Question

If A= [ 1 0 0 , 5 1 0 , 1 3 1] then find matrix B such that AB = I . verify that BA = I .

Answer 1

If A= [ 1 0 0 , 5 1 0 , 1 3 1] then find matrix B such that AB = I . verify that BA = I .

Answer 2

Answer:

If AB = I then B = $A^{-1}$ which is inverse of A

Therefore matrix B =$A^{-1}$

$B=A^{-1} =\left[\begin{array}{ccc}1&0&0\\-5&1&0\\14&-3&1\end{array}\right]$

since A has the inverse AB = BA= I

Step-by-step explanation:

If A is a square matrix of order m and B is a square matrix of same order m such that AB = BA = I then Matrix B is called inverse of Matrix A

given that AB = I therefore B =  $A^{-1}$

To find inverse of A by elementary operation,

In order to use elementary row operation we may write A=IA

$\left[\begin{array}{ccc}1&0&0\\5&1&0\\1&3&1\end{array}\right] =\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right] A$(applying $R_{2}$→$R_{2} -5R_{1}$)

$\left[\begin{array}{ccc}1&0&0\\0&1&0\\1&3&1\end{array}\right] =\left[\begin{array}{ccc}1&0&0\\-5&1&0\\0&0&1\end{array}\right] A$(applying$R_{3}$→$R_{3} -R_{1}$)

$\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&3&1\end{array}\right] =\left[\begin{array}{ccc}1&0&0\\-5&1&0\\-1&0&1\end{array}\right] A$(applying$R_{3}$→$R_{3} -3R_{2}$)

$\\\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right] =\left[\begin{array}{ccc}1&0&0\\-5&1&0\\14&-3&1\end{array}\right] A$

Hence

$B=A^{-1} =\left[\begin{array}{ccc}1&0&0\\-5&1&0\\14&-3&1\end{array}\right]$

To verify BA = I,

consider

$BA=\left[\begin{array}{ccc}1&0&0\\-5&1&0\\14&-3&1\end{array}\right]\left[\begin{array}{ccc}1&0&0\\5&1&0\\1&3&1\end{array}\right]$

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

      $=I$

Hence BA = I