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[SOLVED] 311.1.7.11 Find the value(s) of h for which the vectors are linearly dependent. Justify

karush

Well-known member
Jan 31, 2012
3,066
$\tiny{311.1.7.11}$
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:
 

topsquark

Well-known member
MHB Math Helper
Aug 30, 2012
1,273
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
 

Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
821
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!
 

karush

Well-known member
Jan 31, 2012
3,066
row reduction tends to be arithmetic torture
 

Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
821
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.