To understand matrix equations, you should know how to multiply two matrices. Let's say you are showed the result of two matrices multiplication (A . B = C) and you are asked to find the matrix B and you are provided matrix A and C, how will you find it? With the help of the matrix equation. In a matrix equation, the unknown is a matrix. This means that you will denote the unknown matrix as matrix X.

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 A = \bigl(\begin{smallmatrix} 2 & 3 \\ 1 & 2 \end{smallmatrix}\bigr) \quad B = \bigl(\begin{smallmatrix} 3 & 1 \\ 2 & -5 \end{smallmatrix}\bigr). Solve the equation: A · X = B

\begin{vmatrix} 2 & 3 \\ 1 & 2 \end{vmatrix}

Find the determinant of the above matrix.

\left | A \right | = ad - bc

\begin{vmatrix} 2 & 3 \\ 1 & 2 \end{vmatrix} = (2)(2) - (3)(1) = 4 - 3 = 1

\left | A \right | = 1 \neq 0, this means that there is an inverse { A }^{ -1 }

{ A }^{ -1 } = \frac { 1 }{ \left | A \right | } \times \begin{vmatrix} d & -b \\ -c & a \end{vmatrix}

{ A }^{ -1 } = \frac { 1 }{ 1 } \times \begin{vmatrix} 2 & -3 \\ -1 & 2 \end{vmatrix}

{ A }^{ -1 } = \begin{pmatrix} 2 & -3 \\ -1 & 2 \end{pmatrix}

 

{ A }^{ -1 } (A . X) = { A }^{ -1 } . B

({ A }^{ -1 } . A) . X = { A }^{ -1 } . B

I . X  = { A }^{ -1 } . B

X = { A }^{ -1 } . B

X = \begin{pmatrix} 2 & -3 \\ -1 & 2 \end{pmatrix} \begin{pmatrix} 3 & 1 \\ 2 & -5 \end{pmatrix} = \begin{pmatrix} 0 & 17 \\ 1 & -11 \end{pmatrix}

 

2. Given the matrices A = \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix} \quad B = \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix} \quad C = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}. Solve the equation: X · A + B = C

 

\left | A \right | = ad - bc

\begin{vmatrix} 1 & 1 \\ 1 & 2 \end{vmatrix} = (1)(2) - (1)(1) = 2 - 1 = 1

\left | A \right | = 1 \neq 0, this means that there is an inverse { A }^{ -1 }

{ A }^{ -1 } = \frac { 1 }{ \left | A \right | } \times \begin{vmatrix} d & -b \\ -c & a \end{vmatrix}

{ A }^{ -1 } = \frac { 1 }{ 1 } \times \begin{vmatrix} 2 & -1 \\ -1 & 1 \end{vmatrix}

{ A }^{ -1 } = \begin{pmatrix} 2 & -1 \\ -1 & 1 \end{pmatrix}

 

(X . A + B) - B = C - B

X . A + (B - B) = 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 }

X = [\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} - \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix}] \begin{pmatrix} 2 & -1 \\ -1 & 1 \end{pmatrix}

X = \begin{pmatrix} -1 & 1 \\ 1 & -2 \end{pmatrix} \begin{pmatrix} 2 & -1 \\ -1 & 1 \end{pmatrix} = \begin{pmatrix} -3 & 2 \\ 4 & -3 \end{pmatrix}

 

3.Solve the matrix equation:

A · X + 2 · B = 3 · C

A = \begin{pmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 1 & 1 \end{pmatrix} \quad B = \begin{pmatrix} 0 & 1 & 1 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix} \quad C = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{pmatrix}

 

\left | A \right | = 1 \neq 0, this means that there is an inverse { A }^{ -1 }

{ A }^{ -1 } = \begin{pmatrix} 1 & 0 & 0 \\ -1 & 1 & 0 \\ 0 & -1 & 1 \end{pmatrix}

 

(A . X + 2 . B) - 2 . B = 3 . C - 2B

A . X + (2 . B - 2 .B) = 3 . C - 2B

A . X = 3 . C - 2B

A . X . { A }^{ -1 } = (3 . C - 2B) . { A }^{ -1 }

X . (A . { A }^{ -1 }) = (3 . C - 2B) . { A }^{ -1 }

X . I = (3 . C - 2B) . { A }^{ -1 }

X = (3 . C - 2B) . { A }^{ -1 }

 

