How To Find Unknown Matrix

To empathise matrix equations, you should know how to multiply ii matrices. Allow'southward say you are showed the issue of 2 matrices multiplication ( $A . B = C$ ) and you are asked to discover the matrix B and you are provided matrix A and C, how will y'all observe it? With the help of the matrix equation. In a matrix equation, the unknown is a matrix. This means that you volition denote the unknown matrix asmatrix X.

A · X = B

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

If the equation is of blazon X · A = B, the members must postmultiply (multiply to the right) because matrix multiplication is non 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 higher up 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 in that location is an inverse ${ A }^{ -1 }$

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

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

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

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

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

$X . A = C - B$

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

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

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

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

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

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

iii.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 ways that there is an changed ${ A }^{ -1 }$

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

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

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

$A . X = 3 . C - 2B$

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

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

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

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

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

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

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

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

4.Solve the matrix equation:

$AX + 2B = 3C$

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

$AX = 3C - 2B$

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

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

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

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

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

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

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

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

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

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

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

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

$A . X = C$

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

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

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

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

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

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

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

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