Algebra Examples

$[\begin{array}{c} 4 & 2 \\ 3 & 1 \end{array}]$

Step 1

Assign the matrix the name $A$ to simplify the descriptions throughout the problem.

$A = [\begin{array}{c} 4 & 2 \\ 3 & 1 \end{array}]$

Step 2

Set up the formula to find the characteristic equation $p (λ)$ .

$p (λ) = determinant (A + (- λ I_{2}))$

Step 3

Substitute the known values in the formula.

$p (λ) = determinant ([\begin{array}{c} 4 & 2 \\ 3 & 1 \end{array}] - λ [\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}])$

Step 4

Subtract the eigenvalue times the identity matrix from the original matrix.

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Multiply $- λ$ by each element of the matrix.

$p (λ) = [\begin{array}{c} 4 & 2 \\ 3 & 1 \end{array}] + [\begin{array}{c} - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 \end{array}]$

Simplify each element in the matrix.

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Rearrange $- λ \cdot 1$ .

$p (λ) = [\begin{array}{c} 4 & 2 \\ 3 & 1 \end{array}] + [\begin{array}{c} - λ & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 \end{array}]$

Rearrange $- λ \cdot 0$ .

$p (λ) = [\begin{array}{c} 4 & 2 \\ 3 & 1 \end{array}] + [\begin{array}{c} - λ & 0 \\ - λ \cdot 0 & - λ \cdot 1 \end{array}]$

Rearrange $- λ \cdot 0$ .

$p (λ) = [\begin{array}{c} 4 & 2 \\ 3 & 1 \end{array}] + [\begin{array}{c} - λ & 0 \\ 0 & - λ \cdot 1 \end{array}]$

Rearrange $- λ \cdot 1$ .

$p (λ) = [\begin{array}{c} 4 & 2 \\ 3 & 1 \end{array}] + [\begin{array}{c} - λ & 0 \\ 0 & - λ \end{array}]$

Add the corresponding elements.

$p (λ) = [\begin{array}{c} 4 - λ & 2 + 0 \\ 3 + 0 & 1 - λ \end{array}]$

Simplify each element of the matrix $[\begin{array}{c} 4 - λ & 2 + 0 \\ 3 + 0 & 1 - λ \end{array}]$ .

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Simplify $2 + 0$ .

$p (λ) = [\begin{array}{c} 4 - λ & 2 \\ 3 + 0 & 1 - λ \end{array}]$

Simplify $3 + 0$ .

$p (λ) = [\begin{array}{c} 4 - λ & 2 \\ 3 & 1 - λ \end{array}]$

Step 5

Find the determinant of $[\begin{array}{c} 4 - λ & 2 \\ 3 & 1 - λ \end{array}]$ .

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These are both valid notations for the determinant of a matrix.

$determinant [\begin{array}{c} 4 - λ & 2 \\ 3 & 1 - λ \end{array}] = | \begin{array}{c} 4 - λ & 2 \\ 3 & 1 - λ \end{array} |$

The determinant of a $2 \times 2$ matrix can be found using the formula $| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$p (λ) = (4 - λ) (1 - λ) - 3 \cdot 2$

Simplify the determinant.

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Simplify each term.

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Expand $(4 - λ) (1 - λ)$ using the FOIL Method.

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Apply the distributive property.

$p (λ) = 4 (1 - λ) - λ (1 - λ) - 3 \cdot 2$

Apply the distributive property.

$p (λ) = 4 \cdot 1 + 4 (- λ) - λ (1 - λ) - 3 \cdot 2$

Apply the distributive property.

$p (λ) = 4 \cdot 1 + 4 (- λ) - λ \cdot 1 - λ (- λ) - 3 \cdot 2$

Simplify and combine like terms.

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Simplify each term.

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Multiply $4$ by $1$ .

$p (λ) = 4 + 4 (- λ) - λ \cdot 1 - λ (- λ) - 3 \cdot 2$

Multiply $- 1$ by $4$ .

$p (λ) = 4 - 4 λ - λ \cdot 1 - λ (- λ) - 3 \cdot 2$

Multiply $- 1$ by $1$ .

$p (λ) = 4 - 4 λ - λ - λ (- λ) - 3 \cdot 2$

Rewrite using the commutative property of multiplication.

