## Elementary Linear Algebra 7th Edition

Published by Cengage Learning

# Chapter 4 - Vector Spaces - Review Exercises - Page 221: 36

#### Answer

$S$ is a basis for $M_{2,2}$.

#### Work Step by Step

Let $S$ be given by $$S=\left\{\left[\begin{array}{cc}{1} & {0} \\ {0} & {1}\end{array}\right],\left[\begin{array}{cc}{-1} & {0} \\ {1} & {1}\end{array}\right],\left[\begin{array}{cc}{2} & {1} \\ {1} & {0}\end{array}\right],\left[\begin{array}{cc}{1} & {1} \\ {0} & {1}\end{array}\right]\right\}.$$ Consider the combination $$a \left[\begin{array}{cc}{1} & {0} \\ {0} & {1}\end{array}\right]+b\left[\begin{array}{cc}{-1} & {0} \\ {1} & {1}\end{array}\right]+c\left[\begin{array}{cc}{2} & {1} \\ {1} & {0}\end{array}\right]+d \left[\begin{array}{cc}{1} & {1} \\ {0} & {1}\end{array}\right]=0, \quad a,b,cd,\in R.$$ Which yields the following system of equations \begin{align*} a-b+2c+d&=0\\ c+d&=0\\ b+c&=0\\ a+b+d&=0. \end{align*} The coefficient matrix of the above system is given by $$\left[ \begin {array}{cccc} 1&-1&2&1\\ 0&0&1&1 \\ 0&1&1&0\\ 1&1&0&1\end {array} \right] .$$ One can see that the determinant of the coefficient matrix is non zero, hence the system has only the trivial solution and hence, $S$ is linearly independent set of vectors. Since, $M_{2,2}2$ has dimension $4$, then by Theorem 4.12 $S$ is a basis for $M_{2,2}$.

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