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How do we find a co-ordinate vector in Algebra?

For example, given:

$$ \begin{align*} \left(\begin{matrix} 2 & -3 \\ 0 & -4 \end{matrix}\right) & \left(\begin{matrix} v_1 \\ v_2 \end{matrix}\right) = 0 \end{align*} $$

How do we calculate $ v_1 $ and $ v_2 $?

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Do you know how to multiply matrices? Do you know what it means to say two matrices are equal? Furthermore do you understand that the RHS is actually short for $\begin{pmatrix} 0\\0\end{pmatrix}$? –  Git Gud Apr 22 '13 at 19:05

2 Answers 2

up vote 5 down vote accepted

$$ \begin{align*} \left(\begin{matrix} 2 & -3 \\ 0 & -4 \end{matrix}\right) & \left(\begin{matrix} v_1 \\ v_2 \end{matrix}\right) = 0 \end{align*}\Rightarrow $$ $$\begin{align*} \left( \begin{matrix} 2v_1-3v_2 \\0v_1-4v_2 \end{matrix} \right)=\left(\begin{matrix} 0 \\ 0 \end{matrix}\right) \end{align*}\Rightarrow$$ $$\begin{cases}2v_1-3v_2=0\\-4v_2=0\end{cases}\Rightarrow$$ $$\begin{cases}v_1=0\\v_2=0\end{cases}$$

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The hard way

Whenever you have a linear system of the form: $$A\vec x = \vec b$$ where $A$ is a square matrix and $\vec x$ and $\vec b$ are column vectors, you can solve this by finding $A^{-1}$: $$\vec x = A^{-1}\vec b$$ To find $A^{-1}$, we augment $A$ with the identity, and row reduce.

So, for your example: $$A = \left(\begin{matrix} 2 & -3 \\ 0 & -4 \end{matrix}\right)$$ $$\vec{b} = \vec 0$$ $$\vec{x} = \binom{v_1}{v_2}$$ So, to find $A^{-1}$ $$\left(\begin{array}{cc|cc} 2 & -3 & 1 & 0\\ 0 & -4 & 0 & 1\end{array}\right) \implies \left(\begin{array}{cc|cc} 2 & -3 & 1 & 0\\ 0 & 1 & 0 & \frac{-1}{4}\end{array}\right) \implies\left(\begin{array}{cc|cc} 1 & 0 & \frac{1}{2} & \frac{-3}{8}\\ 0 & 1 & 0 & \frac{-1}{4}\end{array}\right)$$ So, $A^{-1} = \pmatrix{\frac{1}{2} & \frac{-3}{8}\\ 0 & \frac{-1}{4}}$.

Now, we simply plug into our formula: $$\pmatrix{v_1 \\v_2} = \pmatrix{\frac{1}{2} & \frac{-3}{8}\\ 0 & \frac{-1}{4}}\pmatrix{0 \\ 0}$$

Multiplying matrices, we have: $$\pmatrix{v_1 \\v_2} = \pmatrix{0 \\ 0}$$

The easy way

We can also note that this is a homogeneous linear system with linearly independent rows. We can conclude, therefore, that the only solution of this system is the trivial one; that is, $v_1 = 0, v_2 = 0$.

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