linear algebra inverse of matrix problem

[bronze laurel]

Here i can see that the inverse matrix determinant will be 11 which is the same in all options. Next step is here i do minor of a11 which is 3 all options wrong

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[bronze laurel]

Question seems wrong

[steep rapids]

A and B are inverses of one another iff AB = BA = I

the question is wrong if none of the options satisfy this requirement

[obsidian cairn]

It's not, WA verifies it's a)

[steep rapids]

what I'm confused about is this

why are we given this information?

sure you can manipulate it

$$ A(A^2-6A+5I) = -11I $$

[balmy vale]

aL

[steep rapids]

which also shows what the inverse is

the inverse is unique, so it can be determined from it

[obsidian cairn]

The charpoly here is $\chi_A = -X^3 + 6X^2 - 5X - 11$. This tells us that $\det(A) = -11$. Now $A^{-1} = \frac{1}{\det(A)} \tilde A$ where $\tilde A$ is the adjugate matrix. \ \hr \ Now what's left is just to compute $\tilde{a_{ij}}$ for an entry that differs in the answers. Seems that is $\tilde{a_{32}}$. So: [\tilde{a_{32}} = (-1)^{3 + 2} \cdot \det(A^{23}) = - \det \mrm{1 & 1 \ 2 & -1} = 3.] So $a_{32}^{-1} = \frac{1}{-11} \tilde {a_{32}} = \frac{1}{11} \cdot (-3)$. Thus it has to be (a).

[bronze laurel]

Failed

[obsidian cairn]

?

[bronze laurel]

??? Where is latex drawn?

[obsidian cairn]

Here

[bronze laurel]

By the way we can see that det is -11

And the Minus goes inside

Then a31 which is 5 A

+close