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Let the price of one pen, one notepad, and one eraser be $x$, $y$, and $z$ respectively. Based on the purchases: $$4x + 3y + 2z = 60$$ $$2x + 4y + 6z = 90$$ $$6x + 2y + 3z = 70$$ In matrix form $AX = B$: $$\begin{pmatrix} 4 & 3 & 2 \\ 2 & 4 & 6 \\ 6 & 2 & 3 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 60 \\ 90 \\ 70 \end{pmatrix}$$
Calculate $|A|$: $$|A| = 4(12 - 12) - 3(6 - 36) + 2(4 - 24)$$ $$|A| = 4(0) - 3(-30) + 2(-20) = 0 + 90 - 40 = 50$$ Since $|A| \neq 0$, $A^{-1}$ exists.
The formula for $X$ is $X = A^{-1}B$. To find $A^{-1}$, we use $A^{-1} = \frac{1}{|A|} adj(A)$. Calculating the cofactor matrix and transposing it yields the adjoint. Given the complexity, the primary requirement is the formula $X = A^{-1}B$.
Final Answer: $|A| = 50$, $X = A^{-1}B$
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