X = [3 \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{pmatrix} - 2 \begin{pmatrix} 0 & 1 & 1 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix} ] . \begin{pmatrix} 1 & 0 & 0 \\ -1 & 1 & 0 \\ 0 & -1 & 1 \end{pmatrix}

X = [ \begin{pmatrix} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 3 & 0 & 3 \end{pmatrix} - \begin{pmatrix} 0 & 2 & 2 \\ 2 & 0 & 0 \\ 0 & 0 & 2 \end{pmatrix} ] . \begin{pmatrix} 1 & 0 & 0 \\ -1 & 1 & 0 \\ 0 & -1 & 1 \end{pmatrix}

X = \begin{pmatrix} 3 & -2 & -2 \\ -2 & 3 & 0 \\ 3 & 0 & 1 \end{pmatrix} . \begin{pmatrix} 1 & 0 & 0 \\ -1 & 1 & 0 \\ 0 & -1 & 1 \end{pmatrix}

X = \begin{pmatrix} 3 & -2 & -2 \\ -5 & 5 & 2 \\ 5 & -3 & 1 \end{pmatrix}

 

4.Solve the matrix equation:

AX  + 2B = 3C

A = \begin{pmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 1 & 1 \end{pmatrix} \quad B = \begin{pmatrix} 0 & 1 & 1 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix} \quad C = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{pmatrix}

AX +2B - 2B = 3C - 2B

AX = 3C - 2B

A . X . { A }^{ -1 } = (3C - 2B) . { A }^{ -1 }

X . (A . { A }^{ -1 }) = (3C - 2B) . { A }^{ -1 }

X . I = (3C - 2B) . { A }^{ -1 }

X = (3C - 2B) . { A }^{ -1 }

 

{ A }^{ -1 } = \begin{pmatrix} 1 & 0 & 0 \\ -1 & 1 & 0 \\ 0 & -1 & 1 \end{pmatrix}

X = [3 \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{pmatrix} - 2 \begin{pmatrix} 0 & 1 & 1 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}] . \begin{pmatrix} 1 & 0 & 0 \\ -1 & 1 & 0 \\ 0 & -1 & 1 \end{pmatrix}

X = [ \begin{pmatrix} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 3 & 0 & 3 \end{pmatrix} - \begin{pmatrix} 0 & 2 & 2 \\ 2 & 0 & 0 \\ 0 & 0 & 2 \end{pmatrix}] . \begin{pmatrix} 1 & 0 & 0 \\ -1 & 1 & 0 \\ 0 & -1 & 1 \end{pmatrix}

X = \begin{pmatrix} 3 & -2 & -2 \\ -2 & 3 & 0 \\ 3 & 0 & 1 \end{pmatrix}  . \begin{pmatrix} 1 & 0 & 0 \\ -1 & 1 & 0 \\ 0 & -1 & 1 \end{pmatrix}

X = \begin{pmatrix} 3 & -2 & -2 \\ -5 & 5 & 2 \\ 5 & -3 & 1 \end{pmatrix}

 

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

\left\{\begin{matrix} x + y + z = 6 \\ x + 2y + 5z = 12 \\ x + 4y + 25z = 36 \end{matrix}\right

\begin{pmatrix} 1 & 1 & 1 \\ 1 & 2 & 5 \\ 1 & 4 & 25 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 6 \\ 12 \\ 36 \end{pmatrix}

A . X = C

A . { A }^{ -1 } . X = C . { A }^{ -1 }

I . X = C . { A }^{ -1 }

X = C . { A }^{ -1 }

{ A }^{ -1 } = \frac { 1 }{ 12 } \begin{pmatrix} 30 & -21 & 3 \\ -20 & 24 & -4 \\ 2 & -3 & 1 \end{pmatrix}

\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \frac { 1 }{ 12 } \begin{pmatrix} 30 & -21 & 3 \\ -20 & 24 & -4 \\ 2 & -3 & 1 \end{pmatrix} \begin{pmatrix} 6 \\ 12 \\ 36 \end{pmatrix} = \frac { 1 }{ 12 } \begin{pmatrix} 36 \\ 24 \\ 12 \end{pmatrix}

\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 3 \\ 2 \\ 1 \end{pmatrix}

x = 3, \quad x = 2, \quad z = 1

Need a Maths teacher?

Did you like the article?

1 Star2 Stars3 Stars4 Stars5 Stars 3.67/5 - 3 vote(s)
Loading...

Hamza

Hi! I am Hamza and I am from Pakistan. My hobbies are reading, writing and playing chess. Currently, I am a student enrolled in the Chemical Engineering Bachelor program.