 # Determine if b is a linear combination of the vectors formed from the columns of the matrix A.

$A=\begin{bmatrix} 1&-4&2 \\ 0&3&5 \\ -2&8&-4 \end{bmatrix},\space b = \begin{bmatrix} 3 \\ -7 \\ -3 \end{bmatrix}$

This problem aims to familiarize us with vector equations, linear combinations of a vector, and echelon form. The concepts required to solve this problem are related to basic matrices, which include linear combinations, augmented vectors, and row-reduced forms.

Linear combinations are acquired by multiplying matrices by scalars and by adding them all together. Let’s start by looking at a formal definition:

Let $A_1,….., A_n$ be matrices carrying dimension $K\times L$. A $K\times L$ matrix is called a linear combination of $A_1,….., A_n$ only if they manage to have scalars, known as coefficients of the linear combination, such that:

$B = \alpha_1 A_1 +….+ \alpha_n A_n$

We will start by looking into the matrix $\vec{b}$, which can be written as a linear combination of vector $\vec{A}$, $\implies$ the following vector has some solution, such that:

$\vec{u}= \begin{bmatrix} 1 \\ 0 \\ -2 \end{bmatrix},\space\vec{v}= \begin{bmatrix} -4 \\ 3 \\ 5 \end{bmatrix},and\space\vec{w}= \begin{bmatrix} -2 \\ 8 \\ -4 \end{bmatrix}$

The vector equation: $\vec{b} = x\vec{u} + y\vec{v} + z\vec{w}$, where $x,y,z$ are scalar unknowns.

Since we have taken each column of $\vec{A}$ as a separate vector, we can simply form the equation using them:

$\implies \begin{bmatrix} 3 \\ -7 \\ -3 \end{bmatrix} = x\begin{bmatrix} 1 \\ 0 \\ -2 \end{bmatrix}+ y\begin{bmatrix} -4 \\ 3 \\ 5 \end{bmatrix}+ z\begin{bmatrix} -2 \\ 8 \\ -4 \end{bmatrix}$

$\implies \begin{pmatrix} 3 \\ -7 \\ -3 \end{pmatrix} = \begin{pmatrix} x \\ 0 \\ -2x \end{pmatrix}+ \begin{pmatrix} -4y \\ 3y \\ 5y \end{pmatrix}+ \begin{pmatrix} -2z \\ 8z \\ -4z \end{pmatrix}$

$\implies \begin{pmatrix} 3 \\ -7 \\ -3 \end{pmatrix} = \begin{pmatrix} x-4y-2z \\ 3y+8z \\ -2x+5y-4z \end{pmatrix}$

Now, we get the corresponding system of equations:

$\begin{matrix} x-4y-2z = 3\\ 0x+3y+8z = -7 \\ -2x+5y-4z =-3 \end{matrix}$

And its corresponding augmented matrix comes out to be:

$\begin{pmatrix} 1&-4&-2&3\\ 0&3&8&-7 \\ -2&5&-4&-3 \end{pmatrix}$

Now we are going to reduce it to reduced Echelon form as follows:

$\begin{pmatrix} 1&-4&-2&3\\ 0&3&8&-7 \\ -2&5&-4&-3 \end{pmatrix}$

By $R_1 \leftrightarrow R_2$:

$\begin{pmatrix} 0&3&8&-7 \\ 1&-4&-2&3\\ -2&5&-4&-3 \end{pmatrix}$

By $R_3 + \dfrac{1}{2}R_1 \implies R_3$:

$\begin{pmatrix} -2&8&-4&-3 \\ 0&3&5&-7 \\ 0&0&0&\dfrac{3}{2} \end{pmatrix}$

Since we have row reduced it, the equivalent system of equations becomes:

$\begin{matrix} x-4y+2z = 3\\ 0x+3y+5z = -7 \\ 0= 3 \end{matrix}$

Since the last equation does not hold valid $0 \neq 3$, thus the system has no solution.

## Numerical Result

The system has no solution since the equation $0\neq 3$ does not hold as a valid one.

## Example

Let $A_1$ and $A_2$ be $2$ vectors:

$A_1 = \begin{bmatrix} 2 \\ 1 \end{bmatrix}, \space A_2 =\begin{bmatrix} 0 \\ 1 \end{bmatrix}$

Compute the value of linear combination $3A_1 -2A_2$.

It can be started as follows:

$3A_1 -2A_2 = 3\times \begin{bmatrix} 2 \\ 1 \end{bmatrix}-2\times\begin{bmatrix} 0 \\ 1 \end{bmatrix}$

$=\begin{bmatrix} 3.2 \\ 3.1 \end{bmatrix}+\begin{bmatrix} -2.0 \\ -2.1 \end{bmatrix}$

$=\begin{bmatrix} 6 \\ 3 \end{bmatrix}+\begin{bmatrix} 0 \\ -2 \end{bmatrix}$

$=\begin{bmatrix} 6 \\ 1 \end{bmatrix}$