106k views
0 votes
Determine the​ value(s) of h such that the matrix is the augmented matrix of a consistent linear system

2 Answers

7 votes

Final answer:

To determine the value(s) of h for a consistent linear system, the augmented matrix must not have any row that implies a contradiction, like 0x + 0y = c where c is non-zero. The exact values of h depend on the matrix itself and usually involve row reduction to identify non-contradictory solutions. The provided examples do not appear to relate directly to a linear system matrix.

Step-by-step explanation:

To determine the value(s) of h that result in the matrix representing a consistent linear system, we must ensure that there are no contradictions within the augmented matrix. If the matrix represents a system of linear equations, consistency is achieved when the system has at least one solution. A contradiction would occur if a row in the augmented matrix had all zeros in the coefficient part and a non-zero value in the last column, corresponding to an equation like 0x + 0y = c (where c is non-zero), which has no solution.

The exact values of h would be determined by the context of the specific matrix and its elements, usually involving row reduction to echelon form to identify possible contradictions. In some cases, specific values of h can lead to a row of all zeros, which does not affect consistency, but a particular non-zero value in the rightmost column would create inconsistency. Without the explicit matrix, we cannot provide the exact values of h.

In the examples provided, it seems there is a mix of mathematical equations and expressions that are not directly related to a specific matrix or linear system. As such, the value of h in those expressions may refer to concepts like the uncertainty principle in quantum mechanics or gravitational force in physics, and thus do not pertain to solving a system of linear equations.

User Mariglen
by
6.3k points
4 votes
Given the matrix


\left[\begin{array}{ccc}1&h&4\\3&6&8\end{array}\right]

We perform the following matrix operations


\left[\begin{array}{ccc}1&h&4\\3&6&8\end{array}\right] -3R_1+R_2\rightarrow R_2 \\ \\ \left[\begin{array}{ccc}1&h&4\\0&6-3h&-4\end{array}\right] (1)/(6-3h)R_2\rightarrow R_2 \\ \\ \left[\begin{array}{ccc}1&h&4\\0&1&(-4)/(6-3h)\end{array}\right] -hR_2+R_1\rightarrow R_1 \\ \\ \left[\begin{array}{ccc}1&0& (24-8h)/(6-3h) \\0&1&(-4)/(6-3h)\end{array}\right]

Thus, the matrix is consistent for

h\\eq6-3h \\ \\ h+3h\\eq6 \\ \\ 4h\\eq6 \\ \\ h\\eq (6)/(4) \\ \\ h\\eq (3)/(2)
User Chinatsu
by
5.7k points