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The three vectors are linearly dependent if there exists a nontrivial solution to \begin{align*} c_{1} \mathbf{x}^{(1)}+c_{2} \mathbf{x}^{(2)}+c_{3} \mathbf{x}^{(3)}=\mathbf{0} \end{align*} for \(c_{1}, c_{2}\), and \(c_{3}\). Rewrite this equation.

\begin{equation*} \begin{array}{l} c_{1}\left(\begin{array}{l} 2 \\ 1 \\ \end{array}\right)+c_{2}\left(\begin{array}{l} 0 \\ 1 \\ \end{array}\right)+c_{3}\left(\begin{array}{r} -1 \\ 2 \\ \end{array}\right)=\left(\begin{array}{l} 0 \\ 0 \\ \end{array}\right) \\\\ \left(\begin{array}{llr} 2 & 0 & -1 \\ 1 & 1 & 2 \\ 0 & 0 & 0 \end{array}\right)\left(\begin{array}{l} c_{1} \\ c_{2} \\ c_{3} \end{array}\right)=\left(\begin{array}{l} 0 \\ 0 \\ \end{array}\right) \end{array} \end{equation*}

Calculate the determinant of the coefficient matrix. \begin{align*} \operatorname{det}\left(\begin{array}{rrr} 2 & 0 & -1 \\ 1 & 1 & 2 \\ 0 & 0 & 0 \end{array}\right)=0\left(\begin{array}{rr} 0 & -1 \\ 1 & 2 \end{array}\right)-0\left(\begin{array}{rr} 2 & -1 \\ 1 & 2 \end{array}\right)+0\left(\begin{array}{ll} 2 & 0 \\ 1 & 1 \end{array}\right)=0 \end{align*}

Since it’s zero, there are infinitely many solutions for \(c_{1}, c_{2}\), and \(c_{3}\).

\begin{align*} \begin{array}{r} 2 c_{1}-c_{3}=0 \\ c_{1}+c_{2}+2 c_{3}=0 \end{array} \end{align*}

Solve this first equation for \(c_{3}\) \begin{align*} c_{3}=2 c_{1} \end{align*}

and plug it into the second one.

\begin{align*} c_{1}+c_{2}+2\left(2 c_{1}\right)=0 \end{align*}

Solve for \(c_{2}\) \begin{align*} c_{2}=-5 c_{1} \end{align*} In terms of the free variable \(c_{1}\), the solution to the system of equations is \begin{align*} \left\{c_{1},-5 c_{1}, 2 c_{1}\right\} \end{align*} For example, choose \(c_{1}=1\). Then \begin{align*} \mathbf{x}^{(1)}-5 \mathbf{x}^{(2)}+2 \mathbf{x}^{(3)}=\mathbf{0} \end{align*} Therefore, the three given vectors are linearly dependent.

The aim is to solve the eigenvalue problem, \begin{align*} \mathbf{A} \mathbf{x}=\lambda \mathbf{x} \end{align*} where \(\mathbf{A}\) is the given matrix. Bring \(\lambda \mathbf{x}\) to the left side and combine the terms. \begin{align*} (\mathbf{A}-\lambda \mathbf{I}) \mathbf{x}=\mathbf{0} \end{align*}

The eigenvalues satisfy

\begin{align*} \operatorname{det}(\mathbf{A}-\lambda \mathbf{I})=0 \end{align*}

Evaluate the determinant and solve for \(\lambda\).

\begin{align*} \begin{array}{c} \operatorname{det}\left(\begin{array}{cc} -2-\lambda & 1 \\ 1 & -2-\lambda \end{array}\right)=0 \\\\ (-2-\lambda)(-2-\lambda)-1=0 \\ \lambda^{2}+4 \lambda+3=0 \\ (\lambda+3)(\lambda+1)=0 \\ \lambda=\{-3,-1\} \end{array} \end{align*}

