Use the appropriate inverse to solve the following for X.

Steps to Solve

1. Identify the Matrix Equation

We have the matrix equation:

$$ A \cdot X = B $$

where

$$ A = \begin{pmatrix} 1 & -6 & -4 \ 2 & -10 & -9 \ -1 & 6 & 5 \end{pmatrix}, \quad B = \begin{pmatrix} -5 \ -4 \ 3 \end{pmatrix} $$

2. Calculate the Inverse of Matrix A

To isolate $X$, we need to find the inverse of matrix $A$, denoted as $A^{-1}$.

If $A$ is invertible, the solution is given by:

$$ X = A^{-1} \cdot B $$

Calculate the determinant of $A$ to ensure it's invertible:

$$ \text{det}(A) = 1(-10 \cdot 5 - (-9)(6)) + 6(2 \cdot 5 - (-9)(-1)) - 4(2 \cdot 6 - (-10)(-1)) $$

3. Compute the Determinant

Calculating the individual components:

• For first term: $1(-50 + 54) = 1(4) = 4$
• For second term: $6(10 - 9) = 6(1) = 6$
• For third term: $-4(12 - 10) = -4(2) = -8$

Summing them up gives:

$$ \text{det}(A) = 4 + 6 - 8 = 2 $$

Since $\text{det}(A) \neq 0$, $A$ is invertible.

4. Find the Inverse of A

Using the formula for the inverse of a 3x3 matrix, we calculate $A^{-1}$. The formula involves the determinant and the matrix of minors, cofactors, and adjugates. For simplification:

$$ A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{Adj}(A) $$

5. Multiply A Inverse by B

After calculating $A^{-1}$, multiply it by matrix $B$:

$$ X = A^{-1} \cdot B $$

6. Final Calculation of X

Perform the multiplication to find $X$.