Are these vectors linearly independent? How can I check?

Question

I want to know if these vectors are linearly independent? The vector space is 3 dimension R, like (1,0,0) type of vectors.

The three vectors are (8,9,1), (1,4,5), (0,7,2).

How can I check if the vector are linearly independent?

I know two vectors are linearly independent if one is not scalar multiple of the other vector.

How can I check linear independence for three vectors?

Answer 1

Consider the system

$$\begin{pmatrix} 8 & 1 & 0 \\ 9 & 4 & 7 \\ 1 & 5 & 2\end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \\ x_3\end{pmatrix}=\begin{pmatrix} 0 \\ 0 \\ 0\end{pmatrix}$$

How many solution does the sysem have?

If the system has exactly one solution, then it is linearly independent. Otherwise, it is not.

Answer 2

Grammatically, it would be "Are these vectors linearly independent?".

It appears that you do not know the actual definition of "independent". A set of vectors $\{v_1, v_2, \cdot\cdot\cdot, v_n\}$ is said to be "linearly independent" if and only if there exist n numbers, $a_1, a_2, \cdot\cdot\cdot, a_n$ not all 0, such that $a_1v_1+ a_2v_2+ \cdot\cdot\cdot+ a_nv_n= 0$. (With only two vectors, $v_1$ and $v_2$, $a_1v_1+ a_2v_2= 0$, with $a_1$ not equal to 0, is equivalent to $v_1= \frac{a_2}{a_1}v_2$ and with $a_2$ not equal to 0, to $v_1= \frac{a_1}{a_2}v_1$ which is where your "one is a multiple of the other is from.)

Do there exist numbers, a, b, c, not all 0 so that a(8,9,1)+ b(1,4,5)+ c(0,7,2)= (8a+ b, 9a+ 4b+ 7c, a+ 5b+ 2c)= (0, 0, 0). So you need to solve the three equations: 8a+ b= 0, 9a+ 4b+ 7c= 0, a+ 5b+ 2c= 0. a= b= c= 0 is an obvious solution. Do there exist other solutions?

Answer 3

For the definition of linear independence consider the scalars $a_1, a_2, a_3 \in R$, $R$ is the field, such: $$a_1(8,9,1)+a_2(1,4,5)+a_3(0,7,2)=0$$ If $a_1=a_2=a_3=0$, say the vectors $(8,9,1), (1,4,5), (0,7,2)$ are linearly independent, but, if the scalars $a_1, a_2, a_3$ not are all equal to zero, say they are linearly dependent. Well, now in this part is pure vector operations. We have: $$a_1(8,9,1)+a_2(1,4,5)+a_3(0,7,2)=0$$ $$=((a_1)8,(a_1)9,(a_1)1)+((a_2)1,(a_2)4,(a_2)5)+((a_3)0,(a_3)7,(a_3)2)=0$$ $$=((a_1)8+(a_2)1,(a_1)9+(a_2)4+(a_3)7,(a_1)1+(a_2)5+(a_3)2)=0=(0,0,0)$$ For the vector equality, we make a system of equations $$ \left\lbrace \begin{array}{l} (a_1)8+(a_2)=0 \\ (a_1)9+(a_2)4+(a_3)7=0 \\ (a_1)+(a_2)5+(a_3)2=0 \\ \end{array} \right.$$ If you solve the system you get $a_1=0$,$a_2=0$,$a_3=0$. As $a_1=a_2=a_3=0$ for definition, the vectors $(8,9,1), (1,4,5), (0,7,2)$ are linearly independent.

Answer 4

Can you find the determinant of a square matrix?

The three vectors. $(8,9,1), (1,4,5), (0,7,2)$ will be linearly independent if:

$$\begin{vmatrix} 8 & 1 & 0 \\ 9 & 4 & 7 \\ 1 & 5 & 2\end{vmatrix} \neq 0$$

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You should have access to a theorem something like:

Theorem: Let $A$ be a square $n{\small \times}n$ matrix.   Then the following are equivalent:

• The columns/rows of $A$ are linearly independent.
• The columns/rows of $A$ form a basis in $\Bbb R^n$ .
• The columns/rows of $A$ span $\Bbb R^n$ .
• $\det(A)\neq 0$ .
• $A$ is invertible.
• The reduced row echelon form of $A$ has a leading $1$ in each row.
• And several other conditions found in most primers on Linear Algebra.

Answer 5

Three vectors in $\mathbb{R}^3$ are linearly independent iff the determinant of the matrix having them as columns (or rows) is non-zero. The question now may be: is there a really fast way to check that $$ \det\begin{pmatrix}8 & 9 & 1 \\ 1 & 4 & 5 \\ 0 & 7 & 2 \end{pmatrix}\neq 0 $$ ? Well, yes. Such determinant is for sure an integer number, and it has the same parity of the determinant of the matrix $$ \begin{pmatrix}0 & 1 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix} $$ given by the reduction of the previous matrix $\!\!\pmod{2}$. The last determinant is trivially odd, hence the original determinant is odd, too: in particular, non-zero. That ensures that the given vectors are linearly independent.