$p (λ) = 4 - 4 λ - λ - 1 \cdot (- 1 λ \cdot λ) - 3 \cdot 2$

Multiply $λ$ by $λ$ by adding the exponents.

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Move $λ$ .

$p (λ) = 4 - 4 λ - λ - 1 \cdot (- 1 (λ \cdot λ)) - 3 \cdot 2$

Multiply $λ$ by $λ$ .

$p (λ) = 4 - 4 λ - λ - 1 \cdot (- 1 λ^{2}) - 3 \cdot 2$

Multiply $- 1$ by $- 1$ .

$p (λ) = 4 - 4 λ - λ + 1 λ^{2} - 3 \cdot 2$

Multiply $λ^{2}$ by $1$ .

$p (λ) = 4 - 4 λ - λ + λ^{2} - 3 \cdot 2$

Subtract $λ$ from $- 4 λ$ .

$p (λ) = 4 - 5 λ + λ^{2} - 3 \cdot 2$

Multiply $- 3$ by $2$ .

$p (λ) = 4 - 5 λ + λ^{2} - 6$

Subtract $6$ from $4$ .

$p (λ) = - 5 λ + λ^{2} - 2$

Step 6

Reorder the polynomial.

$p (λ) = λ^{2} - 5 λ - 2$

Step 7

Set the characteristic polynomial equal to $0$ to find the eigenvalues $λ$ .

$λ^{2} - 5 λ - 2 = 0$

Step 8

Find the roots of $λ^{2} - 5 λ - 2 = 0$ by solving for $λ$ .

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Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Substitute the values $a = 1$ , $b = - 5$ , and $c = - 2$ into the quadratic formula and solve for $λ$ .

$\frac{5 \pm \sqrt{{(- 5)}^{2} - 4 \cdot (1 \cdot - 2)}}{2 \cdot 1}$

Simplify.

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Simplify the numerator.

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Raise $- 5$ to the power of $2$ .

$λ = \frac{5 \pm \sqrt{25 - 4 \cdot 1 \cdot - 2}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 2$ .

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Multiply $- 4$ by $1$ .

$λ = \frac{5 \pm \sqrt{25 - 4 \cdot - 2}}{2 \cdot 1}$

Multiply $- 4$ by $- 2$ .

$λ = \frac{5 \pm \sqrt{25 + 8}}{2 \cdot 1}$

Add $25$ and $8$ .

$λ = \frac{5 \pm \sqrt{33}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$λ = \frac{5 \pm \sqrt{33}}{2}$

Simplify the expression to solve for the $+$ portion of the $\pm$ .

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Simplify the numerator.

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Raise $- 5$ to the power of $2$ .

$λ = \frac{5 \pm \sqrt{25 - 4 \cdot 1 \cdot - 2}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 2$ .

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Multiply $- 4$ by $1$ .

$λ = \frac{5 \pm \sqrt{25 - 4 \cdot - 2}}{2 \cdot 1}$

Multiply $- 4$ by $- 2$ .

$λ = \frac{5 \pm \sqrt{25 + 8}}{2 \cdot 1}$

Add $25$ and $8$ .

$λ = \frac{5 \pm \sqrt{33}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$λ = \frac{5 \pm \sqrt{33}}{2}$

Change the $\pm$ to $+$ .

$λ = \frac{5 + \sqrt{33}}{2}$

Simplify the expression to solve for the $-$ portion of the $\pm$ .

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Simplify the numerator.

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Raise $- 5$ to the power of $2$ .

$λ = \frac{5 \pm \sqrt{25 - 4 \cdot 1 \cdot - 2}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 2$ .

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Multiply $- 4$ by $1$ .

$λ = \frac{5 \pm \sqrt{25 - 4 \cdot - 2}}{2 \cdot 1}$

Multiply $- 4$ by $- 2$ .

$λ = \frac{5 \pm \sqrt{25 + 8}}{2 \cdot 1}$

Add $25$ and $8$ .

$λ = \frac{5 \pm \sqrt{33}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$λ = \frac{5 \pm \sqrt{33}}{2}$

Change the $\pm$ to $-$ .

$λ = \frac{5 - \sqrt{33}}{2}$

The final answer is the combination of both solutions.

