a. \begin{bmatrix}g&h&i\\2d&2e&2f\\3a&3b&3c\end{bmatrix}
b. \begin{bmatrix}a&b&c\\d-2a&e-2b&f-2c\\5g&5h&5i\end{bmatrix}
Hello, I am not sure how to go about answering this question. I don't need and exact answer, but I just need to know how to get started on answering part A and B of this question.
On
Lets try solve a similar determinant: $$\det\begin{pmatrix}2d & 2e & 2f \\ a-3g & b - 3h & c - 3i \\4g & 4h & 4i\end{pmatrix} $$
We have that: \begin{align*} \det\begin{pmatrix}2d & 2e & 2f \\ a-3g & b - 3h & c - 3i \\4g & 4h & 4i\end{pmatrix} & = 2\det\begin{pmatrix}d & e & f \\ a-3g & b - 3h & c - 3i \\4g & 4h & 4i\end{pmatrix} \\ & \text{Multiply the first row by a constant} \\ & = 8\det\begin{pmatrix}d & e & f \\ a-3g & b - 3h & c - 3i \\g & h & i\end{pmatrix} \\ & \text{Multiply the third row by a constant} \\ & = 8\det\begin{pmatrix}d & e & f \\ a & b & c \\g & h & i\end{pmatrix} \\ &\text{Add three times the third row to the second row} \\ & = -8\det\begin{pmatrix} a & b & c \\d & e & f \\g & h & i\end{pmatrix} \\ &\text{Switch the first and second rows} \\ & = -8\times (7) = -56 \\ \text{Substitute in the determinant we know} \end{align*} Hopefully you see how you can solve your problems from this.
On
Use the property that $\det(\mathrm{AB}) = \det(\mathrm{A})\det(\mathrm{B})$
Decompose your given matrices into a product of row-operation matrices with your given matrix. For instance:
$$\begin{bmatrix}a&b&c\\2d&2e&2f\\g&h&i\end{bmatrix} = \begin{bmatrix}1&0&0\\0&2&0\\0&0&1\end{bmatrix} \begin{bmatrix}a&b&c\\d&e&f\\g&h&i\end{bmatrix}$$
So that matrix's determinant is 14.
Hint:
a) \begin{align} \begin{vmatrix} a & b & c \\ 2d & 2e & 2f\\ 3g & 3h & 3i \end{vmatrix} =2\begin{vmatrix} a & b & c \\ d & e & f\\ 3g & 3h & 3i \end{vmatrix} \end{align}
b) \begin{align} \begin{vmatrix} a & b & c \\ d-a & e-b & f-c\\ 5g & 5h & 5i \end{vmatrix} =\begin{vmatrix} a & b & c \\ d & e & f\\ 5g & 5h & 5i \end{vmatrix} \end{align}