Therefore, the eigenvalues are \begin{equation*} \lambda_{1}=-3 \text { and } \quad \lambda_{2}=-1 \end{equation*} Substitute \(\lambda_{1}\) and \(\lambda_{2}\) back into equation (1) to determine the corresponding eigenvectors, \(\mathbf{x}_{1}\) and \(\mathbf{x}_{2}\) \begin{equation*} \begin{array}{r} \left(\mathbf{A}-\lambda_{1} \mathbf{I}\right) \mathbf{x}_{1}=\mathbf{0} & \left(\mathbf{A}-\lambda_{2} \mathbf{I}\right) \mathbf{x}_{2}=\mathbf{0} \\ \left(\begin{array}{ll} 1 & 1 \\ 1 & 1 \end{array}\right)\left(\begin{array}{l} x_{1} \\ x_{2} \end{array}\right)\left(\begin{array}{l} 0 \\ \end{array}\right) & \left(\begin{array}{cc} -1 & 1 \\ 1 & -1 \end{array}\right)\left(\begin{array}{l} x_{1} \\ x_{2} \end{array}\right)=\left(\begin{array}{l} 0 \\ \end{array}\right) \\ x_{1}+x_{2}=0 & -x_{1}+x_{2}=0 \\ \left.x_{1}+x_{2}=0\right\} & x_{1}-x_{2}=0 \\ x_{2}-x_{1} & x_{2}=x_{1} \\ \mathbf{x}_{1}=\left(\begin{array}{c} x_{1} \\ -x_{1} \end{array}\right)=x_{1}\left(\begin{array}{c} 1 \\ -1 \end{array}\right) & \mathbf{x}_{2}=\left(\begin{array}{l} x_{1} \\ x_{1} \end{array}\right)=x_{1}\left(\begin{array}{l} 1 \\ 1 \end{array}\right) \end{array} \end{equation*} Note that \(x_{1}\) is a free variable, or arbitrary constant.

The aim is to solve the eigenvalue problem, \begin{align*} \mathbf{A x}=\lambda \mathbf{x} \end{align*} where \(\mathbf{A}\) is the given matrix. Bring \(\lambda \mathbf{x}\) to the left side and combine the terms. \begin{align*} (\mathbf{A}-\lambda \mathbf{I}) \mathbf{x}=\mathbf{0} \end{align*} The eigenvalues satisfy \begin{align*} \operatorname{det}(\mathbf{A}-\lambda \mathbf{I})0 \end{align*} Evaluate the determinant and solve for \(\lambda\). \begin{equation*} \begin{array}{c} \operatorname{det}\left(\begin{array}{cc} 1-\lambda & \sqrt{3} \\ \sqrt{3} & -1-\lambda \end{array}\right)=0 \\\\ (1-\lambda)(-1-\lambda)-3=0 \\ \lambda^{2}-4=0 \\ (\lambda+2)(\lambda-2)=0 \\ \lambda=\{-2,2\} \end{array} \end{equation*} Therefore, the eigenvalues are \begin{equation*} \lambda_{1}-2 \text { and } \quad \lambda_{2}=2 \end{equation*} Substitute \(\lambda_{1}\) and \(\lambda_{2}\) back into equation (1) to determine the corresponding eigenvectors, \(\mathbf{x}_{1}\) and \(\mathbf{x}_{2}\)

\begin{equation*} \begin{array}{c} \left(\mathbf{A}-\lambda_{1} \mathbf{I}\right) \mathbf{x}_{1}=\mathbf{0} \\\\ \left(\begin{array}{cc} 3 & \sqrt{3} \\ \sqrt{3} & 1 \end{array}\right)\left(\begin{array}{l} x_{1} \\ x_{2} \end{array}\right)=\left(\begin{array}{l} 0 \\ \end{array}\right) \end{array} \end{equation*}

\begin{equation*} \left.\begin{array}{r} 3 x_{1}+\sqrt{3} x_{2}=0 \\ \sqrt{3} x_{1}+x_{2}=0 \end{array}\right\} \end{equation*}

\begin{equation*} \begin{array}{r} x_{2}=-\sqrt{3} x_{1} \\\\ \mathbf{x}_{1}=\left(\begin{array}{c} x_{1} \\ -\sqrt{3} x_{1} \end{array}\right)=x_{1}\left(\begin{array}{c} 1 \\ -\sqrt{3} \end{array}\right) \end{array} \end{equation*}

and

\begin{equation*} \begin{array}{c} \left(\mathbf{A}-\lambda_{2} \mathbf{I}\right) \mathbf{x}_{2}=\mathbf{0} \\\\ \left(\begin{array}{ll} -1 & \sqrt{3} \\ \sqrt{3} & -3 \end{array}\right)\left(\begin{array}{l} x_{1} \\ x_{2} \end{array}\right)=\left(\begin{array}{l} 0 \\ \end{array}\right) \end{array} \end{equation*}

\begin{equation*} \left.\begin{array}{r} -x_{1}+\sqrt{3} x_{2}=0 \\ \sqrt{3} x_{1}-3 x_{2}=0 \end{array}\right\} \end{equation*}

\begin{equation*} \begin{array}{c} x_{2}=\frac{1}{\sqrt{3}} x_{1} \\\\ \mathbf{x}_{2}=\left(\begin{array}{c} x_{1} \\ \frac{1}{\sqrt{3}} x_{1} \end{array}\right)=x_{1}\left(\begin{array}{c} 1 \\ \frac{1}{\sqrt{3}} \end{array}\right) \end{array} \end{equation*}

Note that \(x_{1}\) is a free variable, or arbitrary constant.

