Solve Matrix

$$\left. \begin{array} { l } { \text { matriz } A = \begin{bmatrix} \begin{array} { l l } { 1 } & { 2 } \\ { 2 } & { 1 } \end{array} \end{bmatrix} } \\ { A ^ { 4 } \text { . (Nota: } A ^ { 2 } = A } \end{array} \right.$$

Solve for A

$$A=1$$

$$A=0$$

Steps Using Factoring

Subtract A from both sides.

$$A^{2}-A=0$$

Factor out A.

$$A\left(A-1\right)=0$$

To find equation solutions, solve A=0 and A-1=0.

$$A=0 A=1$$

Steps Using the Quadratic Formula

Subtract A from both sides.

$$A^{2}-A=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -1 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$A=\frac{-\left(-1\right)±\sqrt{1}}{2}$$

Take the square root of 1.

$$A=\frac{-\left(-1\right)±1}{2}$$

The opposite of -1 is 1.

$$A=\frac{1±1}{2}$$

Now solve the equation A=\frac{1±1}{2} when ± is plus. Add 1 to 1.

$$A=\frac{2}{2}$$

Divide 2 by 2.

$$A=1$$

Now solve the equation A=\frac{1±1}{2} when ± is minus. Subtract 1 from 1.

$$A=\frac{0}{2}$$

Divide 0 by 2.

$$A=0$$

The equation is now solved.

$$A=1 A=0$$

Steps for Completing the Square

Subtract A from both sides.

$$A^{2}-A=0$$

Divide -1, the coefficient of the x term, by 2 to get -\frac{1}{2}. Then add the square of -\frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$A^{2}-A+\left(-\frac{1}{2}\right)^{2}=\left(-\frac{1}{2}\right)^{2}$$

Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$A^{2}-A+\frac{1}{4}=\frac{1}{4}$$

Factor A^{2}-A+\frac{1}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(A-\frac{1}{2}\right)^{2}=\frac{1}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(A-\frac{1}{2}\right)^{2}}=\sqrt{\frac{1}{4}}$$

Simplify.

$$A-\frac{1}{2}=\frac{1}{2} A-\frac{1}{2}=-\frac{1}{2}$$

Add \frac{1}{2} to both sides of the equation.

$$A=1 A=0$$