Let $$A = \begin{bmatrix}x & 1 \\0 & 1 \end{bmatrix}$$ where $$x$$ is a real number, and $$B = \begin{bmatrix}2\sqrt{2} & 3+ \sqrt{2} \\0 & 1 \end{bmatrix}$$. If $$A^{3} = B$$, then $$x$$ is equal to

Given,

$$A = \begin{bmatrix}x & 1 \\0 & 1 \end{bmatrix}$$

So, $$A^2 = \begin{bmatrix}x & 1 \\0 & 1 \end{bmatrix}$$* $$\begin{bmatrix}x & 1 \\0 & 1 \end{bmatrix}$$

=$$\begin{bmatrix}x^2 & x+1 \\0 & 1 \end{bmatrix}$$

Now, $$A^{3\ }=A^2\cdot A$$

= $$\begin{bmatrix}x^2 & x+1 \\0 & 1 \end{bmatrix}$$ * $$\begin{bmatrix}x & 1 \\0 & 1 \end{bmatrix}$$

= $$\begin{bmatrix}x^3 & x^2+x+1 \\0 & 1 \end{bmatrix}$$

Given, $$A^3=B$$

$$\begin{bmatrix}x^3 & x^2+x+1 \\0 & 1 \end{bmatrix}$$ = $$\begin{bmatrix}2\sqrt{2} & 3+ \sqrt{2} \\0 & 1 \end{bmatrix}$$

For both the matrices to be equal, all elements of the matrix on the LHS need to be equal to all elements on the RHS.

$$x^3\ =\ 2\sqrt{\ 2}$$

$$x^3=\sqrt{\ 8}$$

$$x^3=2^{\frac{3}{2}}$$

$$x=\sqrt{\ 2}$$