$λ = \frac{5 + \sqrt{33}}{2}, \frac{5 - \sqrt{33}}{2}$

Step 9

The eigenvector for $λ = \frac{5 + \sqrt{33}}{2}$ is equal to the null space of the matrix minus the eigenvalue times the identity matrix.

$ε_{A} (\frac{5 + \sqrt{33}}{2}) = N (A - λ I_{2})$

Step 10

Substitute the known values into the formula.

$[\begin{array}{c} 4 & 2 \\ 3 & 1 \end{array}] - \frac{5 + \sqrt{33}}{2 [\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}]}$

Step 11

Simplify the matrix expression.

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Multiply $- \frac{5 + \sqrt{33}}{2}$ by each element of the matrix.

$[\begin{array}{c} 4 & 2 \\ 3 & 1 \end{array}] + [\begin{array}{c} - \frac{5 + \sqrt{33}}{2} \cdot 1 & - \frac{5 + \sqrt{33}}{2} \cdot 0 \\ - \frac{5 + \sqrt{33}}{2} \cdot 0 & - \frac{5 + \sqrt{33}}{2} \cdot 1 \end{array}]$

Simplify each element in the matrix.

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Rearrange $- \frac{5 + \sqrt{33}}{2} \cdot 1$ .

$[\begin{array}{c} 4 & 2 \\ 3 & 1 \end{array}] + [\begin{array}{c} - \frac{5 + \sqrt{33}}{2} & - \frac{5 + \sqrt{33}}{2} \cdot 0 \\ - \frac{5 + \sqrt{33}}{2} \cdot 0 & - \frac{5 + \sqrt{33}}{2} \cdot 1 \end{array}]$

Rearrange $- \frac{5 + \sqrt{33}}{2} \cdot 0$ .

$[\begin{array}{c} 4 & 2 \\ 3 & 1 \end{array}] + [\begin{array}{c} - \frac{5 + \sqrt{33}}{2} & 0 \\ - \frac{5 + \sqrt{33}}{2} \cdot 0 & - \frac{5 + \sqrt{33}}{2} \cdot 1 \end{array}]$

Rearrange $- \frac{5 + \sqrt{33}}{2} \cdot 0$ .

$[\begin{array}{c} 4 & 2 \\ 3 & 1 \end{array}] + [\begin{array}{c} - \frac{5 + \sqrt{33}}{2} & 0 \\ 0 & - \frac{5 + \sqrt{33}}{2} \cdot 1 \end{array}]$

Rearrange $- \frac{5 + \sqrt{33}}{2} \cdot 1$ .

$[\begin{array}{c} 4 & 2 \\ 3 & 1 \end{array}] + [\begin{array}{c} - \frac{5 + \sqrt{33}}{2} & 0 \\ 0 & - \frac{5 + \sqrt{33}}{2} \end{array}]$

Add the corresponding elements.

$[\begin{array}{c} 4 - \frac{5 + \sqrt{33}}{2} & 2 + 0 \\ 3 + 0 & 1 - \frac{5 + \sqrt{33}}{2} \end{array}]$

Simplify each element of the matrix $[\begin{array}{c} 4 - \frac{5 + \sqrt{33}}{2} & 2 + 0 \\ 3 + 0 & 1 - \frac{5 + \sqrt{33}}{2} \end{array}]$ .

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Simplify $4 - \frac{5 + \sqrt{33}}{2}$ .

$[\begin{array}{c} \frac{3 - \sqrt{33}}{2} & 2 + 0 \\ 3 + 0 & 1 - \frac{5 + \sqrt{33}}{2} \end{array}]$

Simplify $2 + 0$ .

$[\begin{array}{c} \frac{3 - \sqrt{33}}{2} & 2 \\ 3 + 0 & 1 - \frac{5 + \sqrt{33}}{2} \end{array}]$

Simplify $3 + 0$ .

$[\begin{array}{c} \frac{3 - \sqrt{33}}{2} & 2 \\ 3 & 1 - \frac{5 + \sqrt{33}}{2} \end{array}]$

Simplify $1 - \frac{5 + \sqrt{33}}{2}$ .

