Linear Algebra and Its Applications (5th Edition)

Published by Pearson
ISBN 10: 0-32198-238-X
ISBN 13: 978-0-32198-238-4

Chapter 1 - Linear Equations in Linear Algebra - 1.2 Exercises - Page 22: 17



Work Step by Step

We know the system of equations is consistent if the last column of the corresponding augmented matrix is not a pivot column. So we first need to convert the matrix, $\left[ \begin{array}{ccc} 2 & 3 & h\\ 4 & 6 & 7 \end{array} \right]$, to echelon form. We replace row 2 with -2(row 1)+(row 2) to get the equivalent matrix $$\left[ \begin{array}{ccc} 2 & 3 & h\\ 0 & 0 & -2h+7 \end{array} \right].$$ Now, in order for the last column not to be a pivot column, we must have$$-2h+7=0.$$ We solve this equation to get $h=\frac{7}{2}.$ Hence if $h=\frac{7}{2}$, then the last column of the augmented matrix is not a pivot column, which means the corresponding system of equations is consistent.
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