UPSC NDA I and II 2020 Math Solved Paper

Concept:Matrix multiplication is possible when the number of columns of the first matrix equals the number of rows of the second matrix.The product is obtained by multiplying each row of the first matrix by the single column of the second matrix and summing the products.Explanation:Given $A = \begin{bmatrix} a & h & g \\ h & b & f \\ g & f & c \end{bmatrix}_{3\times 3}$ and $B = \begin{bmatrix} x \\ y \\ z \end{bmatrix}_{3\times 1}$.The inner dimensions are both $3$, so $AB$ exists and its order is $3 \times 1$.To find the entry in the first row of $AB$, multiply the first row of $A$ by $B$:$a \cdot x + h \cdot y + g \cdot z = ax + hy + gz$.For the second row of $AB$, multiply the second row of $A$ by $B$:$h \cdot x + b \cdot y + f \cdot z = hx + by + fz$.For the third row of $AB$, multiply the third row of $A$ by $B$:$g \cdot x + f \cdot y + c \cdot z = gx + fy + cz$.Thus, $AB = \begin{bmatrix} ax+hy+gz \\ hx+by+fz \\ gx+fy+cz \end{bmatrix}$.Answer:$AB = \begin{bmatrix} ax+hy+gz \\ hx+by+fz \\ gx+fy+cz \end{bmatrix}$This matches option C.