The aim is to solve the eigenvalue problem,

\begin{align*} \mathbf{A} \mathbf{x}=\lambda \mathbf{x} \end{align*}

where \(\mathbf{A}\) is the given matrix. Bring \(\lambda \mathbf{x}\) to the left side and combine the terms.

\begin{align*} (\mathbf{A}-\lambda \mathbf{I}) \mathbf{x}=\mathbf{0} \end{align*}

The eigenvalues satisfy

\begin{align*} \operatorname{det}(\mathbf{A}-\lambda \mathbf{I})=0 \end{align*}

Evaluate the determinant and solve for \(\lambda\).

\begin{align*} \begin{array}{c} \operatorname{det}\left(\begin{array}{ccc} 1-\lambda & 0 & 0 \\ 2 & 1-\lambda & -2 \\ 3 & 2 & 1-\lambda \end{array}\right)=0 \\\\ (1-\lambda)\left|\begin{array}{cc} 1-\lambda & -2 \\ 2 & 1-\lambda \end{array}\right|=0 \\ (1-\lambda)[(1-\lambda)(1-\lambda)+4]=0 \\ 1-\lambda=0 \quad \text { or } \quad \lambda^{2}-2 \lambda+5=0 \\ \quad 5-7 \lambda+3 \lambda^{2}-\lambda^{3}=0 \\ (1-\lambda)\left(\lambda^{2}-2 \lambda+5\right)=0 \\ \lambda=1 \quad \text { or } \quad \lambda=\frac{2 \pm \sqrt{4-20}}{2}=1 \pm 2 i \end{array} \end{align*}

Therefore, the eigenvalues are

\begin{align*} \begin{array}{|l|l|l|l|} \hline \lambda_{1}=1 & \lambda_{2}=1-2 i & \lambda_{3}=1+2 i \end{array} \end{align*}

Substitute \(\lambda_{1}\) back into equation (1) to determine the corresponding eigenvector \(\mathbf{x}_{1}\).

\begin{align*} \begin{array}{c} \left(\mathbf{A}-\lambda_{1} \mathbf{I}\right) \mathbf{x}_{1}=\mathbf{0}\\\\ \left(\begin{array}{ccc} 1-(1) & 0 & 0 \\ 2 & 1-(1) & -2 \\ 3 & 2 & 1-(1) \end{array}\right)\left(\begin{array}{l} x_{1} \\ x_{2} \\ x_{3} \end{array}\right)=\left(\begin{array}{c} 0 \\ 0 \\ \end{array}\right) \\\\ \left(\begin{array}{ccc} 0 & 0 & 0 \\ 2 & 0 & -2 \\ 3 & 2 & 0 \end{array}\right)\left(\begin{array}{l} x_{1} \\ x_{2} \\ x_{3} \end{array}\right)=\left(\begin{array}{l} 0 \\ 0 \\ \end{array}\right) \end{array} \end{align*}

Write the implied system of equations.

\begin{align*} \left.\begin{array}{l} 2 x_{1}-2 x_{3}=0 \\ 3 x_{1}+2 x_{2}=0 \end{array}\right\} \end{align*}

Solve for \(x_{2}\) and \(x_{3}\) in terms of the free variable \(x_{1}\).

\begin{align*} \begin{array}{l} x_{3}=x_{1} \\ x_{2}=-\frac{3}{2} x_{1} \end{array} \end{align*}

This means

\begin{equation*} \mathbf{x}_{1}=\left(\begin{array}{l} x_{1} \\ x_{2} \\ x_{3} \end{array}\right)=\left(\begin{array}{c} x_{1} \\ -\frac{3}{2} x_{1} \\ x_{1} \end{array}\right)=x_{1}\left(\begin{array}{c} 1 \\ -\frac{3}{2} \\ 1 \end{array}\right) \end{equation*}

Since \(x_1\) is arbitrary, the eigenvector can be multiplied by 2 to get rid of the fraction.

