
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
163.2k+ 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.
Recently Updated Pages
JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

JEE Energetics Important Concepts and Tips for Exam Preparation

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Degree of Dissociation and Its Formula With Solved Example for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

JoSAA JEE Main & Advanced 2025 Counselling: Registration Dates, Documents, Fees, Seat Allotment & Cut‑offs

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

NEET 2025 – Every New Update You Need to Know

Verb Forms Guide: V1, V2, V3, V4, V5 Explained

NEET Total Marks 2025

1 Billion in Rupees