$[\begin{array}{c} \frac{3 - \sqrt{33}}{2} & 2 \\ 3 & - \frac{3 + \sqrt{33}}{2} \end{array}]$

Step 12

Find the reduced row echelon form of the matrix.

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Perform the row operation $R_{1} = - \frac{3 + \sqrt{33}}{12} R_{1}$ on $R_{1}$ (row $1$ ) in order to convert some elements in the row to $1$ .

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Replace $R_{1}$ (row $1$ ) with the row operation $R_{1} = - \frac{3 + \sqrt{33}}{12} R_{1}$ in order to convert some elements in the row to the desired value $1$ .

$[\begin{array}{c} - \frac{3 + \sqrt{33}}{12} R_{1} & - \frac{3 + \sqrt{33}}{12} R_{1} \\ 3 & - \frac{3 + \sqrt{33}}{2} \end{array}]$

$R_{1} = - \frac{3 + \sqrt{33}}{12} R_{1}$

Replace $R_{1}$ (row $1$ ) with the actual values of the elements for the row operation $R_{1} = - \frac{3 + \sqrt{33}}{12} R_{1}$ .

$[\begin{array}{c} (- \frac{3 + \sqrt{33}}{12}) \cdot (\frac{3 - \sqrt{33}}{2}) & (- \frac{3 + \sqrt{33}}{12}) \cdot (2) \\ 3 & - \frac{3 + \sqrt{33}}{2} \end{array}]$

$R_{1} = - \frac{3 + \sqrt{33}}{12} R_{1}$

Simplify $R_{1}$ (row $1$ ).

$[\begin{array}{c} 1 & - \frac{3 + \sqrt{33}}{6} \\ 3 & - \frac{3 + \sqrt{33}}{2} \end{array}]$

Perform the row operation $R_{2} = - 3 \cdot R_{1} + R_{2}$ on $R_{2}$ (row $2$ ) in order to convert some elements in the row to $0$ .

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Replace $R_{2}$ (row $2$ ) with the row operation $R_{2} = - 3 \cdot R_{1} + R_{2}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} 1 & - \frac{3 + \sqrt{33}}{6} \\ - 3 \cdot R_{1} + R_{2} & - 3 \cdot R_{1} + R_{2} \end{array}]$

$R_{2} = - 3 \cdot R_{1} + R_{2}$

Replace $R_{2}$ (row $2$ ) with the actual values of the elements for the row operation $R_{2} = - 3 \cdot R_{1} + R_{2}$ .

$[\begin{array}{c} 1 & - \frac{3 + \sqrt{33}}{6} \\ (- 3) \cdot (1) + 3 & (- 3) \cdot (- \frac{3 + \sqrt{33}}{6}) - \frac{3 + \sqrt{33}}{2} \end{array}]$

$R_{2} = - 3 \cdot R_{1} + R_{2}$

Simplify $R_{2}$ (row $2$ ).

$[\begin{array}{c} 1 & - \frac{3 + \sqrt{33}}{6} \\ 0 & 0 \end{array}]$

Step 13

Use the result matrix to declare the final solutions to the system of equations.

$x_{1} - \frac{(3 + \sqrt{33}) x_{2}}{6} = 0$

Step 14

This expression is the solution set for the system of equations.

${(\frac{(3 + \sqrt{33}) x_{2}}{6}, x_{2}) | x_{2} \in ℝ}$

Step 15

Decompose a solution vector by re-arranging each equation represented in the row-reduced form of the augmented matrix by solving for the dependent variable in each row yields the vector equality.

$X = [\begin{array}{c} x_{1} \\ x_{2} \end{array}] = [\begin{array}{c} \frac{(3 + \sqrt{33}) x_{2}}{6} \\ x_{2} \end{array}]$

Step 16

Express the vector as a linear combination of column vector using the properties of vector column addition.

$X = [\begin{array}{c} x_{1} \\ x_{2} \end{array}] = x_{2} [\begin{array}{c} \frac{3 + \sqrt{33}}{6} \\ 1 \end{array}]$

Step 17

The null space of the set is the set of vectors created from the free variables of the system.

${[\begin{array}{c} \frac{3 + \sqrt{33}}{6} \\ 1 \end{array}]}$

Step 18

The eigenvector for $λ = \frac{5 - \sqrt{33}}{2}$ is equal to the null space of the matrix minus the eigenvalue times the identity matrix.

