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Question: Convert \(A = \left[ {\begin{array}{*{20}{c}} 1&{ - 1} \\\ 2&3 \end{array}} \right]\)...

Convert
A = \left[ {\begin{array}{*{20}{c}} 1&{ - 1} \\\ 2&3 \end{array}} \right] into an identity matrix by suitable row transformations.

Explanation

Solution

We have given a 2×22 \times 2 matrix. We asked to convert the given matrix into an identity matrix by substituting suitable row transformations. To get the solution students need to use basic row operations in the given matrix. That is, addition of rows, subtracting the rows and multiplying a number with the row and getting to operate with the other row. In this way we are going to find an identity matrix.

Formula used: AA−1=IA{A^{ - 1}} = I where AA is the given matrix, A−1{A^{ - 1}} is the inverse matrix and II is the identity matrix.

Complete step-by-step solution:
Let
A = \left[ {\begin{array}{*{20}{c}} 1&{ - 1} \\\ 2&3 \end{array}} \right]
∵AA−1=I\because A{A^{ - 1}} = I
Substitute the value of AA in the formula
\left[ {\begin{array}{*{20}{c}} 1&{ - 1} \\\ 2&3 \end{array}} \right]{A^{ - 1}} = I
Here our aim is to convert AA into an identity matrix applying Elementary Row operation.
\left[ {\begin{array}{*{20}{c}} 1&{ - 1} \\\ 2&3 \end{array}} \right]{A^{ - 1}} = \left[ {\begin{array}{*{20}{c}} 1&0 \\\ 0&1 \end{array}} \right]
Applying row operation on R1{R_1} implies,
R1→R1+13R2{R_1} \to {R_1} + \dfrac{1}{3}{R_2}
We get,
{A^{ - 1}}\left[ {\begin{array}{*{20}{c}} {\dfrac{5}{3}}&0 \\\ 2&3 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&{\dfrac{1}{3}} \\\ 0&1 \end{array}} \right]
Again apply row operation on R1{R_1} implies,
R1→35R1{R_1} \to \dfrac{3}{5}{R_1}
\Rightarrow {A^{ - 1}}\left[ {\begin{array}{*{20}{c}} 1&0 \\\ 2&3 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {\dfrac{3}{5}}&{\dfrac{1}{5}} \\\ 0&1 \end{array}} \right]
Now, applying row operation on R2{R_2} implies,
R2→R2−2R1{R_2} \to {R_2} - 2{R_1}
We get,
{A^{ - 1}}\left[ {\begin{array}{*{20}{c}} 1&0 \\\ 0&3 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {\dfrac{3}{5}}&{\dfrac{1}{5}} \\\ {\dfrac{{ - 6}}{5}}&{\dfrac{3}{5}} \end{array}} \right]
Again applying row operation on R2{R_2} implies,
R2→13R2{R_2} \to \dfrac{1}{3}{R_2}

1&0 \\\ 0&1 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {\dfrac{3}{5}}&{\dfrac{1}{5}} \\\ {\dfrac{{ - 2}}{5}}&{\dfrac{1}{5}} \end{array}} \right]$$ $ \Rightarrow {A^{ - 1}}I = {A^{ - 1}}$ That is multiplication of any matrix with the identity matrix results in the matrix itself. Therefore we converted the given matrix into an identity matrix. As we started the problem with $A{A^{ - 1}} = I$ now we concluded the problem with the result ${A^{ - 1}}I = {A^{ - 1}}$. So we changed$A{\text{ into }}I$. Additional information: An identity matrix is a square matrix having 1s on the main diagonal, and 0s everywhere else. For example identity matrix for $3 \times 3$is$\left[ {\begin{array}{*{20}{c}} 1&0&0 \\\ 0&1&0 \\\ 0&0&1 \end{array}} \right]$. **Note:** There is an alternative method to solve this problem. Also in this method we are going to use row transformation to solve this problem. Let $A = \left[ {\begin{array}{*{20}{c}} 1&{ - 1} \\\ 2&3 \end{array}} \right]$ ${R_2} \to {R_2} - 2{R_1}$ $ \Rightarrow A = \left[ {\begin{array}{*{20}{c}} 1&{ - 1} \\\ 0&5 \end{array}} \right]$ Now again apply row operation in the second row, we get ${R_2} \to \dfrac{{{R_2}}}{5}$ $ \Rightarrow A = \left[ {\begin{array}{*{20}{c}} 1&{ - 1} \\\ 0&1 \end{array}} \right]$ Now again apply row operation in the first row, we get ${R_1} \to {R_1} + {R_2}$ $ \Rightarrow A = \left[ {\begin{array}{*{20}{c}} 1&0 \\\ 0&1 \end{array}} \right]$ $\therefore A = I$ Therefore, we converted the given matrix into an identity matrix.