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I know the theory of determinants, but I have no idea how to apply it to this problem.

Suppose $$\det\begin{bmatrix}a&b&c\\ d&e&f\\ g&h&i \end{bmatrix} = 6$$ What is the value of $$\det\begin{bmatrix}g - 2a&h - 2b&i - 2c\\ 3a&3b&3c\\2d&2e&2f \end{bmatrix}$$

The options for the answers are:

  1. 4
  2. 36
  3. 24
  4. -24
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Elementary transformation on matrix do not change the determinant. –  Willard Zhan Aug 18 '13 at 11:35
Ah, thanks 😆 I remember reading that. What i did not add onto this question though is the fact that this is a multiple choice question. I will edit the question with the options. Of course, the options could be wrong. –  Leon Aug 18 '13 at 11:47

5 Answers 5

up vote 11 down vote accepted

Also, for the fast way not the theoretical approach, you can set $a=2,e=3,i=1$ and other arrays to be zero and just examine the value of determinant you're asked. So, it gives you $36$.

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This method is usefull to verify a long calculus (+1). –  user63181 Aug 18 '13 at 11:49
Nice work, @Babak! –  amWhy Aug 19 '13 at 0:38

$$\det\begin{bmatrix}g - 2a&h - 2b&i - 2c\\ 3a&3b&3c\\2d&2e&2f \end{bmatrix}$$

$$=\det\begin{bmatrix}g&h&i\\ 3a&3b&3c\\2d&2e&2f \end{bmatrix}$$ (Applying $R_1'=R_1+\frac23 R_2$)

$$=3\cdot2\cdot \det\begin{bmatrix}g&h&i\\ a&b&c\\d&e&f \end{bmatrix}$$ (Taking out $3,2$ as common factors from the $R_2,R_3$ respectively)

$$=(-1)3\cdot2\cdot \det\begin{bmatrix}a&b&c\\ g&h&i\\d&e&f \end{bmatrix}$$ (Exchanging $R_1,R_2$ resulting '-' sign )

$$=(-1)(-1)3\cdot2\cdot \det\begin{bmatrix}a&b&c\\d&e&f\\ g&h&i \end{bmatrix}$$

(Exchanging $R_2,R_3$ resulting '-' sign again)

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We switch the first and the second row and the second and third row so

$$\det\begin{bmatrix}g - 2a&h - 2b&i - 2c\\ 3a&3b&3c\\2d&2e&2f \end{bmatrix}=3\times 2\times \det\begin{bmatrix}a&b&c\\ d&e&f\\ g - 2a&h - 2b&i - 2c\end{bmatrix}$$ and we add $2\times$ the first row to the third row we find $$\det\begin{bmatrix}g - 2a&h - 2b&i - 2c\\ 3a&3b&3c\\2d&2e&2f \end{bmatrix}=3\times 2\times \det\begin{bmatrix}a&b&c\\ d&e&f\\ g &h &i \end{bmatrix}=6\times 6=36$$

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det$\begin{bmatrix}g - 2a&h - 2b&i - 2c\\ 3a&3b&3c\\2d&2e&2f \end{bmatrix}$

=3$\times 2\times \det\begin{bmatrix}g - 2a&h - 2b&i - 2c\\ a&b&c\\d&e&f \end{bmatrix}$

=6$\times -1\times \det\begin{bmatrix}a&b&c\\g - 2a&h - 2b&i - 2c \\d&e&f \end{bmatrix}$R1 ->R2, R2 ->R1

= 6$\times -1\times -1\times\det\begin{bmatrix}a&b&c\\d&e&f \\g - 2a&h - 2b&i - 2c \end{bmatrix}$R2 ->R3, R3 ->R2

= 6$\times\det\begin{bmatrix}a&b&c\\d&e&f \\g - 2a&h - 2b&i - 2c \end{bmatrix}$ = 6$\times\det\begin{bmatrix}a&b&c\\d&e&f \\g&h&i\end{bmatrix}$ R3 -> 2R1+R3

As, $\det\begin{bmatrix}a&b&c\\d&e&f \\g&h&i\end{bmatrix}$=6

= 6$\times6=36$

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determined of square matrix is the multiplication of all term present on diagonal. so you can set $a,e,i$ such that is gives $6$ and all other terms $0$. Now put all values in second matrix and evaluate the result it is $36$. let $a=1, e=1,i=6$

$$\det\begin{bmatrix} -2&0&6\\ 3&0&0\\0&2&0\end{bmatrix}$$ it gives $36$

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