Question

Let $A=\left[\begin{array}{cc}x& 1\\ 1& 0\end{array}\right]$, $x\in \mathrm{ℝ}$ and $A^4=\left[a_{ij}\right]$. If $a_{11}=109$, then $a_{22}$ is equal to

Answer 1

Step 1: Use matrix multiplication:

A matrix $A$ is said to be conformable for multiplication with a matrix $B$ if the number of columns of $A$ is equal to the number of rows in $B$ .

Number of rows in $A$ = Number of columns $A$

Therefore matrix multiplication is possible.

$$\begin{gathered} A^2=\left[\begin{array}{cc}x& 1\\ 1& 0\end{array}\right]\left[\begin{array}{cc}x& 1\\ 1& 0\end{array}\right] \\ =\left[\begin{array}{cc}{x}^2+1& x\\ x& 1\end{array}\right] \\ {A}^3=A^{2}A \\ =\left[\begin{array}{cc}{x}^2+1& x\\ x& 1\end{array}\right]\left[\begin{array}{cc}x& 1\\ 1& 0\end{array}\right] \\ =\left[\begin{array}{cc}\left(x^2+1\right)x+1.x& 1(x^2+1)\\ x.x+1.1& 1.x\end{array}\right] \\ =\left[\begin{array}{cc}{x}^3+x+x& x^2+1\\ x^2+1& x\end{array}\right] \\ =\left[\begin{array}{cc}{x}^3+2x& x^2+1\\ x^2+1& x\end{array}\right] \\ {A}^4=A^{3}A \\ =\left[\begin{array}{cc}{x}^3+2x& x^2+1\\ x^2+1& x\end{array}\right]\left[\begin{array}{cc}x& 1\\ 1& 0\end{array}\right] \\ =\left[\begin{array}{cc}\left(x^3+2x\right)x+(x^2+1).1& \left(x^3+2x\right)1\\ (x^2+1).x+x.1& \left(x^2+1\right).1\end{array}\right] \\ =\left[\begin{array}{cc}{x}^4+2x^2+x^2+1& x^3+2x\\ x^3+x+x& x^2+1\end{array}\right] \\ =\left[\begin{array}{cc}{x}^4+3x^2+1& x^3+2x\\ x^3+2x& x^2+1\end{array}\right] \end{gathered}$$

Step 2: To find $a_{22}$ in $A^4=\left[a_{ij}\right]$.

$$\begin{gathered} \Rightarrow a_{11}=109 \\ \Rightarrow x^4+3x^2+1=109 \\ \Rightarrow x^4+3x^2-108=0 \\ \Rightarrow \left(x^2+12\right)\left(x^2-9\right)=0 \\ \Rightarrow x^2=9or-12 \\ \Rightarrow x=\pm 3\left[\because x^2=-12isnotpossible,xbecomescomplexnumber\right] \\ {a}_{22}=x^2+1 \\ Atx=\pm 3 \\ \Rightarrow a_{22}={\left(\pm 3\right)}^2+1 \\ =9+1 \\ =10 \end{gathered}$$

Hence, the value of $a_{22}=10$.