Question

If A = \left[ {\begin{array}{*{20}{c}} 2&1 \\\ 0&1 \end{array}} \right] and $AB = I$ then $B =$
A. \left[ {\begin{array}{*{20}{c}} 1&2 \\\ 1&0 \end{array}} \right]
B. \left[ {\begin{array}{*{20}{c}} {\dfrac{1}{2}}&0 \\\ { - \dfrac{1}{2}}&1 \end{array}} \right]
C. \left[ {\begin{array}{*{20}{c}} 1&{ - \dfrac{1}{2}} \\\ 0&{\dfrac{1}{2}} \end{array}} \right]
D. \left[ {\begin{array}{*{20}{c}} {\dfrac{1}{2}}&{ - \dfrac{1}{2}} \\\ 0&1 \end{array}} \right]

Answer 1

In this problem, first we will assume the matrix $B$ as an arbitrary $2 \times 2$ matrix. Then, we will multiply the given matrix $A$ with assumed matrix $B$. That is, we will find $AB$. It is given that $AB = I$ where $I$ is $2 \times 2$ identity matrix. We will use this given information and equate the terms of the matrix on both sides. We will get the required matrix $B$.

Complete step-by-step solution:
Let us assume that the required matrix is B = \left[ {\begin{array}{*{20}{c}} a&b; \\\ c&d; \end{array}} \right]. Therefore, now we have to find the values of $a,b,c,d$. In this problem, the matrix $A$ is given as A = \left[ {\begin{array}{*{20}{c}} 2&1 \\\ 0&1 \end{array}} \right]. Let us find $AB$ by multiplying the matrix $A$ with matrix $B$. Therefore, we get
AB = \left[ {\begin{array}{*{20}{c}} 2&1 \\\ 0&1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} a&b; \\\ c&d; \end{array}} \right] \\\ \Rightarrow AB = \left[ {\begin{array}{*{20}{c}} {2\left( a \right) + 1\left( c \right)}&{2\left( b \right) + 1\left( d \right)} \\\ {0\left( a \right) + 1\left( c \right)}&{0\left( b \right) + 1\left( d \right)} \end{array}} \right] \\\ \Rightarrow AB = \left[ {\begin{array}{*{20}{c}} {2a + c}&{2b + d} \\\ c&d; \end{array}} \right] \cdots \cdots \left( 1 \right) \\\
In this problem, it is also given that $AB = I \cdots \cdots \left( 2 \right)$ where $I$ is $2 \times 2$ identity matrix. Note that here the matrix $I$ can be written as I = \left[ {\begin{array}{*{20}{c}} 1&0 \\\ 0&1 \end{array}} \right] \cdots \cdots \left( 3 \right). Now from $\left( 1 \right),\left( 2 \right)$ and $\left( 3 \right)$, we can write AB = I \\\ \Rightarrow \left[ {\begin{array}{*{20}{c}} {2a + c}&{2b + d} \\\ c&d; \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&0 \\\ 0&1 \end{array}} \right] \cdots \cdots \left( 4 \right) \\\
Let us compare the elements of both matrices. So, from $\left( 4 \right)$ we can write
$2a + c = 1 \cdots \cdots \left( 5 \right) \\\ 2b + d = 0 \cdots \cdots \left( 6 \right) \\\ c = 0 \\\ d = 1 \\\$
Let us put the value of $c$ in the equation $\left( 5 \right)$. Therefore, we get $2a + 0 = 1 \Rightarrow 2a = 1 \Rightarrow a = \dfrac{1}{2}$
Let us put the value of $d$ in the equation $\left( 6 \right)$. Therefore, we get $2b + 1 = 0 \Rightarrow 2b = - 1 \Rightarrow b = - \dfrac{1}{2}$
Now we will put all these values of $a,b,c,d$ in the assumed matrix B = \left[ {\begin{array}{*{20}{c}} a&b; \\\ c&d; \end{array}} \right]. So, we get B = \left[ {\begin{array}{*{20}{c}} {\dfrac{1}{2}}&{ - \dfrac{1}{2}} \\\ 0&1 \end{array}} \right]. Therefore, we can say that if A = \left[ {\begin{array}{*{20}{c}} 2&1 \\\ 0&1 \end{array}} \right] and $AB = I$ then B = \left[ {\begin{array}{*{20}{c}} {\dfrac{1}{2}}&{ - \dfrac{1}{2}} \\\ 0&1 \end{array}} \right]. Hence, option D is correct.

Note: To solve the given problem, we can use the different method. It is given that $AB = I$. Let us pre-multiply ${A^{ - 1}}$ on both sides. So, we get
${A^{ - 1}}\left( {AB} \right) = {A^{ - 1}}I \\\ \Rightarrow \left( {{A^{ - 1}}A} \right)B = {A^{ - 1}}I \cdots \cdots \left( 1 \right) \\\$
Now we know that ${A^{ - 1}}A = I$ and ${A^{ - 1}}I = {A^{ - 1}}$. Use this information in equation $\left( 1 \right)$, we get
$IB = {A^{ - 1}} \Rightarrow B = {A^{ - 1}}\quad \left[ {\because AI = IA = A} \right]$. Therefore, in this problem to find the matrix $B$, we will find the matrix ${A^{ - 1}}$. The inverse of matrix $A$ is denoted by ${A^{ - 1}}$ and it is obtained by using the formula ${A^{ - 1}} = \dfrac{{adj\left( A \right)}}{{\left| A \right|}}$ where $\left| A \right| \ne 0$.