\begin{equation*} \mathbf{x}_{1}=x_{1}^{\prime}\left(\begin{array}{c} 2 \\ -3 \\ 2 \end{array}\right) \end{equation*}

The aim is to solve the eigenvalue problem, \begin{align*} \mathbf{A} \mathbf{x}=\lambda \mathbf{x} \end{align*} where \(\mathbf{A}\) is the given matrix. Bring \(\lambda \mathbf{x}\) to the left side and combine the terms. \begin{align*} (\mathbf{A}-\lambda \mathbf{I}) \mathbf{x}=\mathbf{0} \end{align*} The eigenvalues satisfy \begin{align*} \operatorname{det}(\mathbf{A}-\lambda \mathbf{I})=0 \end{align*} Evaluate the determinant and solve for \(\lambda\).

Therefore, the eigenvalues are

\begin{equation*} \begin{array}{|l|l|l|} \hline \lambda_{1}=1 & \lambda_{2}=2 & \lambda_{3}=-1 \end{array} \end{equation*}

Substitute \(\lambda_{1}\) back into equation (1) to determine the corresponding eigenvector \(\mathbf{x}_{1}\).

\begin{equation*} \begin{array}{c} \left(\mathbf{A}-\lambda_{1} \mathbf{I}\right) \mathbf{x}_{1}=\mathbf{0} \\\\ \left(\begin{array}{ccc} 11 / 9-(1) & -2 / 9 & 8 / 9 \\ -2 / 9 & 2 / 9-(1) & 10 / 9 \\ 8 / 9 & 10 / 9 & 5 / 9-(1) \end{array}\right)\left(\begin{array}{l} x_{1} \\ x_{2} \\ x_{3} \end{array}\right)=\left(\begin{array}{c} 0 \\ 0 \\ \end{array}\right) \\ \\ \left(\begin{array}{ccc} 2 / 9 & -2 / 9 & 8 / 9 \\ -2 / 9 & -7 / 9 & 10 / 9 \\ 8 / 9 & 10 / 9 & -4 / 9 \end{array}\right)\left(\begin{array}{l} x_{1} \\ x_{2} \\ x_{3} \end{array}\right)=\left(\begin{array}{c} 0 \\ 0 \\ \end{array}\right) \end{array} \end{equation*}

Write the augmented matrix. \begin{align*} \left(\begin{array}{rrr|r} 2 / 9 & -2 / 9 & 8 / 9 & 0 \\ -2 / 9 & -7 / 9 & 10 / 9 & 0 \\ 8 / 9 & 10 / 9 & -4 / 9 & 0 \end{array}\right) \end{align*} Multiply each row by \(9\) \begin{align*} \left(\begin{array}{rrr|r} 2 & -2 & 8 & 0 \\ -2 & -7 & 10 & 0 \\ 8 & 10 & -4 & 0 \end{array}\right) \end{align*} Multiply the first row by \(-4\) and add it to the third row. \begin{align*} \left(\begin{array}{rrr|r} 2 & -2 & 8 & 0 \\ -2 & -7 & 10 & 0 \\ 0 & 18 & -36 & 0 \end{array}\right) \end{align*} Add the first row to the second row. \begin{align*} \left(\begin{array}{rrr|r} 2 & -2 & 8 & 0 \\ 0 & -9 & 18 & 0 \\ 0 & 18 & -36 & 0 \end{array}\right) \end{align*}

Write the implied system of equations and solve for \(x_{1}\) and \(x_{2}\) in terms of the free variable \(x_{3}\) \begin{align*} \left.\begin{array}{r} 2 x_{1}-2 x_{2}+8 x_{3}=0 \\ -9 x_{2}+18 x_{3}=0 \\ 18 x_{2}-36 x_{3}=0 \end{array}\right\} \quad \rightarrow \quad \begin{array}{l} x_{1}=-2 x_{3} \\ x_{2}=2 x_{3} \end{array} \end{align*} This means \begin{align*} \mathbf{x}_{1}=\left(\begin{array}{l} x_{1} \\ x_{2} \\ x_{3} \end{array}\right)=\left(\begin{array}{c} -2 x_{3} \\ 2 x_{3} \\ x_{3} \end{array}\right) \end{align*} Therefore, \begin{align*} \mathbf{x}_{1}=x_{3}\left(\begin{array}{c} -2 \\ 2 \\ 1 \end{array}\right) \end{align*} Substitute \(\lambda_{2}\) back into equation (1) to determine the corresponding eigenvector \(\mathbf{x}_{2}\).

