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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.

Ben Grossmann
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WhipCored
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3 Answers3

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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}

Jacky Chong
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  • Can you explain this hint? I don't understand what that means. I am having trouble understanding what the question is even asking me to do. – WhipCored Oct 23 '16 at 22:23
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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.

Mark Schultz-Wu
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  • So Right now this is what I have done for part A and would like to know if this is right.

    \begin{bmatrix}1&0&2\0&7&5\0&0&1\end{bmatrix} = \begin{bmatrix}a&b&c\d&e&f\g&h&i\end{bmatrix} =

     – WhipCored Oct 23 '16 at 23:01
  • \begin{bmatrix}0&0&1\0&14&10\3&0&6\end{bmatrix} = – WhipCored Oct 23 '16 at 23:03
  • @WhipCored while this is a method to solve it, I'd recommend against it. The power of this method is that it will work for ANY matrix with $\det A = 7$, not just the specific one you wrote down. Think about it in terms of row operations (you switched the first and last rows, then scaled the middle row by 2 and the last row by 3) and how that impacts a general matrix,not a specific one. If you can understand this, it will greatly help your understanding. – Mark Schultz-Wu Oct 24 '16 at 02:35
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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.

NovaDenizen
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  • Ok, I think I understand what you have written here but can you also explain how you got to this point? I am not sure where you got the other matrices from. – WhipCored Oct 23 '16 at 22:38
  • @WhipCored they got to this point by understanding what "Elementary" matrices do. To get an elementary matrix, take the identity matrix and change it however you want your matrix you multiplied to be changed. If you want to switch the first and last rows, your elementary matrix is $$\Big(\begin{smallmatrix}0 & 0 & 1 \ 0 & 1 & 0 \ 1 & 0 & 0\end{smallmatrix}\Big)$$. If you want to scale the first row by two, you get $$\Big(\begin{smallmatrix}2 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{smallmatrix}\Big)$$ They wanted to scale the second row by two, so wrote the corresponding elementary matrix. – Mark Schultz-Wu Oct 24 '16 at 02:32
  • Those are useful ones. You also need ones like $\Big(\begin{smallmatrix}1 & 0 & 0 \ 3 & 1 & 0 \ 0 & 0 & 1 \end{smallmatrix}\Big)$, which adds 3 times the first row to the second row. – NovaDenizen Oct 24 '16 at 02:54