Question

Let A be a $ 3 \times 3 $ matrix such that $ A \times \left( {\begin{array}{*{20}{c}}
  1&2&3 \\
  0&2&3 \\
  0&1&1
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  0&0&1 \\
  1&0&0 \\
  0&1&0
\end{array}} \right) $ . Then $ {A^{ - 1}} $ is:

Answer 1

Hint: To solve this matrix question, first we will convert the right hand side of the above equation into an identity matrix by changing columns and rows with each other. Thus we will get another equation. We know that according to the inverse matrix rule, $ A \times {A^{ - 1}} = I $ . Comparing both the equation we will get $ {A^{ - 1}} $

Complete step-by-step answer:
According to the question, A be a $ 3 \times 3 $ matrix.
And $ A \times \left( {\begin{array}{*{20}{c}}
  1&2&3 \\
  0&2&3 \\
  0&1&1
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  0&0&1 \\
  1&0&0 \\
  0&1&0
\end{array}} \right) $ …………….. (1)
To get the inverse of matrix A, $ {A^{ - 1}} $ , we have to convert the right hand side of the equation into identity matrix I.
Taking $ {C_1} \leftrightarrow {C_3} $ in the both of side of equation (1) we get,
 $ A \times \left( {\begin{array}{*{20}{c}}
  3&2&1 \\
  3&2&0 \\
  1&1&0
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  1&0&0 \\
  0&0&1 \\
  0&1&0
\end{array}} \right) $
Again taking $ {C_2} \leftrightarrow {C_3} $ on the both of the side of the above equation we get,
 $ A \times \left( {\begin{array}{*{20}{c}}
  3&1&2 \\
  3&0&2 \\
  1&0&1
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  1&0&0 \\
  0&1&0 \\
  0&0&1
\end{array}} \right) $
We can rewrite the above equation as,
 $ A \times \left( {\begin{array}{*{20}{c}}
  3&1&2 \\
  3&0&2 \\
  1&0&1
\end{array}} \right) = I $ …………… (2)
As per the inverse matrix rule we know that,
 $ A \times {A^{ - 1}} = I $ ………………….. (3)
Comparing equation (2) and (3) we get,
 $ {A^{ - 1}} = \left( {\begin{array}{*{20}{c}}
  3&1&2 \\
  3&0&2 \\
  1&0&1
\end{array}} \right) $
Therefore, $ {A^{ - 1}} = \left( {\begin{array}{*{20}{c}}
  3&1&2 \\
  3&0&2 \\
  1&0&1
\end{array}} \right) $

Note: Matrix is a rectangular array of numbers, symbols or expressions arranged in rows and columns
Let A be a $ 3 \times 3 $ matrix, $ A = \left( {\begin{array}{*{20}{c}}
  a&b&c \\
  d&e&f \\
  g&h&i
\end{array}} \right) $
It is called an identity matrix or unit matrix. A square matrix in which all the elements of principal diagonal are ones and all other elements are zeros is called identity matrix. It is given by,
 $ I = \left( {\begin{array}{*{20}{c}}
  1&0&0 \\
  0&1&0 \\
  0&0&1
\end{array}} \right) $
When we multiply a matrix with its inverse matrix, we get an identity matrix.
i.e. $ A \times {A^{ - 1}} = I $
You should remember all the rules, formulae and properties of matrices.
We can also solve the same question in alternative method in which we will assume the given equation as $ A \times B = C $ , from which we will derive matrix A by dividing matrix B with matrix C i.e. first we will derive inverse of matrix B and will multiply it with matrix C to get A. again from matrix A, we will derive its inverse. It is a very large method to solve the above question.