The question says lớn find the value of each matrix expression where A & B are the invertible 3 x 3 matrices such that $$A^-1 = left(eginarrayccc1& 2& 3\ 2& 0& 1\ 1& 1& -1endarray ight)$$ and $$B^-1=left(eginarrayccc2 &-1 &3\ 0& 0 &4\ 3& -2 và 1endarray ight)$$

The actual question is to lớn find $ (AB)^-1$.

$ (AB)^-1$ is just $ A^-1B^-1$ & we already know matrices $ A^-1$ and $ B^-1$ so taking the hàng hóa should give us the matrix $$left(eginarrayccc11 &-7 &14\ 7& -4 &7\ -1& 1 & 6endarray ight)$$yet the answer is $$left(eginarrayccc 3 &7 &2 \ 4& 4 &-4\ 0 và 7 và 6 endarray ight)$$

What am I not understanding about the problem or what am I doing wrong? Isn"t this just matrix multiplication?


linear-algebra matrices inverse matrix-equations
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edited Feb 11, năm ngoái at 13:42
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Martin Sleziak
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asked Feb 24, năm trước at 9:58
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Actually the inverse of matrix hàng hóa does not work in that way. Suppose that we have two invertible matrices, $A$ và $B$. Then it holds:$$(AB)^-1=B^-1A^-1,$$and, in general:$$left(prod_k=0^NA_k ight)^-1=prod_k=0^NA^-1_N-k$$


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edited May 11, 2017 at 19:34
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answered Feb 24, 2014 at 10:05
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7raiden77raiden7
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Note that the matrix multiplication is not commutative, i.e, you"ll not always have: $AB = BA$.

Bạn đang xem: Step

Now, say the matrix $A$ has the inverse $A^-1$ (i.e $A cdot A^-1 = A^-1cdot A = I$); và $B^-1$ is the inverse of $B$ (i.e $Bcdot B^-1 = B^-1 cdot B = I$).

Claim

$B^-1A^-1$ is the inverse of $AB$. So basically, what I need to prove is: $(B^-1A^-1)(AB) = (AB)(B^-1A^-1) = I$.

Note that, although matrix multiplication is not commutative, it is however, associative. So:

$(B^-1A^-1)(AB) = B^-1(A^-1A)B = B^-1IB = (B^-1I)B = B^-1B=I$

$(AB)(B^-1A^-1) = A(BB^-1)A^-1 = A^-1IA = (A^-1I)A = A^-1A=I$

So, the inverse if $AB$ is indeed $B^-1A^-1$, and NOT $A^-1B^-1$.


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edited May 22, 2021 at 19:15
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Not really. Matrices vày not follow exponential laws. In fact, $(AB)^-1=B^-1A^-1$. Here is the proof:

Let $I$ be a 3 by 3 identity matrix. If $A$ & $B$ are 3 by 3 invertible matrices, then:$$eginalign*(AB)(AB)^-1&=I\(A^-1AB)(AB)^-1&=A^-1I\(IB)(AB)^-1&=A^-1\B(AB)^-1&=A^-1\B^-1B(AB)^-1&=B^-1A^-1\I(AB)^-1&=B^-1A^-1\(AB)^-1&=B^-1A^-1endalign*$$


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answered Feb 24, 2014 at 10:08
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$(AB)^-1$ is not equal to lớn $A^-1B^-1$, but it is equal to lớn $B^-1A^-1$.


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answered Feb 24, năm trước at 10:03
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Intuitively, think of matrices as linear operators. To lớn reverse a composition of operators you have khổng lồ re-run it backwards. So if you want the inverse of the operator $A(B(x))$ on vector $x$, you need to first reverse $A$ và then reverse $B$, so that $A^-1(A(B(x))) = B(x)$, so $B^-1(A^-1(A(B(x)))) = B^-1Bx = x$


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answered May 25, 2021 at 16:22
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I have some personal opinions which might perfect it (many students made mistakes about this), in your case it works fine$$(AB)^-1=B^-1A^-1$$Because you already have the fact that $A,B$ are both square matrix and invertible, but suppose $A$ is $m imes n$ matrix, and $B$ is $n imes m$ matrix, then $AB$ is $m imes m$ matrix which might be invertible, but in this case we don"t even have square matrix so we could never have such fomular.


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edited Jun 15, 2018 at 1:24
answered Jun 14, 2018 at 22:16
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