Solve {cc}{0}&{1}{-3}&{2}{l}{a}{b}={c}{-3}{0}

$$\left( \begin{array} { c c } { 0 } & { 1 } \\ { - 3 } & { 2 } \end{array} \right) \left( \begin{array} { l } { a } \\ { b } \end{array} \right) = \left( \begin{array} { c } { - 3 } \\ { 0 } \end{array} \right)$$

Solve for b, a

$$b=-3$$

$$a=-2$$

Solution Steps

Consider the second equation. Insert the known values of variables into the equation.

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

Multiply 2 and -3 to get -6.

$$-3a-6=0$$

Add 6 to both sides. Anything plus zero gives itself.

$$-3a=6$$

Divide both sides by -3.

$$a=\frac{6}{-3}$$

Divide 6 by -3 to get -2.

$$a=-2$$

The system is now solved.

$$b=-3 a=-2$$