Solve {l}{a-3b=0}{2a+6b-b=0}

$$\left\{ \begin{array} { l } { a - 3 b = 0 } \\ { 2 a + 6 b - b = 0 } \end{array} \right.$$

Solve for a, b

$$a=0$$

$$b=0$$

Steps Using Substitution

Consider the second equation. Combine 6b and -b to get 5b.

$$2a+5b=0$$

To solve a pair of equations using substitution, first solve one of the equations for one of the variables. Then substitute the result for that variable in the other equation.

$$a-3b=0,2a+5b=0$$

Choose one of the equations and solve it for a by isolating a on the left hand side of the equal sign.

$$a-3b=0$$

Add 3b to both sides of the equation.

$$a=3b$$

Substitute 3b for a in the other equation, 2a+5b=0.

$$2\times 3b+5b=0$$

Multiply 2 times 3b.

$$6b+5b=0$$

Add 6b to 5b.

$$11b=0$$

Divide both sides by 11.

$$b=0$$

Substitute 0 for b in a=3b. Because the resulting equation contains only one variable, you can solve for a directly.

$$a=0$$

The system is now solved.

$$a=0,b=0$$

Steps Using Matrices

Consider the second equation. Combine 6b and -b to get 5b.

$$2a+5b=0$$

Put the equations in standard form and then use matrices to solve the system of equations.

$$a-3b=0,2a+5b=0$$

Write the equations in matrix form.

$$\left(\begin{matrix}1&-3\\2&5\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}0\\0\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&-3\\2&5\end{matrix}\right).

$$inverse(\left(\begin{matrix}1&-3\\2&5\end{matrix}\right))\left(\begin{matrix}1&-3\\2&5\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=inverse(\left(\begin{matrix}1&-3\\2&5\end{matrix}\right))\left(\begin{matrix}0\\0\end{matrix}\right)$$

The product of a matrix and its inverse is the identity matrix.

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=inverse(\left(\begin{matrix}1&-3\\2&5\end{matrix}\right))\left(\begin{matrix}0\\0\end{matrix}\right)$$

Multiply the matrices on the left hand side of the equal sign.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=inverse(\left(\begin{matrix}1&-3\\2&5\end{matrix}\right))\left(\begin{matrix}0\\0\end{matrix}\right)$$

For the 2\times 2 matrix \left(\begin{matrix}a&b\\c&d\end{matrix}\right), the inverse matrix is \left(\begin{matrix}\frac{d}{ad-bc}&\frac{-b}{ad-bc}\\\frac{-c}{ad-bc}&\frac{a}{ad-bc}\end{matrix}\right), so the matrix equation can be rewritten as a matrix multiplication problem.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}\frac{5}{5-\left(-3\times 2\right)}&-\frac{-3}{5-\left(-3\times 2\right)}\\-\frac{2}{5-\left(-3\times 2\right)}&\frac{1}{5-\left(-3\times 2\right)}\end{matrix}\right)\left(\begin{matrix}0\\0\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}\frac{5}{11}&\frac{3}{11}\\-\frac{2}{11}&\frac{1}{11}\end{matrix}\right)\left(\begin{matrix}0\\0\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}0\\0\end{matrix}\right)$$

Extract the matrix elements a and b.

$$a=0,b=0$$

Steps Using Elimination

Consider the second equation. Combine 6b and -b to get 5b.

$$2a+5b=0$$

In order to solve by elimination, coefficients of one of the variables must be the same in both equations so that the variable will cancel out when one equation is subtracted from the other.

$$a-3b=0,2a+5b=0$$

To make a and 2a equal, multiply all terms on each side of the first equation by 2 and all terms on each side of the second by 1.

$$2a+2\left(-3\right)b=0,2a+5b=0$$

Simplify.

$$2a-6b=0,2a+5b=0$$

Subtract 2a+5b=0 from 2a-6b=0 by subtracting like terms on each side of the equal sign.

$$2a-2a-6b-5b=0$$

Add 2a to -2a. Terms 2a and -2a cancel out, leaving an equation with only one variable that can be solved.

$$-6b-5b=0$$

Add -6b to -5b.

$$-11b=0$$

Divide both sides by -11.

$$b=0$$

Substitute 0 for b in 2a+5b=0. Because the resulting equation contains only one variable, you can solve for a directly.

$$2a=0$$

Divide both sides by 2.

$$a=0$$

The system is now solved.

$$a=0,b=0$$