Linear Algebra Examples

$\sqrt{4 - 5 x} = 2 x - 1$ , $\sqrt{4 - 5 x} = 2 x - 1$

Step 1

Subtract $2 x$ from both sides of the equation.

$\sqrt{4 - 5 x} - 2 x = - 1, \sqrt{4 - 5 x} = 2 x - 1$

Step 2

Subtract $2 x$ from both sides of the equation.

$\sqrt{4 - 5 x} - 2 x = - 1, \sqrt{4 - 5 x} - 2 x = - 1$

Step 3

Write the system of equations in matrix form.

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

Step 4

Row reduce to eliminate one of the variables.

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Step 4.1

Multiply each element of $R_{1}$ by $- \frac{1}{2}$ to make the entry at $1, 1$ a $1$ .

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Step 4.1.1

Multiply each element of $R_{1}$ by $- \frac{1}{2}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} - \frac{1}{2} \cdot - 2 & - \frac{1}{2} \cdot - 1 \\ - 2 & - 1 \end{array}]$

Step 4.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & \frac{1}{2} \\ - 2 & - 1 \end{array}]$

Step 4.2

Perform the row operation $R_{2} = R_{2} + 2 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Step 4.2.1

Perform the row operation $R_{2} = R_{2} + 2 R_{1}$ to make the entry at $2, 1$ a $0$ .

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

Step 4.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{1}{2} \\ 0 & 0 \end{array}]$

Step 5

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

$x = \frac{1}{2}$

Step 6

The solution is the set of ordered pairs that makes the system true.

$\frac{1}{2}$

Step 7

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.

$X = [\begin{array}{c} x \end{array}] = [\begin{array}{c} \frac{1}{2} \end{array}]$