If \[ A=\left[\begin{array}{cc} 3 & 6 \\ 2 & -5 \end{array}\right], \] then \[ A^{-1}=\left[\begin{array}{cc|cc} \frac{5}{27} & & \frac{6}{27} & \\ \frac{2}{27} & \frac{1}{9} & \\ \hline \end{array}\right] \] Given \( \vec{b}=\left[\begin{array}{l}-3 \\ -4\end{array}\right] \), solve \( A \vec{x}=\vec{b} \). \[ \vec{x}=\left[\begin{array}{r} \sigma_{0} \end{array}\right] \]

To solve the linear system \( A \vec{x} = \vec{b} \) where \[ A = \begin{bmatrix} 3 & 6 \\ 2 & -5 \end{bmatrix}, \quad \vec{b} = \begin{bmatrix} -3 \\ -4 \end{bmatrix}, \] we can use the inverse of matrix \( A \) to find the solution vector \( \vec{x} \). ### Step 1: Compute the Inverse of \( A \) First, calculate the determinant of \( A \): \[ \text{det}(A) = (3)(-5) - (6)(2) = -15 - 12 = -27 \] The inverse of \( A \) is then: \[ A^{-1} = \frac{1}{\text{det}(A)} \begin{bmatrix} -5 & -6 \\ -2 & 3 \end{bmatrix} = \frac{1}{-27} \begin{bmatrix} -5 & -6 \\ -2 & 3 \end{bmatrix} = \begin{bmatrix} \frac{5}{27} & \frac{6}{27} \\ \frac{2}{27} & -\frac{1}{9} \end{bmatrix} \] ### Step 2: Multiply \( A^{-1} \) by \( \vec{b} \) Now, compute \( \vec{x} = A^{-1} \vec{b} \): \[ \vec{x} = \begin{bmatrix} \frac{5}{27} & \frac{6}{27} \\ \frac{2}{27} & -\frac{1}{9} \end{bmatrix} \begin{bmatrix} -3 \\ -4 \end{bmatrix} \] Calculate each component of \( \vec{x} \): \[ x_1 = \left(\frac{5}{27}\right)(-3) + \left(\frac{6}{27}\right)(-4) = -\frac{15}{27} - \frac{24}{27} = -\frac{39}{27} = -\frac{13}{9} \] \[ x_2 = \left(\frac{2}{27}\right)(-3) + \left(-\frac{1}{9}\right)(-4) = -\frac{6}{27} + \frac{4}{9} = -\frac{2}{9} + \frac{4}{9} = \frac{2}{9} \] ### Final Solution \[ \vec{x} = \begin{bmatrix} -\dfrac{13}{9} \\ \dfrac{2}{9} \end{bmatrix} \]