# Chapter 4 - Vector Spaces - 4.4 Spanning Sets and Linear Independence - 4.4 Exercises - Page 178: 8

$$\left[ \begin {array}{cc} 0&0\\ 0&0 \end {array} \right]=(0)\left[ \begin {array}{cc} 2&-3\\ 4&1 \end {array} \right]+(0)\left[ \begin {array}{cc} 0&5\\ 1&-2 \end {array} \right].$$

#### Work Step by Step

Assume the combination $$\left[ \begin {array}{cc} 0&0\\ 0&0 \end {array} \right]=a\left[ \begin {array}{cc} 2&-3\\ 4&1 \end {array} \right]+b\left[ \begin {array}{cc} 0&5\\ 1&-2 \end {array} \right].$$ We have the system \begin{align*} 2a&=0\\ -3a+5b&=0\\ 4a+b&=0\\ a-2b&=0. \end{align*} We get the solution $$a=0, \quad b=0.$$ Consequently, $$\left[ \begin {array}{cc} 0&0\\ 0&0 \end {array} \right]=(0)\left[ \begin {array}{cc} 2&-3\\ 4&1 \end {array} \right]+(0)\left[ \begin {array}{cc} 0&5\\ 1&-2 \end {array} \right].$$

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