\begin{equation*} \begin{array}{c} \left(\mathbf{A}-\lambda_{2} \mathbf{I}\right) \mathbf{x}_{2}=\mathbf{0}\\\\ \left(\begin{array}{ccc} 11 / 9-(2) & -2 / 9 & 8 / 9 \\ -2 / 9 & 2 / 9-(2) & 10 / 9 \\ 8 / 9 & 10 / 9 & 5 / 9-(2) \end{array}\right)\left(\begin{array}{l} x_{1} \\ x_{2} \\ x_{3} \end{array}\right)=\left(\begin{array}{c} 0 \\ 0 \\ \end{array}\right) \\\\ \left(\begin{array}{ccc} -7 / 9 & -2 / 9 & 8 / 9 \\ -2 / 9 & -16 / 9 & 10 / 9 \\ 8 / 9 & 10 / 9 & -13 / 9 \end{array}\right)\left(\begin{array}{l} x_{1} \\ x_{2} \\ x_{3} \end{array}\right)=\left(\begin{array}{l} 0 \\ 0 \\ \end{array}\right) \end{array} \end{equation*}

Write the augmented matrix. \begin{align*} \left(\begin{array}{rrr|r} -7 / 9 & -2 / 9 & 8 / 9 & 0 \\ -2 / 9 & -16 / 9 & 10 / 9 & 0 \\ 8 / 9 & 10 / 9 & -13 / 9 & 0 \end{array}\right) \end{align*} Multiply each row by \(9\) \begin{align*} \left(\begin{array}{rrr|r} -7 & -2 & 8 & 0 \\ -2 & -16 & 10 & 0 \\ 8 & 10 & -13 & 0 \end{array}\right) \end{align*} Multiply the second row by 4 and add it to the third row. \begin{align*} \left(\begin{array}{rrr|r} -7 & -2 & 8 & 0 \\ -2 & -16 & 10 & 0 \\ 0 & -54 & 27 & 0 \end{array}\right) \end{align*} Multiply the first row by \(-8\) and add it to the second row. \begin{align*} \left(\begin{array}{rrr|r} -7 & -2 & 8 & 0 \\ 54 & 0 & -54 & 0 \\ 0 & -54 & 27 & 0 \end{array}\right) \end{align*}

Write the implied system of equations and solve for \(x_1\) and \(x_2\) in terms of the free variable \(x_3\)

\begin{equation*} \left.\begin{array}{r} -7 x_{1}-2 x_{2}+8 x_{3}=0 \\ 54 x_{1}-54 x_{3}=0 \\ -54 x_{2}+27 x_{3}=0 \end{array}\right\} \quad \rightarrow \quad \begin{aligned} & x_{1}=x_{3} \\ & x_{2}=\frac{1}{2} x_{3} \end{aligned} \end{equation*}

This means

\begin{equation*} \mathbf{x}_{2}=\left(\begin{array}{l} x_{1} \\ x_{2} \\ x_{3} \end{array}\right)=\left(\begin{array}{c} x_{3} \\ \frac{1}{2} x_{3} \\ x_{3} \end{array}\right)=x_{3}\left(\begin{array}{c} 1 \\ \frac{1}{2} \\ 1 \end{array}\right) \end{equation*}

Since \(x_{3}\) is arbitrary, the eigenvector can be multiplied by 2 to get rid of the fraction. \(\mathbf{x}_{2}=x_{3}^{\prime}\left(\begin{array}{l}2 \\ 1 \\ 2\end{array}\right)\)

Substitute \(\lambda_{3}\) back into equation (1) to determine the corresponding eigenvector \(\mathbf{x}_{3}\).

\begin{equation*} \begin{array}{c} \left(\mathbf{A}-\lambda_{3} \mathbf{I}\right) \mathbf{x}_{3}=\mathbf{0}\\\\ \left(\begin{array}{ccc} 11 / 9-(-1) & -2 / 9 & 8 / 9 \\ -2 / 9 & 2 / 9-(-1) & 10 / 9 \\ 8 / 9 & 10 / 9 & 5 / 9-(-1) \end{array}\right)\left(\begin{array}{l} x_{1} \\ x_{2} \\ x_{3} \end{array}\right)=\left(\begin{array}{l} 0 \\ 0 \\ \end{array}\right) \\\\ \left(\begin{array}{ccc} 20 / 9 & -2 / 9 & 8 / 9 \\ -2 / 9 & 11 / 9 & 10 / 9 \\ 8 / 9 & 10 / 9 & 14 / 9 \end{array}\right)\left(\begin{array}{l} x_{1} \\ x_{2} \\ x_{3} \end{array}\right)=\left(\begin{array}{c} 0 \\ 0 \\ \end{array}\right) \end{array} \end{equation*}

