
$$

I'm entirely unsure how to go about solving this.

This how far I got:

$$
\sim\begin{bmatrix}
1 && h && 4 \\
0 && (6-3h) && -4
\end{bmatrix} R_2' = R_2-3R_1
$$

$$
6-3h=-4 \\
6+4=3h \\
10=3h \\
h=3/10 \text{ ?}
$$

New progress

$$
\sim\begin{bmatrix}
3 && 6 && 8 \\
1 && h && 4
\end{bmatrix}
R_1 \leftarrow\rightarrow R_2
$$

$$
\sim\begin{bmatrix}
3 && 6 && 8 \\
0 && (h-2) && 4/3
\end{bmatrix}
R_2' = – \frac13 R_1
$$

$$
\sim\begin{bmatrix}
3 && 6 && 8 \\
0 && 3(h-2) && 4
\end{bmatrix}
R_2' = 3R_2
$$

$$
\sim\begin{bmatrix}
3 && 6 && 8 \\ \\
0 && 1 && \frac{4}{3(h-2)}
\end{bmatrix}
R_2' = R_2 \div 3(h-2)
$$

$$
\sim\begin{bmatrix}
-3 && 0 && z \\ \\
0 && 1 && \frac{4}{3(h-2)}
\end{bmatrix}
R_1' = 6R_2 – R_1 \\
z = -8 + \frac{24}{3(h-2)} = \frac{-32(h-2) + 24}{3(h-2)}
$$

$$
\sim\begin{bmatrix}
1 && 0 && z \\ \\
0 && 1 && \frac{4}{3(h-2)}
\end{bmatrix}
R_1' = -\frac13 R_1 \\
z = – \frac{-32(h-2) + 24}{(h-2)}
$$

Best Answer

Do these row operations:

  • Swap $R_1$ and $R_2$
  • $R_2 \rightarrow -R_1/3 + R_2$
  • $3R_2$
  • Divide $R_2/(3(h-2))$
  • $R_1 \rightarrow 6R_2 - R_1$
  • Divide $R_1/3$

This should provide the value of $h$ that makes the system inconsistent.

Spoiler - Do Not Peek

$\begin{bmatrix} 1 && 0 && \dfrac{8(h-3)}{3(h-2)} \\ 0 && 1 && \dfrac{4}{3(h-2)}\end{bmatrix}$

Notes:

  • Sometimes you can also look at the determinant of the system, in this case it is $6-3h$ (what value of $h$ would make this inconsistent).
  • You could have also reduced and solved the system as another approach. You get $y = \dfrac{4}{3(2-h)}$ and can substitute back.
  • There are other approaches.
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