
If\[2X - \left[ {\begin{array}{*{20}{c}}
1&2 \\
7&4
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
3&2 \\
0&{ - 2}
\end{array}} \right]\], then \[X\]is equal to
A. \[\left[ {\begin{array}{*{20}{c}}
2&2 \\
7&4
\end{array}} \right]\]
B. \[\left[ {\begin{array}{*{20}{c}}
1&2 \\
{\dfrac{7}{2}}&2
\end{array}} \right]\]
C. \[\left[ {\begin{array}{*{20}{c}}
2&2 \\
{\dfrac{7}{2}}&1
\end{array}} \right]\]
D. None of these
Answer
161.7k+ views
Hint: We can find the value of 2X by taking the second term to the R.H.S and adding the two matrices. The matrix X can then be found by dividing the matrix thus obtained by 2. When the matrix is divided by 2, each and every term of the matrix gets divided by 2.
Formula used:
If \[A = \left[ {\begin{array}{*{20}{c}}
a&b \\
c&d
\end{array}} \right]\]and \[B = \left[ {\begin{array}{*{20}{c}}
e&f \\
g&h
\end{array}} \right]\] then,
\[
A + B = \left[ {\begin{array}{*{20}{c}}
a&b \\
c&d
\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}
e&f \\
g&h
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{a + e}&{b + f} \\
{c + g}&{d + h}
\end{array}} \right] \\
\\
\]
Complete step by step solution:
We are given that,
\[2X - \left[ {\begin{array}{*{20}{c}}
1&2 \\
7&4
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
3&2 \\
0&{ - 2}
\end{array}} \right]\]
\[ \Rightarrow 2X = \left[ {\begin{array}{*{20}{c}}
3&2 \\
0&{ - 2}
\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}
1&2 \\
7&4
\end{array}} \right]\]
On adding the two matrices on the R.H.S we get,
\[2X = \left[ {\begin{array}{*{20}{c}}
4&4 \\
7&2
\end{array}} \right]\]
On dividing by 2 on both sides we get,
\[X = \left[ {\begin{array}{*{20}{c}}
2&2 \\
{\dfrac{7}{2}}&1
\end{array}} \right]\]
Option C. is the correct answer.
Note:
To solve the given problem, one must know to add two matrices. When adding two matrices one must make sure the order of the two matrices is the same. If a matrix is divided by a scalar quantity, then the scalar is to be divided by each and every term of the matrix.
Formula used:
If \[A = \left[ {\begin{array}{*{20}{c}}
a&b \\
c&d
\end{array}} \right]\]and \[B = \left[ {\begin{array}{*{20}{c}}
e&f \\
g&h
\end{array}} \right]\] then,
\[
A + B = \left[ {\begin{array}{*{20}{c}}
a&b \\
c&d
\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}
e&f \\
g&h
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{a + e}&{b + f} \\
{c + g}&{d + h}
\end{array}} \right] \\
\\
\]
Complete step by step solution:
We are given that,
\[2X - \left[ {\begin{array}{*{20}{c}}
1&2 \\
7&4
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
3&2 \\
0&{ - 2}
\end{array}} \right]\]
\[ \Rightarrow 2X = \left[ {\begin{array}{*{20}{c}}
3&2 \\
0&{ - 2}
\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}
1&2 \\
7&4
\end{array}} \right]\]
On adding the two matrices on the R.H.S we get,
\[2X = \left[ {\begin{array}{*{20}{c}}
4&4 \\
7&2
\end{array}} \right]\]
On dividing by 2 on both sides we get,
\[X = \left[ {\begin{array}{*{20}{c}}
2&2 \\
{\dfrac{7}{2}}&1
\end{array}} \right]\]
Option C. is the correct answer.
Note:
To solve the given problem, one must know to add two matrices. When adding two matrices one must make sure the order of the two matrices is the same. If a matrix is divided by a scalar quantity, then the scalar is to be divided by each and every term of the matrix.
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