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Find the value(s) of $h$ for which the vectors are linearly dependent. Justify each answer.$\left[\begin{array}{r}{1} \\ {-1} \\ {4}\end{array}\right],\left[\begin{array}{r}{3} \\ {-5} \\ {7}\end{array}\right],\left[\begin{array}{r}{-1} \\ {5} \\ {h}\end{array}\right]$

Thus, the vectors are linearly dependent if and only if $h = 6 .$

Algebra

Chapter 1

Linear Equations in Linear Algebra

Section 7

Linear Independence

Introduction to Matrices

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In this example, we have three different vectors that are provided where the third vector also depends on the value of H here. One question we could ask about this set of vectors is when with the offset be linearly independent, for example, for example, with that value depend on h. Was it always be linearly dependent, or will it always be linearly dependent to determine the answer to such questions? We can start off with the Matrix A where a has calm is formed from the vectors given to be V one, V two and V three. So this would be equal to one negative 14 in the first column. Three negative 57 for calm to and then negative 15 and h so for this given matrix, we're going to roll reduce to echelon form and that will tell us where the pivots are. First we start here and we're going to eliminate the entry Negative one and positive for below. So let me copy row one to start out, then add row one to row to replacing Row two. That gives us zero negative two and four, then multiply row one by negative four and added to Row three will obtain zero negative five and H plus four. So our new pivot position is found on the two to entry here in the Matrix, and we're just going to eliminate the entry below this time since we just require echelon form. So let me copy both row ones and two. So there's our Row one and row to Let's first Divide row to buy the quantity. Negative two will obtain 01 negative, too. Now our operation is going to be to multiply row two by positive five so that we can add it to Row three and obtain a zero here. Upon performing, that operation will obtain zero zero and H minus six. So let's look at the situation so far for pivots. Calm one and calm to are automatically pivot columns and calm. Three. Could be a pivot column, but we require H minus six to not be zero. So if H is not equal to positive six, then A has a pivot in every column. What this would mean is then, if we consider the Matrix equation a X equals zero vector. This equation has no free variables When h is not equal to positive six. And we know that whenever we're dealing with no free variables, we have linearly independent columns. So we're ready to state our our conclusion will say when h is not equal to six. The columns of a which are these vectors from the start of the problem, are linearly independent.

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