# Linearly Dependent Vectors: Find h Value and Justify

• MHB
• karush
In summary, the conversation discusses finding the values of h for which a set of vectors are linearly dependent. The process involves using row reduction or the determinant to solve for h. It is noted that h cannot equal -4 in order for there to be multiple solutions and for the vectors to be dependent.
karush
Gold Member
MHB
$\tiny{311.1.7.11}$
ok I am going to do several of these till I get it...
Find the value(s) of h for which the vectors are linearly dependent. Justify
$\left[\begin{array}{rrrrrr} 2\\-2\\4 \end{array}\right], \left[\begin{array}{rrrrrr} 4\\-6\\7 \end{array}\right], \left[\begin{array}{rrrrrr} -2\\2\\h \end{array}\right]$
so the first step would be, I like the augment line
$\left[\begin{array}{rrr|r} 2&4&-2&0\\ -2&-6&2&0\\ 4&7&h&0 \end{array}\right]$
EMH returned...
$\text{rref}=\left[ \begin{array}{cccc}1&0&0&0\\0&1&0&0\\0&0&1&0\end{array} \right]$

$h$ disappeared :censored:

Remember that we are looking for constants a, b, and c such that
$$\displaystyle a \vec{v_1} + b \vec{v_2} + c \vec{v_3} = \vec{0}$$

So you are looking for values a, b, c, and h such that
$$\displaystyle \left [ \begin{matrix} 2a + 4b - 2c \\ -2a - 6b + 2c \\ 4a + 7b + ch \end{matrix} \right ] = \left [ \begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right ]$$

-Dan

Do you know how to do the "row reduction" yourself?
Starting with
$$\displaystyle \begin{bmatrix}2 & 4 & -2 & 0 \\ -2 & -6 & 2 & 0 \\ 4 & 7 & h & 0 \end{bmatrix}$$
I would (1) add the first row to the second row, (2) subtract twice the first row from the third row and (3) divide the first two by 2:
$$\displaystyle \begin{bmatrix}1 & 2 & -1 & 0 \\ 0 & -2 & 0 & 0 \\ 0 & -1 & h+ 4 & 0 \end{bmatrix}$$

Now, add the second row to the first row, divide the second row by -2, and then add this new second row to the third row:
$$\displaystyle \begin{bmatrix}1 & 0 & -1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & h+ 4 & 0 \end{bmatrix}$$

Now, what your "EMH" did was divide the third row by h+4 and then add the new third row to the first row to get
$$\displaystyle \begin{bmatrix}1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{bmatrix}$$
which, as you say, has no h!

But you are supposed to be SMARTER than some machine and recognize that you CAN'T divide by h+4 if h= -4 because then h+ 4= 0!

row reduction tends to be arithmetic torture

Would you prefer to use the determinant? Expanding on the third column,
$\left|\begin{array}{ccc} 2 & 4& -2\\ -2& -6 & 2 \\ 4 & 7 & h \end{array}\right|= -2\left|\begin{array}{cc}-2& -6 \\ 4 & 7\end{array}\right|- 2\left|\begin{array}{cc}2& 4 \\ 4& 7\end{array}\right|+ h\left|\begin{array}{cc}2& 4 \\ -2 & -6\end{array}\right|$= -2(10)- 2(-2)- 4h=-16- 4h.

In order that a= b= c= 0 NOT be the only solution (so that the vectors be dependent) the determinant must be 0: -16- 4h= 0 so 4h= -16 and h=-4 again.

## 1. What are linearly dependent vectors?

Linearly dependent vectors are a set of vectors that can be written as a linear combination of each other. This means that one of the vectors in the set can be expressed as a combination of the other vectors, using scalar multiplication and addition.

## 2. How do you determine if vectors are linearly dependent?

To determine if vectors are linearly dependent, you can use the h value method. This involves setting up a system of equations using the vectors and solving for the h value. If the h value is equal to 0, the vectors are linearly dependent. If the h value is not equal to 0, the vectors are linearly independent.

## 3. What is the h value method?

The h value method is a method used to determine if vectors are linearly dependent or independent. It involves setting up a system of equations using the vectors and solving for the h value. If the h value is equal to 0, the vectors are linearly dependent. If the h value is not equal to 0, the vectors are linearly independent.

## 4. Why is it important to determine if vectors are linearly dependent?

It is important to determine if vectors are linearly dependent because it affects the properties and behavior of the vectors. Linearly dependent vectors can cause issues in calculations and can also be redundant in certain situations. It is also important in applications such as linear algebra and physics.

## 5. Can a set of two vectors be linearly dependent?

Yes, a set of two vectors can be linearly dependent. This means that one of the vectors can be expressed as a combination of the other vector, using scalar multiplication and addition. It is also possible for a set of three or more vectors to be linearly dependent.

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