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 Martin Sleziak
asked Feb 24, năm trước at 9:58 user131127user131127
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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 CommunityBot
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answered Feb 24, 2014 at 10:05 7raiden77raiden7
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Note that the matrix multiplication is not commutative, i.e, you"ll not always have: \$AB = BA\$.

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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 PNT
answered Feb 24, năm trước at 10:11
user49685user49685
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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
user127151user127151
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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
CasteelsCasteels
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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
ViliVili
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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
H-HH-H
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