Question

Find values for the the entries in the second matrix to satisfy this equation:
(check comments for matrcies - edit 1) sorry, the first entry is suppose to be a +1 NOT -1. Sorry )

-based from the following matricies I deduce we use the following; AX=B, by finding A inverse from 1/det A [adj A] = A inverse. However when checking my answer, it doesnt equal to B. What am I doing wrong?

Answer 1

\[\left[\begin{matrix}-1 & -2 \\ -3 & 7 \end{matrix}\right]\] & [X] = \[\left[\begin{matrix}-5 & 2 \\ 16 & -6\end{matrix}\right]\]

Answer 2

matrices, lol.

Answer 3

what is B?

Answer 4

B is the last matrix , \[\left[\begin{matrix}-5 & 2 \\ 16 & -6\end{matrix}\right]\]

Answer 5

Ax =B , so, x = A^- B ,now find A^- show me your work, please

Answer 6

I am assuming you mean x = A inverse B? Correct?

Answer 7

no need to assume, you are right, just show me A inverse

Answer 8

If so, I have A ^ -1 =
1/ det A *[ adj A] =
1 *\[\left[\begin{matrix}7 & 2 \\ 3 & 1\end{matrix}\right]\].

X = 1* \[\left[\begin{matrix}7 & 2 \\ 3 & 1\end{matrix}\right]\]. * \[\left[\begin{matrix}-5 & 2 \\ 16 & -6\end{matrix}\right]\]

Answer 9

ok, I work with you. there is something wrong, first off, det A =?

Answer 10

det A should be ad-bc. In this case, 1/ (1*7)- (-2*-3)
det A = 1/7-6 = det A = 1

Answer 11

not, det A = (-1)*7 - ( (-3)*(-2)) = -7 -6 = -13

Answer 12

- 1?!

Answer 13

-13 not -1

Answer 14

yes, I know, but isn't ad-bc? in this case, it should be (1*7)-((-3*-2))

Answer 15

but the first entry is -1 , not 1

Answer 16

pellet. Sorry Wow, I am completely blind. It should be \[\left[\begin{matrix}1 & -2 \\-3 & 7\end{matrix}\right]\]

Answer 17

ok, let start over, so the det = 1, ok now the adj A=?

Answer 18

hey, no need to do , just exchange the place of entries.

Answer 19

your A^- =