Question

If $A=\left[\begin{array}{cc}2& -3\\ p& q\end{array}\right]$, find $p$and $q$ so that $A^2=I$.

Answer 1

Step1: Finding $A^2$

Multiply the matrix $A$ by itself to obtain $A^2$.

$\begin{array}{rcl}{A}^2& =& A\times A\\ & \Rightarrow & A^2=\left[\begin{array}{cc}2& -3\\ p& q\end{array}\right]\times \left[\begin{array}{cc}2& -3\\ p& q\end{array}\right]\\ & \Rightarrow & A^2=\left[\begin{array}{cc}2·2+\left(-3\right)·p& 2·\left(-3\right)+\left(-3\right)·q\\ p·2+q·p& p·\left(-3\right)+q·q\end{array}\right]\\ \therefore A^2& =& \left[\begin{array}{cc}4-3p& -6-3q\\ 2p+qp& -3p+q^2\end{array}\right]\end{array}$

Step2: Comparison of $\begin{array}{r}{A}^2\end{array}$ and $I$.

The matrix $I$ represents the identity matrix $\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$.

If two matrices are equal, then their corresponding elements are also equal.

Since

$$\begin{gathered} A^2=Ii.e., \\ \begin{array}{l}\left[\begin{array}{cc}4-3p& -6-3q\\ 2p+qp& -3p+q^2\end{array}\right]\end{array}=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right] \end{gathered}$$

Therefore, $4-3p=1$ and $-6-3q=0$..

Step3: Solving for $p$.

Solve the equation $4-3p=1$ for $p$.

$$\begin{gathered} 4-3p=1 \\ \Rightarrow -3p=1-4(Subtracting4frombothsides) \\ \Rightarrow -3p=-3 \\ \Rightarrow p=\frac{-3}{-3}(Dividingbothsidesby-3) \\ \therefore p=1 \end{gathered}$$

Step4: Solving for $q$.

Solve the equation $-6-3q=0$ for $q$.

$$\begin{gathered} -6-3q=0 \\ \Rightarrow 3q=-6(Subtracting6frombothsides) \\ \Rightarrow q=\frac{-6}{3}(Dividingbothsidesby3) \\ \therefore q=-2 \end{gathered}$$

Final answer: The values are $p=1$ and $q=-2$.