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Montano1993 [528]
3 years ago
8

How to determine if the columns of a matrix are linearly independent?

Mathematics
1 answer:
astra-53 [7]3 years ago
8 0
Let's consider an arbitrary 2x2 matrix as an example,

\mathbf A=\begin{bmatrix}\mathbf x&\mathbf y\end{bmatrix}=\begin{bmatrix}x_1&y_1\\x_2&y_2\end{bmatrix}

The columns of \mathbf A are linearly independent if and only if the column vectors \mathbf x,\mathbf y are linearly independent.

This is the case if the only way we can make a linear combination of \mathbf x,\mathbf y reduce to the zero vector is to multiply the vectors by 0; that is,

c_1\mathbf x+c_2\mathbf y=\mathbf 0

only by letting c_1=c_2=0.

A more concrete example: suppose

\mathbf A=\begin{bmatrix}1&2\\4&8\end{bmatrix}

Here, \mathbf x=\begin{bmatrix}1\\4\end{bmatrix} and \amthbf y=\begin{bmatrix}2\\8\end{bmatrix}. Notice that we can get the zero vector by taking c_1=-2 and c_2=1:

-2\begin{bmatrix}1\\4\end{bmatrix}+\begin{bmatrix}2\\8\end{bmatrix}=\begin{bmatrix}-2+2\\-8+8\end{bmatrix}=\mathbf 0

so the columns of \mathbf A are not linearly independent, or linearly dependent.
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How many solutions does the equation x_1 +x_2+x_3+x_4+x_5=21 have where x_1, x_2, x_3, x_4, and x_5 are nonnegative integers and
omeli [17]

Step-by-step explanation:

x_{1} + x_{2} + x_{3} + x_{4} + x_{5} = 21\\     (given)

Let us consider :

x_{1} = t_{1} + 1

x_{2} = t_{2}

x_{3} = t_{3}

x_{4}  = t_{4}

x_{5} = t_{5}

Now, by substituting the above considerations in the above equation, we get:

t_{1} + 1 + t_{2} + t_{3} + t_{4} + t_{5} = 21\\

t_{1}  + t_{2} + t_{3} + t_{4} + t_{5} = 20\\

where,

t_{i} \geq 1

then it follows

n = 20

r = 4

then no. of solutions for the eqn = _{r}^{n + r}\textrm{C}

                                                      = _{4}^{24}\textrm{C}

                                                      = 10626

Answer :

no. of solutions for the eqn 10626

4 0
3 years ago
Factor 3/4 out of 3/4z + 6
Alenkasestr [34]
ANSWER

\frac{3}{4} z + 6 =  \frac{3}{4} (z +  8  )


EXPLANATION

We want to factor
\frac{3}{4}
out of

\frac{3}{4} z + 6

The first term is already having a factor of
\frac{3}{4}
The constant term which is the second term is not having a factor of
\frac{3}{4}
so we need to use a trick of multiplying and dividing by the same factor to obtain,



\frac{3}{4} z + 6 =  \frac{3}{4} z +  \frac{3}{4}  \times  \frac{6}{ \frac{3}{4} }


We can now factor
\frac{3}{4}
out of the right hand side to obtain,

\frac{3}{4} z + 6 =  \frac{3}{4} (z +  \frac{6}{ \frac{3}{4} } )
Let us rewrite the right most fraction using the normal division symbol.

\frac{3}{4} z + 6 =  \frac{3}{4} (z +  6 \div  \frac{3}{4}  )




We simplify to obtain,

\frac{3}{4} z + 6 =  \frac{3}{4} (z +  6  \times  \frac{4}{3}  )



This further gives us,


\frac{3}{4} z + 6 =  \frac{3}{4} (z +  2 \times  4  )



\frac{3}{4} z + 6 =  \frac{3}{4} (z +  8  )


5 0
3 years ago
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Allisa [31]

fx=25x-x²

dy/dx=0

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(=)X=14,5

4 0
3 years ago
Can some one help me in this problem
strojnjashka [21]
I believe x=116 and z=26. 4z-40=64
4z=104
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cluponka [151]

Answer:

Try googling it

Step-by-step explanation:

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2 years ago
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