15. Sean A y C dos matrices n × n invertibles y D una matriz n × n. Para cada una de las siguien

Step 1: To solve for Y in each equation, we need to isolate Y. We can do this by multiplying bot

15. Sean A y C dos matrices n × n invertibles y D una matriz...

15. Sean A y C dos matrices $n \times n$ invertibles y D una matriz $n \times n$. Para cada una de las siguientes ecuaciones en Y a) $C \cdot Y \cdot A = D$ b) $A \cdot C \cdot Y = D$ c) $Y \cdot A \cdot C = D$ d) $C \cdot A \cdot Y = D$ e) $Y \cdot C \cdot A = D$ indique la opción que contiene la solución dentro de la lista 1) $Y = A^{-1} \cdot C^{-1} \cdot D$ 2) $Y = A^{-1} \cdot D \cdot C^{-1}$ 3) $Y = D \cdot A^{-1} \cdot C^{-1}$ 4) $Y = C^{-1} \cdot A^{-1} \cdot D$ 5) $Y = C^{-1} \cdot D \cdot A^{-1}$ 6) $Y = D \cdot C^{-1} \cdot A^{-1}$

Transcript

00:01 In this problem, we are going to solve for a matrix d given a certain matrix equation.

00:06 Now, here it is assumed that all the matrices are of order n and they are invertible.

00:12 And the given matrix equation is c transpose b inverse a square b, a, c inverse b, a to the power minus 2, d transpose c to the power minus 2, this is equals to c transpose.

00:42 Now we need to solve for the matrix b.

00:46 Now in order to do this, we will pre -multiply the left -hand side of the equation with the inverse of this matrix and we will post -multiply with the inverse of this matrix.

01:01 So what we get is c -t -b -inverse a -square b -a -c -inverse times c -t -b -inverse a -square -b -a -c -inverse times d times a -to -the -power -minus 2 b -t -c -to -the -power -2 -inverse.

01:27 Times a to the power minus 2, b t, c to the power minus 2 inverse.

01:34 And because of this we need to do the same operation on the right hand side.

01:38 So the right hand side will become c t b inverse a square b, a, c inverse, times what was on the right hand side originally, which is c to the power t or rather c transpose.

01:58 And with that we post multiply a to the power minus 2, b transpose, c minus 2, inverse.

02:06 Now on the left hand side, since we have that the product of a matrix with its inverse is equals to the identity matrix, so the left hand side reduces to the identity matrix times the matrix d times the identity matrix.

02:23 On the right hand side, we need to simplify this.

02:27 Now, this will be obtained as the first one will be the inverses, the product of the inverses, but in the reverse order.

02:37 So first of all, we have c inverse, inverse, then a inverse, then b inverse, and so on...

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