Question

If A = ${\displaystyle \left[\begin{array}{ccc}-3& -2& -4\\ 2& 1& 2\\ 2& 1& 3\end{array}\right]}$, B = ${\displaystyle \left[\begin{array}{ccc}1& 2& 0\\ -2& -1& -2\\ 0& -1& 1\end{array}\right]}$ then find AB and use it to solve the following system of equations:

x – 2y = 3

2x – y – z = 2

–2y + z = 3

Answer 1

Given, A = ${\displaystyle \left[\begin{array}{ccc}-3& -2& -4\\ 2& 1& 2\\ 2& 1& 3\end{array}\right]}$

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

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

= ${\displaystyle \left[\begin{array}{ccc}-3+4+0& -6+2+4& 0+4-4\\ 2-2+0& 4-1-2& 0-2+2\\ 2-2+0& 4-1-3& 0-2+3\end{array}\right]}$

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

∴ AB = I

(AB)B^–1 = IB^–1

A = B^–1  ...(i)

Now, equations are

x – 2y = 3

2x – y – z = 2

– 2y + z = 3

∴ ${\displaystyle \left[\begin{array}{ccc}1& -2& 0\\ 2& -1& -1\\ 0& -2& 1\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}3\\ 2\\ 3\end{array}\right]}$

CX = D

Here, C = B^T  ...(ii)

∴  C^–1(CX) = C^–1 D

${\displaystyle \Rightarrow }$ IX = C^–1 D 

${\displaystyle \Rightarrow }$ X = C^–1 D

= [B^T]^–1 D  ...[From (ii)]

= [B^–1]^T D 

= [A]^T D  ...[From (i)]

X = ${\displaystyle {\left[\begin{array}{ccc}-3& -2& -4\\ 2& 1& 2\\ 2& 1& 3\end{array}\right]}^T\left[\begin{array}{c}3\\ 2\\ 3\end{array}\right]}$

X = ${\displaystyle \left[\begin{array}{ccc}-3& 2& 2\\ -2& 1& 1\\ -4& 2& 3\end{array}\right]\left[\begin{array}{c}3\\ 2\\ 3\end{array}\right]}$

= ${\displaystyle \left[\begin{array}{c}-9+4+6\\ -6+2+3\\ -12+4+9\end{array}\right]}$

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

Hence x = 1, y = – 1 and z = 1.