Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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
share|cite|improve this question
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
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 $36$.

share|cite|improve this answer
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)

share|cite|improve this answer

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 $$\det\begin{bmatrix}g - 2a&h - 2b&i - 2c\\ 3a&3b&3c\\2d&2e&2f \end{bmatrix}=3\times 2\times \det\begin{bmatrix}a&b&c\\ d&e&f\\ g &h &i \end{bmatrix}=6\times 6=36$$

share|cite|improve this answer

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$\times6=36$

share|cite|improve this answer

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 $36$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.