Question

Use the product of matrices ${\displaystyle \left[\begin{array}{ccc}1& 2& -3\\ 3& 2& -2\\ 2& -1& 1\end{array}\right]\left[\begin{array}{ccc}0& 1& 2\\ -7& 7& -7\\ -7& 5& -4\end{array}\right]}$ to solve the following system of equations:

x + 2y – 3z = 6

3x + 2y – 2z = 3

2x  –  y + z = 2

Answer 1

Let ${\displaystyle A=\left[\begin{array}{ccc}1& 2& -3\\ 3& 2& -2\\ 2& -1& 1\end{array}\right]}$

and ${\displaystyle B=\left[\begin{array}{ccc}0& 1& 2\\ -7& 7& -7\\ -7& 5& -4\end{array}\right]}$

${\displaystyle AB=\left[\begin{array}{ccc}1& 2& -3\\ 3& 2& -2\\ 2& -1& 1\end{array}\right]\left[\begin{array}{ccc}0& 1& 2\\ -7& 7& -7\\ -7& 5& -4\end{array}\right]}$

${\displaystyle =\left[\begin{array}{ccc}0-14+21& 1+14-15& 2-14+12\\ 0-14+14& 3+14-10& 6-14+8\\ 0+7-7& 2-7+5& 4+7-4\end{array}\right]}$

${\displaystyle =\left[\begin{array}{ccc}7& 0& 0\\ 0& 7& 0\\ 0& 0& 7\end{array}\right]}$

${\displaystyle =7\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]}$

AB = 7I

${\displaystyle \frac{1}{7}\left(AB\right)=I}$

${\displaystyle \frac{1}{7}B=A^{-1}}$

${\displaystyle A^{-1}=\frac{1}{7}\left[\begin{array}{ccc}0& 1& 2\\ -7& 7& -7\\ -7& 5& -4\end{array}\right]}$

Now, The given system of equation is

x + 2y − 3z = 6

3x + 2y − 2z = 3

2x − y + z = 2

${\displaystyle \left[\begin{array}{ccc}1& 2& -3\\ 3& 2& -2\\ 2& -1& 1\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}6\\ 3\\ 2\end{array}\right]}$

AX = C

X = A^-1 C

${\displaystyle x=\frac{1}{7}\left[\begin{array}{ccc}0& 1& 2\\ -7& 7& -7\\ -7& 5& -4\end{array}\right]\left[\begin{array}{c}6\\ 3\\ 2\end{array}\right]}$   ...(from eqn 1)

${\displaystyle x=\frac{1}{7}\left[\begin{array}{c}0+3+4\\ -42+21-14\\ -42+15-8\end{array}\right]}$

${\displaystyle x=\frac{1}{7}\left[\begin{array}{c}7\\ -35\\ -35\end{array}\right]}$

${\displaystyle \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}1\\ -5\\ -5\end{array}\right]}$

⇒ x = 1, y = −5, z = −5,