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QuestionFind values of hh for which the matrix is consistent: [14−32h−6] \left[\begin{array}{rrr} 1 & 4 & -3 \\ 2 & h & -6 \end{array}\right] Choices: A. h≠h \neq B. h=h= C. for all hh D. for no hh.

Studdy Solution

STEP 1

Assumptions1. We have a matrix[14−3h−6]\left[\begin{array}{rrr} 1 &4 & -3 \\ & h & -6\end{array}\right] . We need to find the value(s) of hh such that the matrix is the augmented matrix of a consistent linear system.

STEP 2

First, we need to understand what it means for a system to be consistent. A system of linear equations is consistent if it has at least one solution. In terms of an augmented matrix, a system is consistent if the last column is not a pivot column. In other words, we cannot have a row of the form [0 ...0 | b] with b ≠0 in the row echelon form of the matrix.

STEP 3

To find the row echelon form of the matrix, we can perform row operations. Let's subtract2 times the first row from the second row to eliminate the leading2 in the second row.
[1−30h−80]\left[\begin{array}{rrr} 1 & & -3 \\ 0 & h-8 &0\end{array}\right]

STEP 4

Now, we can see that for the system to be consistent, the second row cannot be of the form [00 | b] with b ≠0. This means that h−8h-8 must be equal to0.

STEP 5

olving the equation h−8=0h-8=0 for hh gives ush=8h =8So, the matrix is the augmented matrix of a consistent linear system if h=8h =8.

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