Write the augmented matrix. \begin{align*} \left(\begin{array}{rrr|r} 20 / 9 & -2 / 9 & 8 / 9 & 0 \\ -2 / 9 & 11 / 9 & 10 / 9 & 0 \\ 8 / 9 & 10 / 9 & 14 / 9 & 0 \end{array}\right) \end{align*} Multiply each row by \(9\) \begin{align*} \left(\begin{array}{rrr|r} 20 & -2 & 8 & 0 \\ -2 & 11 & 10 & 0 \\ 8 & 10 & 14 & 0 \end{array}\right) \end{align*} Multiply the second row by 4 and add it to the third row. \begin{align*} \left(\begin{array}{rrr|r} 20 & -2 & 8 & 0 \\ -2 & 11 & 10 & 0 \\ 0 & 54 & 54 & 0 \end{array}\right) \end{align*} Multiply the second row by 10 and add it to the first row. \begin{align*} \left(\begin{array}{rrr|r} 0 & 108 & 108 & 0 \\ -2 & 11 & 10 & 0 \\ 0 & 54 & 54 & 0 \end{array}\right) \end{align*}

Write the implied system of equations and solve for \(x_{1}\) and \(x_{3}\) in terms of the free variable \(x_{2}\) \begin{align*} \left.\begin{array}{r} 108 x_{2}+108 x_{3}=0 \\ -2 x_{1}+11 x_{2}+10 x_{3}=0 \\ 54 x_{2}+54 x_{3}=0 \end{array}\right\} \quad \rightarrow \quad \begin{array}{l} x_{1}=\frac{1}{2} x_{2} \\ x_{3}=-x_{2} \end{array} \end{align*} This means \begin{align*} \mathbf{x}_{3}=\left(\begin{array}{l} x_{1} \\ x_{2} \\ x_{3} \end{array}\right)=\left(\begin{array}{c} \frac{1}{2} x_{2} \\ x_{2} \\ -x_{2} \end{array}\right)=x_{2}\left(\begin{array}{c} \frac{1}{2} \\ 1 \\ -1 \end{array}\right) \end{align*}

Since \(x_{2}\) is arbitrary, the eigenvector can be multiplied by 2 to get rid of the fraction. \begin{align*} \mathbf{x}_{3}=x_{2}^{\prime}\left(\begin{array}{c} 1 \\ 2 \\ -2 \end{array}\right) \end{align*}

\begin{align*} t \begin{pmatrix} 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 2 & -1 \\ 3 & -2 \end{pmatrix} \begin{pmatrix} t \\ t \end{pmatrix} = \begin{pmatrix} 2t-t \\ 3t-2t \end{pmatrix} = \begin{pmatrix} t \\ t \end{pmatrix} \end{align*}

\begin{align*} t \begin{pmatrix} -t^{-2} \\ -3t^{-2} \end{pmatrix} = \begin{pmatrix} 2 & -1 \\ 3 & -2 \end{pmatrix} \begin{pmatrix} t^{-1} \\ 3t^{-1} \end{pmatrix} = \begin{pmatrix} 2t^{-1}-3t^{-1} \\ 3t^{-1}-6t^{-1} \end{pmatrix} = \begin{pmatrix} -t^{-1} \\ -3t^{-1} \end{pmatrix} \end{align*}

\begin{align*} t \begin{pmatrix} -t^{-2} \\ -2t^{-2} \end{pmatrix} = \begin{pmatrix} 3 & -2 \\ 2 & -2 \end{pmatrix} \begin{pmatrix} t^{-1} \\ 2t^{-1} \end{pmatrix} = \begin{pmatrix} 3t^{-1}-4t^{-1} \\ 2t^{-1}-4t^{-1} \end{pmatrix} = \begin{pmatrix} -t^{-1} \\ -2t^{-1} \end{pmatrix} \end{align*}

\begin{align*} t \begin{pmatrix} 4t \\ 2t \end{pmatrix} = \begin{pmatrix} 3 & -2 \\ 2 & -2 \end{pmatrix} \begin{pmatrix} 2t^2 \\ t^2 \end{pmatrix} = \begin{pmatrix} 6t^2 - 2t^2\\ 4t^2 - 2t^2 \end{pmatrix} = \begin{pmatrix} 4t^2\\ 2t^2 \end{pmatrix} \end{align*}