$ε_{A} (\frac{5 - \sqrt{33}}{2}) = N (A - λ I_{2})$

Step 19

Substitute the known values into the formula.

$[\begin{array}{c} 4 & 2 \\ 3 & 1 \end{array}] - \frac{5 - \sqrt{33}}{2 [\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}]}$

Step 20

Simplify the matrix expression.

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Multiply $- \frac{5 - \sqrt{33}}{2}$ by each element of the matrix.

$[\begin{array}{c} 4 & 2 \\ 3 & 1 \end{array}] + [\begin{array}{c} - \frac{5 - \sqrt{33}}{2} \cdot 1 & - \frac{5 - \sqrt{33}}{2} \cdot 0 \\ - \frac{5 - \sqrt{33}}{2} \cdot 0 & - \frac{5 - \sqrt{33}}{2} \cdot 1 \end{array}]$

Simplify each element in the matrix.

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Rearrange $- \frac{5 - \sqrt{33}}{2} \cdot 1$ .

$[\begin{array}{c} 4 & 2 \\ 3 & 1 \end{array}] + [\begin{array}{c} - \frac{5 - \sqrt{33}}{2} & - \frac{5 - \sqrt{33}}{2} \cdot 0 \\ - \frac{5 - \sqrt{33}}{2} \cdot 0 & - \frac{5 - \sqrt{33}}{2} \cdot 1 \end{array}]$

Rearrange $- \frac{5 - \sqrt{33}}{2} \cdot 0$ .

$[\begin{array}{c} 4 & 2 \\ 3 & 1 \end{array}] + [\begin{array}{c} - \frac{5 - \sqrt{33}}{2} & 0 \\ - \frac{5 - \sqrt{33}}{2} \cdot 0 & - \frac{5 - \sqrt{33}}{2} \cdot 1 \end{array}]$

Rearrange $- \frac{5 - \sqrt{33}}{2} \cdot 0$ .

$[\begin{array}{c} 4 & 2 \\ 3 & 1 \end{array}] + [\begin{array}{c} - \frac{5 - \sqrt{33}}{2} & 0 \\ 0 & - \frac{5 - \sqrt{33}}{2} \cdot 1 \end{array}]$

Rearrange $- \frac{5 - \sqrt{33}}{2} \cdot 1$ .

$[\begin{array}{c} 4 & 2 \\ 3 & 1 \end{array}] + [\begin{array}{c} - \frac{5 - \sqrt{33}}{2} & 0 \\ 0 & - \frac{5 - \sqrt{33}}{2} \end{array}]$

Add the corresponding elements.

$[\begin{array}{c} 4 - \frac{5 - \sqrt{33}}{2} & 2 + 0 \\ 3 + 0 & 1 - \frac{5 - \sqrt{33}}{2} \end{array}]$

Simplify each element of the matrix $[\begin{array}{c} 4 - \frac{5 - \sqrt{33}}{2} & 2 + 0 \\ 3 + 0 & 1 - \frac{5 - \sqrt{33}}{2} \end{array}]$ .

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Simplify $4 - \frac{5 - \sqrt{33}}{2}$ .

$[\begin{array}{c} \frac{3 + \sqrt{33}}{2} & 2 + 0 \\ 3 + 0 & 1 - \frac{5 - \sqrt{33}}{2} \end{array}]$

Simplify $2 + 0$ .

$[\begin{array}{c} \frac{3 + \sqrt{33}}{2} & 2 \\ 3 + 0 & 1 - \frac{5 - \sqrt{33}}{2} \end{array}]$

Simplify $3 + 0$ .

$[\begin{array}{c} \frac{3 + \sqrt{33}}{2} & 2 \\ 3 & 1 - \frac{5 - \sqrt{33}}{2} \end{array}]$

Simplify $1 - \frac{5 - \sqrt{33}}{2}$ .

$[\begin{array}{c} \frac{3 + \sqrt{33}}{2} & 2 \\ 3 & - \frac{3 - \sqrt{33}}{2} \end{array}]$

Step 21

Find the reduced row echelon form of the matrix.

