1.1.17

a) Solving $$\begin{bmatrix} b_{11} & b_{12} & b_{13} \\ b_{21} & b_{22} & b_{23} \end{bmatrix}

\begin{bmatrix} 2 & 3 \\ 1 & 2 \\ 2 & 5 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$ is equivalent to solving $$\begin{bmatrix} 2 & 2 & 1 & 0 & 0 & 0 \\ 3 & 2 & 5 & 0 & 0 & 0 \\ 0 & 0 & 0 & 2 & 1 & 2 \\ 0 & 0 & 0 & 3 & 2 & 5 \end{bmatrix}

\begin{bmatrix} b_{11} \\ b_{12} \\ b_{13} \\ b_{21} \\ b_{22} \\ b_{23} \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ 0 \\ 1 \end{bmatrix}.$$ We only reordered some numbers, but now we have the equation in a standard form that we know how to solve. Therefore, $$B = \begin{bmatrix} 2 + r & -3 - 4r & r \\ -1 + t & 2- 4t & t \end{bmatrix}$$, for any $$r$$ and $$t$$.

b) Analogously to the first part, rewrite the equation system $$AC = I_3$$ to $$\begin{bmatrix} 2 & 0 & 0 & 3 & 0 & 0 \\ 0 & 2 & 0 & 0 & 3 & 0 \\ 0 & 0 & 2 & 0 & 0 & 3 \\ 1 & 0 & 0 & 2 & 0 & 0 \\ 0 & 1 & 0 & 0 & 2 & 0 \\ 0 & 0 & 1 & 0 & 0 & 2 \\ 2 & 0 & 0 & 5 & 0 & 0 \\ 0 & 2 & 0 & 0 & 5 & 0 \\ 0 & 0 & 2 & 0 & 0 & 5 \end{bmatrix}

\begin{bmatrix} c_{11} \\ c_{12} \\ c_{13} \\ c_{21} \\ c_{22} \\ c_{23} \\ \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \\ 1 \\ 0 \\ 0 \\ 0 \\ 1 \end{bmatrix}.$$ Using standard methods we see that there is no solution to this system of linear equations.