In a matrix equation, the unknown is a matrix.

A · X = B

To solve, check that the matrix is invertible, if it is, premultiply (multiply to the left) both sides by the matrix inverse of A.

If the equation is of type X · A = B, the members must postmultiply (multiply to the right) because matrix multiplication is not commutative.

1. Given the matrices . Solve the equation: A · X = B

|A|=1 ≠ 0, there is the inverse A-1 .

A-1 (A · X) = A-1 · B

(A-1 · A) · X = A-1 · B

I · X = A-1 · B

X = A-1 · B

2. Given the matrices . Solve the equation: X · A + B = C

|A| = 1 ≠ 0

(X · A + B) − B = C B

X · A + (B B) = C B

X · A + 0 = C B

X · A = C B

X · A · A-1 = (C B) · A-1

X (A · A-1 ) = (C B) · A-1

X · I = (C B) · A-1

X = (C B) · A-1

3.Solve the matrix equation:

A · X + 2 · B = 3 · C

|A| = 1 ≠ 0

(A · X +2 · B) − 2 · B = 3 · C − 2B

A · X + ( 2 · B − 2 · B) = 3 · C − 2B

A · X + 0= 3 · C − 2B

A · X = 3 · C − 2B

(A−1 · A) · X = A−1 · (3 · C − 2B)

I · X = A−1 · (3 · C − 2B)

X = A−1 · (3 · C − 2B)

4.Solve the matrix equation:

To solve a system of linear equations, it can be transformed into a matrix equation and then solved.

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Emma

I am passionate about travelling and currently live and work in Paris. I like to spend my time reading, gardening, running, learning languages and exploring new places.

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