
What must be the matrix if $2X + \left[ {\begin{array}{*{20}{c}}
1&2 \\
3&4
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
3&8 \\
7&2
\end{array}} \right]$
Option:
A. $\left[ {\begin{array}{*{20}{c}}
1&3 \\
2&{ - 1}
\end{array}} \right]$
B. $\left[ {\begin{array}{*{20}{c}}
1&{ - 3} \\
2&{ - 1}
\end{array}} \right]$
C. $\left[ {\begin{array}{*{20}{c}}
2&6 \\
4&{ - 2}
\end{array}} \right]$
D. $\left[ {\begin{array}{*{20}{c}}
2&{ - 6} \\
4&{ - 2}
\end{array}} \right]$
Answer
164.4k+ views
Hint: We will be using the concept of matrix subtraction to solve the equation. For the matrix subtraction, there should be an equal number of rows and columns. A null matrix is produced when a matrix is subtracted from itself, or when $A - A = 0$ . Matrix subtraction is the addition of a matrix's negative to another matrix, i.e., $A - B = A + ( - B)$ .
Formula Used: The difference between two matrices, $A$ and $B$ or $A - B$ is defined as: if there are two matrices, $A = \left[ {{a_{ij}}} \right]$ and $B = \left[ {{b_{ij}}} \right]$ of the same order $m \times n$ then:
$D = \left[ {{d_{ij}}} \right]$ and $A - B = \left[ {{a_{ij}}} \right] - \left[ {{b_{ij}}} \right]$
$\because $ $\left[ {{d_{ij}}} \right] = \left[ {{a_{ij}}} \right] - \left[ {{b_{ij}}} \right]$
Complete step by step solution: We have $2X + \left[ {\begin{array}{*{20}{c}}
1&2 \\
3&4
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
3&8 \\
7&2
\end{array}} \right]$ .
To determine the value of the matrix $X$ , we have to solve the equation. To solve the equation, subtract both the given matrices as shown below:
$2X = \left[ {\begin{array}{*{20}{c}}
3&8 \\
7&2
\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}
1&2 \\
3&4
\end{array}} \right]$
We get
$2X = \left[ {\begin{array}{*{20}{c}}
2&6 \\
4&{ - 2}
\end{array}} \right]$
$ \Rightarrow X = \dfrac{1}{2}\left[ {\begin{array}{*{20}{c}}
2&6 \\
4&{ - 2}
\end{array}} \right]$
On further evaluating, we get matrix
$ \Rightarrow X = \left[ {\begin{array}{*{20}{c}}
1&3 \\
2&{ - 1}
\end{array}} \right]$ .
Option ‘A’ is correct
Note: Only when the two matrices are in the same order do they get subtracted. The two matrices cannot be subtracted from one another if the order is different. Because the 3 x 3 and 2 x 2 matrices have different orders or dimensions, it would not be possible to subtract one from the other. The order of the two matrices must match in order to subtract them.
Formula Used: The difference between two matrices, $A$ and $B$ or $A - B$ is defined as: if there are two matrices, $A = \left[ {{a_{ij}}} \right]$ and $B = \left[ {{b_{ij}}} \right]$ of the same order $m \times n$ then:
$D = \left[ {{d_{ij}}} \right]$ and $A - B = \left[ {{a_{ij}}} \right] - \left[ {{b_{ij}}} \right]$
$\because $ $\left[ {{d_{ij}}} \right] = \left[ {{a_{ij}}} \right] - \left[ {{b_{ij}}} \right]$
Complete step by step solution: We have $2X + \left[ {\begin{array}{*{20}{c}}
1&2 \\
3&4
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
3&8 \\
7&2
\end{array}} \right]$ .
To determine the value of the matrix $X$ , we have to solve the equation. To solve the equation, subtract both the given matrices as shown below:
$2X = \left[ {\begin{array}{*{20}{c}}
3&8 \\
7&2
\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}
1&2 \\
3&4
\end{array}} \right]$
We get
$2X = \left[ {\begin{array}{*{20}{c}}
2&6 \\
4&{ - 2}
\end{array}} \right]$
$ \Rightarrow X = \dfrac{1}{2}\left[ {\begin{array}{*{20}{c}}
2&6 \\
4&{ - 2}
\end{array}} \right]$
On further evaluating, we get matrix
$ \Rightarrow X = \left[ {\begin{array}{*{20}{c}}
1&3 \\
2&{ - 1}
\end{array}} \right]$ .
Option ‘A’ is correct
Note: Only when the two matrices are in the same order do they get subtracted. The two matrices cannot be subtracted from one another if the order is different. Because the 3 x 3 and 2 x 2 matrices have different orders or dimensions, it would not be possible to subtract one from the other. The order of the two matrices must match in order to subtract them.
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