# Thread: Help finding determinants of matrices

1. ## Help finding determinants of matrices

suppose:
|a b c|
|A| = |d e f | = 6
|g h i |

find the determinants of these matrices:
a)
|a c b |
B = |d f e |
|g i h |

b)2A
c) A^-1 (inverse of A)

Is there a way to find a number value for the determinants of these matrices only knowing that |A| = 6 ?
I know how to find a determinant, but this question is confusing to me since there are no values

2. Hint for a:
recall the theorem which state that if if you replace one column with another the det. isn't changes.(wrong!!!)

for b.

det(2A)=2det(A)

for c.

det(A)=det(A^-1)

3. $|A|= a(ei-fh)-b(di-fg)+c(dh-eg) =aei-afh-bdi+bfg+cdh-ceg= 6$

$|B| =a(fh-ei)-c(dh-eg)+b(di-fg) =-aei+afh+bdi-bfg-cdh+ceg=$ $-(aei-afh-bdi+bfg+cdh-ceg)=-|A| = -6$

5. thanks to both of you. just to clarify, zarathustra is right for b and c ?

6. Hello, seanP!

$\text{Suppose: }\;A \:=\:\left|\begin{array}{ccc} a&b&c \\ d&e&f \\ g&h&1\end{array}\right| \;=\;6$

Find the determinants of these matrices:

$a)\;B \;=\;\left|\begin{array}{ccc}a&c&b \\ d&f&e \\ g&i&h \end{array}\right|$

. . the sign of the determinant is changed.

Therefore: . $B \:=\:-6$

$b)\;2A$

Too easy!

$2A \;=\;2(6) \;=\;12$

$c)\;A^{-1}$

$\text{Since }A\cdot A^{-1} \:=\:I\,\text{ and }\,|I| \:=\:1$

. $\text{we have: }\;6\cdot A^{-1} \:=\:1 \quad\Rightarrow\quad A^{-1} \:=\:\dfrac{1}{6}$

7. Hi Soroban, the OP has stated (not so clearly!) that

$|A| \:=\:\left|\begin{array}{ccc} a&b&c \\ d&e&f \\ g&h&1\end{array}\right| \;=\;6$

so

$A \:=\:\left[\begin{array}{ccc} a&b&c \\ d&e&f \\ g&h&1\end{array}\right] \implies 2A \:=\:2\left[\begin{array}{ccc} a&b&c \\ d&e&f \\ g&h&1\end{array}\right]$

Your solution is for $2|A|\neq 2A$