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Let A be an invertible n x n matrix, and let B be an n x p matrix. (a) Show that A^-1 B is a solution of the matrix equation AX = B, and that A^-1 B is the unique solution of AX = B. (b) Use elementary matrices to show that any sequence of elementary row operations that reduces A to I also transforms B into A^-1 B: If [A B] ~ ... ~ [I X], then X = A^-1 B. Comment: In applications, if A is larger than 2 x 2, then finding A^-1 B by row reducing [A B] is much faster than computing both A^-1 and A^-1 B.

          Let A be an invertible n x n matrix, and let B be an n x p matrix.
(a) Show that A^-1 B is a solution of the matrix equation AX = B, and that A^-1 B is the unique solution of AX = B.
(b) Use elementary matrices to show that any sequence of elementary row operations that reduces A to I also transforms B into A^-1 B:
If [A B] ~ ... ~ [I X], then X = A^-1 B.
Comment: In applications, if A is larger than 2 x 2, then finding A^-1 B by row reducing [A B] is much faster than computing both A^-1 and A^-1 B.

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Let A be an invertible n x n matrix, and let B be an n x p matrix.
(a) Show that A^-1 B is a solution of the matrix equation AX = B, and that A^-1 B is the unique solution of AX = B.
(b) Use elementary matrices to show that any sequence of elementary row operations that reduces A to I also transforms B into A^-1 B:
If [A B] ... [I X], then X = A^-1 B.
Comment: In applications, if A is larger than 2 x 2, then finding A^-1 B by row reducing [A B] is much faster than computing both A^-1 and A^-1 B.

Added by Cristina W.

Calculus: Early Transcendentals

James Stewart 8th Edition

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Solved by Expert Adi S

06/05/2023

Step 1

To do this, we can simply multiply both sides of the equation by $A^{-1}$: $$A^{-1}(AX) = A^{-1}B$$ Since $A^{-1}A = I$, where $I$ is the identity matrix, we have: $$IX = A^{-1}B$$ Since $IX = X$, we have: $$X = A^{-1}B$$ So, $A^{-1}B$ is a solution of the Show more…

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Let A be an invertible n x n matrix, and let B be an n x p matrix. (a) Show that A^(-1)B is a solution of the matrix equation AX = B, and that A^(-1)B is the unique solution of AX = B. (b) Use elementary matrices to show that any sequence of elementary row operations that reduces A to I also transforms B into A^(-1)B: If [A B] ~ ... ~ [I X], then X = A^(-1)B. Comment: In applications, if A is larger than 2 x 2, then finding A^(-1)B by row reducing [A B] is much faster than computing both A^(-1) and A^(-1)B.

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