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Perform the row operation $R_{1} = - \frac{3 - \sqrt{33}}{12} R_{1}$ on $R_{1}$ (row $1$ ) in order to convert some elements in the row to $1$ .

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Replace $R_{1}$ (row $1$ ) with the row operation $R_{1} = - \frac{3 - \sqrt{33}}{12} R_{1}$ in order to convert some elements in the row to the desired value $1$ .

$[\begin{array}{c} - \frac{3 - \sqrt{33}}{12} R_{1} & - \frac{3 - \sqrt{33}}{12} R_{1} \\ 3 & - \frac{3 - \sqrt{33}}{2} \end{array}]$

$R_{1} = - \frac{3 - \sqrt{33}}{12} R_{1}$

Replace $R_{1}$ (row $1$ ) with the actual values of the elements for the row operation $R_{1} = - \frac{3 - \sqrt{33}}{12} R_{1}$ .

$[\begin{array}{c} (- \frac{3 - \sqrt{33}}{12}) \cdot (\frac{3 + \sqrt{33}}{2}) & (- \frac{3 - \sqrt{33}}{12}) \cdot (2) \\ 3 & - \frac{3 - \sqrt{33}}{2} \end{array}]$

$R_{1} = - \frac{3 - \sqrt{33}}{12} R_{1}$

Simplify $R_{1}$ (row $1$ ).

$[\begin{array}{c} 1 & - \frac{3 - \sqrt{33}}{6} \\ 3 & - \frac{3 - \sqrt{33}}{2} \end{array}]$

Perform the row operation $R_{2} = - 3 \cdot R_{1} + R_{2}$ on $R_{2}$ (row $2$ ) in order to convert some elements in the row to $0$ .

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Replace $R_{2}$ (row $2$ ) with the row operation $R_{2} = - 3 \cdot R_{1} + R_{2}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} 1 & - \frac{3 - \sqrt{33}}{6} \\ - 3 \cdot R_{1} + R_{2} & - 3 \cdot R_{1} + R_{2} \end{array}]$

$R_{2} = - 3 \cdot R_{1} + R_{2}$

Replace $R_{2}$ (row $2$ ) with the actual values of the elements for the row operation $R_{2} = - 3 \cdot R_{1} + R_{2}$ .

$[\begin{array}{c} 1 & - \frac{3 - \sqrt{33}}{6} \\ (- 3) \cdot (1) + 3 & (- 3) \cdot (- \frac{3 - \sqrt{33}}{6}) - \frac{3 - \sqrt{33}}{2} \end{array}]$

$R_{2} = - 3 \cdot R_{1} + R_{2}$

Simplify $R_{2}$ (row $2$ ).

$[\begin{array}{c} 1 & - \frac{3 - \sqrt{33}}{6} \\ 0 & 0 \end{array}]$

Step 22

Use the result matrix to declare the final solutions to the system of equations.

$x_{1} - \frac{(3 - \sqrt{33}) x_{2}}{6} = 0$

Step 23

This expression is the solution set for the system of equations.

${(\frac{(3 - \sqrt{33}) x_{2}}{6}, x_{2}) | x_{2} \in ℝ}$

Step 24

Decompose a solution vector by re-arranging each equation represented in the row-reduced form of the augmented matrix by solving for the dependent variable in each row yields the vector equality.

$X = [\begin{array}{c} x_{1} \\ x_{2} \end{array}] = [\begin{array}{c} \frac{(3 - \sqrt{33}) x_{2}}{6} \\ x_{2} \end{array}]$

Step 25

Express the vector as a linear combination of column vector using the properties of vector column addition.

$X = [\begin{array}{c} x_{1} \\ x_{2} \end{array}] = x_{2} [\begin{array}{c} \frac{3 - \sqrt{33}}{6} \\ 1 \end{array}]$

Step 26

The null space of the set is the set of vectors created from the free variables of the system.

${[\begin{array}{c} \frac{3 - \sqrt{33}}{6} \\ 1 \end{array}]}$

Step 27

The eigenspace of $A$ is the union of the vector space for each eigenvalue.

${[\begin{array}{c} \frac{3 + \sqrt{33}}{6} \\ 1 \end{array}]} \cup {[\begin{array}{c} \frac{3 - \sqrt{33}}{6} \\ 1 \end{array}]}$

Step 28