# Chapter 4 - Vector Spaces - Review Exercises - Page 221: 35

$S$ is not a basis for $P_2$.

#### Work Step by Step

Let $S$ be given by $$S=\left\{\left[\begin{array}{ll}{1} & {0} \\ {2} & {3}\end{array}\right],\left[\begin{array}{cc}{-2} & {1} \\ {-1} & {0}\end{array}\right],\left[\begin{array}{ll}{3} & {4} \\ {2} & {3}\end{array}\right],\left[\begin{array}{rr}{-3} & {-3} \\ {1} & {3}\end{array}\right]\right\}.$$ Consider the combination $$a \left[\begin{array}{ll}{1} & {0} \\ {2} & {3}\end{array}\right]+b\left[\begin{array}{cc}{-2} & {1} \\ {-1} & {0}\end{array}\right]+c\left[\begin{array}{ll}{3} & {4} \\ {2} & {3}\end{array}\right]+d \left[\begin{array}{rr}{-3} & {-3} \\ {1} & {3}\end{array}\right]=0, \quad a,b,cd,\in R.$$ Which yields the following system of equations \begin{align*} a-2b+3c-3d&=0\\ b+4c-3d&=0\\ 2a-b+2c+d&=0\\ 3a+3c+3d&=0. \end{align*} The coefficient matrix of the above system is given by $$\left[ \begin {array}{cccc} 1&-2&3&-3\\ 0&1&4&-3 \\ 2&-1&2&1\\ 3&0&3&3\end {array} \right].$$ One can see that the determinant of the coefficient matrix is zero, hence the system has non zero solutions and hence, $S$ is not linearly independent set of vectors. Then, $S$ is not a basis for $P_2$.

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