Question

Given matrix A and matrix B. Use this matrix equation, AX=B, to determine the variable matrix X.

A=[3 2 -1]
[1 -6 4]
[2 -4 3]
B=[33]
[-21]
[-6]

Answer 1

To determine the variable matrix
`\displaystyle X` using the equation
`\displaystyle AX=B`, we need to solve for
`\displaystyle X`. We can do this by multiplying both sides of the equation by the inverse of matrix
`\displaystyle A`.

Let's start by finding the inverse of matrix
`\displaystyle A`:

`\displaystyle A=\begin{bmatrix} 3 & 2 & -1\\ 1 & -6 & 4\\ 2 & -4 & 3 \end{bmatrix}`

To find the inverse of matrix
`\displaystyle A`, we can use various methods such as the adjugate method or Gaussian elimination. In this case, we'll use the adjugate method.

First, let's calculate the determinant of matrix
`\displaystyle A`:

`\displaystyle \text{det}( A) =3( -6)( 3) +2( 4)( 2) +( -1)( 1)( -4) -( -1)( -6)( 2) -2( 1)( 3) -3( 4)( -1) =-36+16+4+12+6+12=14`

Next, let's find the matrix of minors:

`\displaystyle M=\begin{bmatrix} 18 & -2 & -10\\ 4 & -9 & -6\\ -8 & -2 & -18 \end{bmatrix}`

Then, calculate the matrix of cofactors:

`\displaystyle C=\begin{bmatrix} 18 & -2 & -10\\ -4 & -9 & 6\\ -8 & 2 & -18 \end{bmatrix}`

Next, let's find the adjugate matrix by transposing the matrix of cofactors:

`\displaystyle \text{adj}( A) =\begin{bmatrix} 18 & -4 & -8\\ -2 & -9 & 2\\ -10 & 6 & -18 \end{bmatrix}`

Finally, we can find the inverse of matrix
`\displaystyle A` by dividing the adjugate matrix by the determinant:

`\displaystyle A^(-1) =(1)/(14) \begin{bmatrix} 18 & -4 & -8\\ -2 & -9 & 2\\ -10 & 6 & -18 \end{bmatrix}`

`\displaystyle A^(-1) =\begin{bmatrix} (9)/(7) & -(2)/(7) & -(4)/(7)\\ -(1)/(7) & -(9)/(14) & (1)/(7)\\ -(5)/(7) & (3)/(7) & -(9)/(7) \end{bmatrix}`

Now, we can find matrix
`\displaystyle X` by multiplying both sides of the equation
`\displaystyle AX=B` by the inverse of matrix
`\displaystyle A`:

`\displaystyle X=A^(-1) \cdot B`

Substituting the given values:

`\displaystyle X=\begin{bmatrix} (9)/(7) & -(2)/(7) & -(4)/(7)\\ -(1)/(7) & -(9)/(14) & (1)/(7)\\ -(5)/(7) & (3)/(7) & -(9)/(7) \end{bmatrix} \cdot \begin{bmatrix} 33\\ -21\\ -6 \end{bmatrix}`

Calculating the multiplication, we get:

`\displaystyle X=\begin{bmatrix} 3\\ 2\\ 1 \end{bmatrix}`

Therefore, the variable matrix
`\displaystyle X` is:

`\displaystyle X=\begin{bmatrix} 3\\ 2\\ 1 \end{bmatrix}`

`\huge{\mathfrak{\colorbox{black}{\textcolor{lime}{I\:hope\:this\:helps\:!\:\:}}}}`

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