Tedious Matrix Multiplications?

Let $L = \begin{pmatrix} 1 & 1 \end{pmatrix}, R = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$

$L \times A = \begin{pmatrix} 1 & 1 \end{pmatrix} \times \begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix} \\ = \begin{pmatrix} 1 \times 1 + 1 \times 2 & 1 \times 2 + 1 \times 1 \end{pmatrix} \\ = \begin{pmatrix} 3 & 3 \end{pmatrix} = 3 \cdot \begin{pmatrix} 1 & 1 \end{pmatrix} = 3L$

$L \times R = \begin{pmatrix} 1 & 1 \end{pmatrix} \times \begin{pmatrix} 1\\ 1 \end{pmatrix} \\= \begin{pmatrix} 1 \times 1 + 1 \times 1 \end{pmatrix} = 2$

$B = L \times A^{10} \times R = 3^{10} \cdot L \times R = 3^{10} \times 2 = 118098$

So the answer is 27.

The eigen value $\lambda$ and eigen vector $\mathrm{v}$ of a square matrix $A$ obey the following definition:

$A \mathrm{v} = \lambda \mathrm{v}$ $\implies (\lambda I - A)\mathrm{v}=0$ Here, $I$ is the identity matrix having the same size as $A$ . For non-trivial solutions, the determinant of the matrix $(\lambda I - A)$ must be zero. Using this fact, the eigen values can be computed, by computing the roots of the resulting characteristic polynomial in $\lambda$ . Using the values of $\lambda$ , the corresponding eigen vectors can be computed.

For the given square matrix of size 2X2, there would be two unique eigen values and eigen vectors respectively. Let each eigen value and eigen vector pair be $(\lambda_1,\mathrm{v}_1)$ and $(\lambda_2,\mathrm{v}_2)$ .

The eigen vectors can be arranged in the matrix as shown below:

$V = \left[\begin{matrix} \mathrm{v}_1&\mathrm{v}_2\end{matrix}\right]$

Multiplying both sides by $A$ :

$AV = \left[\begin{matrix} A\mathrm{v}_1&A\mathrm{v}_2\end{matrix}\right]$

Using the definition of the Eigen value and eigen vector:

$AV = \left[\begin{matrix} \lambda_1\mathrm{v}_1&\lambda_2 \mathrm{v}_2\end{matrix}\right]$

The RHS can be re-arranged as follows:

$AV = \left[\begin{matrix} \mathrm{v}_1&\mathrm{v}_2\end{matrix}\right]\left[\begin{matrix} \lambda_1&0\\0&\lambda_2\end{matrix}\right]$ $\implies AV = V\left[\begin{matrix} \lambda_1&0\\0&\lambda_2\end{matrix}\right]$ $\implies AV = VD$ $\implies A = VDV^{-1}$

Now:

$A^{10} = ( VDV^{-1} )( VDV^{-1} )( VDV^{-1} )( VDV^{-1} )( VDV^{-1} )( VDV^{-1} )( VDV^{-1} )( VDV^{-1} )( VDV^{-1} )( VDV^{-1} )$

Recognizing that:

$VV^{-1} = V^{-1}V = I$

The above relation holds true provided that the eigen vectors are linearly independent. If not, then $V$ will be a singular matrix. Moreover, if the eigen vectors are computed such they are unit vectors, an interesting property holds that the matrix $V$ is orthogonal. In other words:

$V^{-1} = V^T$

After cancelling out (to the identity matrix), $V$ with it's inverse multiple times, one gets:

$\implies \boxed{A^{10} = VD^{10}V^{-1} }$ $\implies \boxed{A^{10} = VD^{10}V^T} \ \because V^{-1} = V^T$

$\implies \boxed{A^{10} = V\left[\begin{matrix} \lambda_1^{10}&0\\0&\lambda_2^{10}\end{matrix}\right]V^T}$ So basically, instead of performing multiple matrix multiplications, all one needs to compute are the eigen values and eigen vectors and replace it in the above expression to get the required result of $A^{10}$ .

After performing all computations, which I have left out of this solution, the required answer comes out to be $\boxed{27}$ .