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(D) We know that $\begin{bmatrix} 1 & a \ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & b \ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & a+b \ 0 & 1 \end{bm

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$\begin{bmatrix} 1 & -13 \ 0 & 1 \end{bmatrix}$

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(D) We know that $\begin{bmatrix} 1 & a \ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & b \ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & a+b \ 0 & 1 \end{bmatrix}$.Applying this property to the given product:$\begin{bmatrix} 1 & 1+2+3+\dots+(n-1) \ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 78 \ 0 & 1 \end{bmatrix}$.The sum of the first $(n-1)$ natural numbers is $\frac{(n-1)n}{2}$.So,$\frac{n(n-1)}{2} = 78 \Rightarrow n^2 - n - 156 = 0$.Solving the quadratic equation: $(n-13)(n+12) = 0$. Since $n$ must be positive,$n = 13$.We need to find the inverse of $A = \begin{bmatrix} 1 & 13 \ 0 & 1 \end{bmatrix}$.For a matrix $A = \begin{bmatrix} 1 & k \ 0 & 1 \end{bmatrix}$,the inverse is $A^{-1} = \begin{bmatrix} 1 & -k \ 0 & 1 \end{bmatrix}$.Thus,the inverse is $\begin{bmatrix} 1 & -13 \ 0 & 1 \end{bmatrix}$.

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