Question

Find the matrix $X$ such that $XA = B$
where A = \left( {\begin{array}{*{20}{c}} {3a}&{2b} \\\ { - a}&b; \end{array}} \right) and B = \left( {\begin{array}{*{20}{c}} { - a}&b; \\\ {2a}&{2b} \end{array}} \right)

Answer 1

According to the question, we have to find the unknown matrix $X$ such that it satisfies the given condition. For this, firstly we will identify the order of the matrix $X$ according to the law of multiplication of the matrices. Then we will assume a variable matrix for $X$ of such order. After that we will frame the different equations using the given condition and solve those equations to find out the variables. Hence, we will get the required matrix $X$

Complete answer:
In the question, we have given two matrices $A$ and $B$ as

{3a}&{2b} \\\ { - a}&b; \end{array}} \right)$$ and $$B = \left( {\begin{array}{*{20}{c}} { - a}&b; \\\ {2a}&{2b} \end{array}} \right)$$ And we have to find a matrix $$X$$ such that $$XA = B$$ Now it is evident that the orders of both $$A$$ and $$B$$ are $$2 \times 2$$ Let the order of the matrix $$X$$ is $$m \times n$$ Then according to the given condition, we have $${\left( X \right)_{m \times n}}{\text{ }} \times {\left( A \right)_{2 \times 2}} = {\left( B \right)_{2 \times 2}}$$ Now according to the law of multiplication of the matrices, we know that the multiplication of two matrices is valid only when the number of columns of the 1st matrix is equal to the number of rows of the 2nd matrix. Therefore, using this condition, we conclude that $$n = 2$$ The same law also states that when the two matrices are multiplied, then the resultant matrix has the same number of rows as the 1st matrix, and the same number of columns as the 2nd matrix. Therefore, using this condition, we conclude that $$m = 2$$ Thus, the order of the matrix $$X$$ is $$2 \times 2$$ Now let the matrix $$X$$ od order $$2 \times 2$$ be $$X = \left( {\begin{array}{*{20}{c}} {{x_1}}&{{x_2}} \\\ {{x_3}}&{{x_4}} \end{array}} \right)$$ So, on substituting the values in the given condition, we have $$\left( {\begin{array}{*{20}{c}} {{x_1}}&{{x_2}} \\\ {{x_3}}&{{x_4}} \end{array}} \right)\left( {\begin{array}{*{20}{c}} {3a}&{2b} \\\ { - a}&b; \end{array}} \right){\text{ }} = \left( {\begin{array}{*{20}{c}} { - a}&b; \\\ {2a}&{2b} \end{array}} \right)$$ On multiplying the matrices, we get $$ \Rightarrow {\text{ }}\left( {\begin{array}{*{20}{c}} {3a{x_1} - a{x_2}}&{2b{x_1} + b{x_2}} \\\ {3a{x_3} - a{x_4}}&{2b{x_3} + b{x_4}} \end{array}} \right){\text{ }} = {\text{ }}\left( {\begin{array}{*{20}{c}} { - a}&b; \\\ {2a}&{2b} \end{array}} \right)$$ On comparing the elements, we get $$3a{x_1} - a{x_2} = - a{\text{ }} - - - \left( i \right)$$ $$2b{x_1} + b{x_2} = b{\text{ }} - - - \left( {ii} \right)$$ $$3a{x_3} - a{x_4} = 2a{\text{ }} - - - \left( {iii} \right)$$ $$2b{x_3} + b{x_4} = 2b{\text{ }} - - - \left( {iv} \right)$$ Now from equation $$\left( i \right)$$ we have $$a\left( {3{x_1} - {x_2}} \right) = - a$$ On cancelling $$a$$ we get $$ \Rightarrow \left( {3{x_1} - {x_2}} \right) = - 1$$ $$ \Rightarrow {x_1} = \dfrac{{{x_2}}}{3} - \dfrac{1}{3}$$ From equation $$\left( {ii} \right)$$ we have $$b\left( {2{x_1} + {x_2}} \right) = b$$ On cancelling $$b$$ we get $$\left( {2{x_1} + {x_2}} \right) = 1$$ $$ \Rightarrow {x_2} = 1 - 2{x_1}$$ Now on putting the value of $${x_1}$$ we get $$ \Rightarrow {x_2} = 1 - 2\left( {\dfrac{{{x_2}}}{3} - \dfrac{1}{3}} \right)$$ $$ \Rightarrow \dfrac{5}{3}{x_2} = \dfrac{5}{3}$$ $$ \Rightarrow {x_2} = 1$$ Therefore, $${x_1} = \dfrac{{{x_2}}}{3} - \dfrac{1}{3}{\text{ }} = {\text{ }}\dfrac{1}{3} - \dfrac{1}{3}{\text{ }} = 0$$ Now from equation $$\left( {iii} \right)$$ we have $$a\left( {3{x_3} - {x_4}} \right) = 2a$$ On cancelling $$a$$ we get $$ \Rightarrow \left( {3{x_3} - {x_4}} \right) = 2$$ $$ \Rightarrow {x_3} = \dfrac{{{x_4}}}{3} + \dfrac{2}{3}$$ From equation $$\left( {iv} \right)$$ we have $$b\left( {2{x_3} + {x_4}} \right) = 2b$$ On cancelling $$b$$ we get $$\left( {2{x_3} + {x_4}} \right) = 2$$ $$ \Rightarrow {x_4} = 2 - 2{x_3}$$ Now on putting the value of $${x_3}$$ we get $$ \Rightarrow {x_4} = 2 - 2\left( {\dfrac{{{x_4}}}{3} + \dfrac{2}{3}} \right)$$ $$ \Rightarrow \dfrac{5}{3}{x_4} = \dfrac{2}{3}$$ $$ \Rightarrow {x_4} = \dfrac{2}{5}$$ Therefore, $${x_3} = \dfrac{{{x_4}}}{3} + \dfrac{2}{3}{\text{ }} = {\text{ }}\dfrac{2}{{15}} + \dfrac{2}{3}{\text{ }} = \dfrac{{12}}{{15}} = \dfrac{4}{5}$$ Thus, gathering the values of $${x_1},{\text{ }}{x_2},{\text{ }}{x_3}$$ and $${x_4}$$ we have matrix $$X$$ as $$X = \left( {\begin{array}{*{20}{c}} 0&1 \\\ {\dfrac{4}{5}}&{\dfrac{2}{5}} \end{array}} \right)$$ Hence, $$X = \left( {\begin{array}{*{20}{c}} 0&1 \\\ {\dfrac{4}{5}}&{\dfrac{2}{5}} \end{array}} \right)$$ such that $$XA = B$$ i.e., $${\left( {\begin{array}{*{20}{c}} 0&1 \\\ {\dfrac{4}{5}}&{\dfrac{2}{5}} \end{array}} \right)_{2 \times 2}} \cdot {\left( {\begin{array}{*{20}{c}} {3a}&{2b} \\\ { - a}&b; \end{array}} \right)_{2 \times 2}} = {\left( {\begin{array}{*{20}{c}} { - a}&b; \\\ {2a}&{2b} \end{array}} \right)_{2 \times 2}}$$ **Note:** In such type of questions, we need to make sure that we have to find the product of $$X$$ and $$A$$ in order to make it equal to $$B$$ Also the order of the matrices should be taken into account when multiplying the matrices. Also note there is another way to solve this question, as We have given $$XA = B$$ Post multiplying by $${A^{ - 1}}$$ we get $$XA{A^{ - 1}} = B{A^{ - 1}}$$ $$ \Rightarrow X = B{A^{ - 1}}$$ where $${A^{ - 1}} = \dfrac{{adjA}}{{|A|}}$$ and $$adjA$$ is the cofactor matrix of the given matrix.