## Precalculus (6th Edition) Blitzer

a) The required form is, $\left[ \begin{array}{*{35}{r}} 3 & 5 \\ 2 & -3 \\ \end{array} \right]\left[ \begin{matrix} x \\ y \\ \end{matrix} \right]=\left[ \begin{array}{*{35}{r}} 9 \\ -13 \\ \end{array} \right]$ b) The inverse of matrix $A=\left[ \begin{array}{*{35}{r}} 3 & 5 \\ 2 & -3 \\ \end{array} \right]$ is, ${{A}^{-1}}=\frac{1}{19}\left[ \begin{array}{*{35}{r}} 3 & 5 \\ 2 & -3 \\ \end{array} \right]$ c) The solution is, $x=-2,y=3$
(a) Consider the system of equations \begin{align} & 3x+5y=9 \\ & 2x-3y=-13 \end{align} Therefore, in matrix form, the system of equations can be written as $AX=B$ Where $A=\left[ \begin{array}{*{35}{r}} 3 & 5 \\ 2 & -3 \\ \end{array} \right];B=\left[ \begin{array}{*{35}{r}} 9 \\ -13 \\ \end{array} \right];X=\left[ \begin{matrix} x \\ y \\ \end{matrix} \right]$ The representation of the system is $\left[ \begin{array}{*{35}{r}} 3 & 5 \\ 2 & -3 \\ \end{array} \right]\left[ \begin{matrix} x \\ y \\ \end{matrix} \right]=\left[ \begin{array}{*{35}{r}} 9 \\ -13 \\ \end{array} \right]$. (b) Consider the given matrix $A=\left[ \begin{array}{*{35}{r}} 3 & 5 \\ 2 & -3 \\ \end{array} \right]$ Consider the determinant of the matrix \begin{align} & \left| A \right|=\left| \begin{array}{*{35}{r}} 3 & 5 \\ 2 & -3 \\ \end{array} \right| \\ & =-9-10 \\ & =-19 \end{align} Consider the adjoint of the matrix \begin{align} & \text{adj}\left( A \right)={{\left[ \begin{array}{*{35}{r}} +\left( -3 \right) & -\left( 2 \right) \\ -\left( 5 \right) & +3 \\ \end{array} \right]}^{t}} \\ & ={{\left[ \begin{array}{*{35}{r}} -3 & -2 \\ -5 & 3 \\ \end{array} \right]}^{t}} \\ & =\left[ \begin{array}{*{35}{r}} -3 & -5 \\ -2 & 3 \\ \end{array} \right] \end{align} Therefore, the inverse of the matrix is given by: \begin{align} & {{A}^{-1}}=\frac{1}{\left| A \right|}\text{adj}\left( A \right) \\ & =-\frac{1}{19}\left[ \begin{array}{*{35}{r}} -3 & -5 \\ -2 & 3 \\ \end{array} \right] \\ & =\frac{1}{19}\left[ \begin{array}{*{35}{r}} 3 & 5 \\ 2 & -3 \\ \end{array} \right] \end{align} The inverse matrix is ${{A}^{-1}}=\frac{1}{19}\left[ \begin{array}{*{35}{r}} 3 & 5 \\ 2 & -3 \\ \end{array} \right]$. (c) Consider the given system of equations in matrix form $AX=B$ Therefore, the solution of the equations is $X={{A}^{-1}}B$ Consider \begin{align} & X=\frac{1}{19}\left[ \begin{array}{*{35}{r}} 3 & 5 \\ 2 & -3 \\ \end{array} \right]\left[ \begin{array}{*{35}{r}} 9 \\ -13 \\ \end{array} \right] \\ & =\frac{1}{19}\left[ \begin{array}{*{35}{r}} 27-65 \\ 18+39 \\ \end{array} \right] \\ & =\frac{1}{19}\left[ \begin{array}{*{35}{r}} -38 \\ 57 \\ \end{array} \right] \\ & =\left[ \begin{array}{*{35}{r}} -2 \\ 3 \\ \end{array} \right] \end{align} The solution of the system is $x=-2,y=3$.