Question

Find the matrix $X$ so that X\left[ {\begin{array}{*{20}{c}} 1&2&3 \\\ 4&5&6 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} { - 7}&{ - 8}&{ - 9} \\\ 2&4&6 \end{array}} \right].

Answer 1

First determine the order of the matrix $X$ according to the law of multiplication of matrices. Then assume a variable matrix for $X$ of such order. Frame different equations using the matrix equation given in the question. Solve those equations to find out the variables.

Complete step-by-step answer:
According to the question, we have been given a matrix equation X\left[ {\begin{array}{*{20}{c}} 1&2&3 \\\ 4&5&6 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} { - 7}&{ - 8}&{ - 9} \\\ 2&4&6 \end{array}} \right].
Let the matrices M = \left[ {\begin{array}{*{20}{c}} 1&2&3 \\\ 4&5&6 \end{array}} \right] and N = \left[ {\begin{array}{*{20}{c}} { - 7}&{ - 8}&{ - 9} \\\ 2&4&6 \end{array}} \right]. It is evident that the orders of both $M$ and $N$ are $2 \times 3$.
Further, let the order of matrix $X$ is $m \times n$, then according to the above matrix equation, we have:
$\Rightarrow {\left( X \right)_{m \times n}} \times {\left( M \right)_{2 \times 3}} = {\left( N \right)_{2 \times 3}}$
As far as the law of matrix multiplication is concerned, we know that the multiplication of two matrices is valid only when the number of columns in the first matrix is equal to the number of rows in the second matrix. Using this condition for the above equation, we conclude that:
$\Rightarrow n = 2$
The same law also states that when the two matrices are multiplied, the number of rows in the resultant matrix is same as that in the first matrix and the number of columns in the resultant matrix is same as that in the second matrix. Again using this condition for the above equation, we conclude that:
$\Rightarrow m = 2$
Thus the order of the matrix $X$ is $2 \times 2$.
So let this be some variable matrix X = \left[ {\begin{array}{*{20}{c}} a&b; \\\ c&d; \end{array}} \right]. From this, our matrix equation will look like:
\Rightarrow \left[ {\begin{array}{*{20}{c}} a&b; \\\ c&d; \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1&2&3 \\\ 4&5&6 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} { - 7}&{ - 8}&{ - 9} \\\ 2&4&6 \end{array}} \right]
On multiplying matrices, we’ll get:
\Rightarrow \left[ {\begin{array}{*{20}{c}} {a + 4b}&{2a + 5b}&{3a + 6b} \\\ {c + 4d}&{2c + 5d}&{3c + 6d} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} { - 7}&{ - 8}&{ - 9} \\\ 2&4&6 \end{array}} \right]{\text{ }}.....{\text{(A)}}
Comparing first two elements in the first row of both the matrix, we have:
$\Rightarrow a + 4b = - 7{\text{ }}.....{\text{(1)}} \\\ \Rightarrow 2a + 5b = - 8{\text{ }}.....{\text{(2)}} \\\$
Multiplying equation (1) by 2 and simultaneously subtracting it from equation (1), we’ll get:
$\Rightarrow 2a + 5b - 2a - 8b = - 8 + 14 \\\ \Rightarrow - 3b = 6 \\\ \Rightarrow b = - 2 \\\$
Putting the value of $b$ in equation (1), we’ll get:
$\Rightarrow a - 8 = - 7 \\\ \Rightarrow a = 1 \\\$
Comparing first two elements in the second row of both the matrix in equation (A), we have:
$\Rightarrow c + 4d = 2{\text{ }}.....{\text{(3)}} \\\ \Rightarrow 2c + 5d = 4{\text{ }}.....{\text{(4)}} \\\$
Multiplying equation (3) by 2 and simultaneously subtracting it from equation (4), we’ll get:
$\Rightarrow 2c + 5d - 2c - 8d = 4 - 4 \\\ \Rightarrow - 3d = 0 \\\ \Rightarrow d = 0 \\\$
Putting the value of $d$ in equation (3), we’ll get:
$\Rightarrow c - 0 = 2 \\\ \Rightarrow c = 2 \\\$
Thus gathering the values of $a,{\text{ }}b,{\text{ }}c{\text{ and }}d$, we have matrix $X$ as:
\Rightarrow X = \left[ {\begin{array}{*{20}{c}} 1&{ - 2} \\\ 2&0 \end{array}} \right]

Note: When in a matrix, the number of rows is the same as the number of columns then this matrix is called a square matrix. And if in a square matrix, all the diagonal elements are 1 and rest of the elements are 0 then such a matrix is called identity matrix. Identity matrix is represented as ${I_n}$, where $n$ is the order of the matrix.
\Rightarrow {I_2} = \left[ {\begin{array}{*{20}{c}} 1&0 \\\ 0&1 \end{array}} \right]{\text{ and }}{I_3} = \left[ {\begin{array}{*{20}{c}} 1&0&0 \\\ 0&1&0 \\\ 0&0&1 \end{array}} \right]