
Inverse matrix of $\left[ \begin{matrix} 4 & 7 \\ 1 & 2 \\ \end{matrix} \right]$ [RPET 1996, 2001]
A. $\left[ \begin{matrix} 2 & -7 \\ -1 & 4 \\ \end{matrix} \right]$
B. $\left[ \begin{matrix} 2 & -1 \\ -7 & 4 \\ \end{matrix} \right]$
C. $\left[ \begin{matrix} -2 & 7 \\ 1 & -4 \\ \end{matrix} \right]$
D. $\left[ \begin{matrix} -2 & 1 \\ 7 & -4 \\ \end{matrix} \right]$
Answer
163.8k+ views
Hint:
Using the determinant and adjoint of the given matrix you can determine the inverse of the matrix.
Formula Used:
Inverse matrix formula $A^{-1}=\dfrac{adjA}{|A|}$
Complete step-by-step solution:
Let $A=\left[ \begin{matrix} 4 & 7 \\ 1 & 2 \\ \end{matrix} \right]$
Determinant;
$|A|=(4\times2)-(7\times1)\\
|A|=8-7\\
|A|=1$
Now, the Adjoint of A can be determined by alternating the main diagonal components. Simply swap the signs of the components in the other diagonal, taking care not to interchange them.
Now, Adjoint of $A=\left[ \begin{matrix} 2 & -7 \\ -1 & 4 \\ \end{matrix} \right]$
Therefore,
$A^{-1}=\dfrac{1}{1}.\left[ \begin{matrix} 2 & -7 \\ -1 & 4 \\ \end{matrix} \right]\\
A^{-1}=\left[ \begin{matrix} 2 & -7 \\ -1 & 4 \\ \end{matrix} \right]$
So, option is A correct.
Note:
${{A}^{-1}}$ exists only when $|A|\ne 0$. We have to remember the formula for ${{A}^{-1}}$. Sometimes students make mistakes while solving $adjA$ and $|A|$ for a matrix. If the inverse of a matrix exists, we can find the adjoint of the given matrix and divide it by the determinant of the matrix.
Using the determinant and adjoint of the given matrix you can determine the inverse of the matrix.
Formula Used:
Inverse matrix formula $A^{-1}=\dfrac{adjA}{|A|}$
Complete step-by-step solution:
Let $A=\left[ \begin{matrix} 4 & 7 \\ 1 & 2 \\ \end{matrix} \right]$
Determinant;
$|A|=(4\times2)-(7\times1)\\
|A|=8-7\\
|A|=1$
Now, the Adjoint of A can be determined by alternating the main diagonal components. Simply swap the signs of the components in the other diagonal, taking care not to interchange them.
Now, Adjoint of $A=\left[ \begin{matrix} 2 & -7 \\ -1 & 4 \\ \end{matrix} \right]$
Therefore,
$A^{-1}=\dfrac{1}{1}.\left[ \begin{matrix} 2 & -7 \\ -1 & 4 \\ \end{matrix} \right]\\
A^{-1}=\left[ \begin{matrix} 2 & -7 \\ -1 & 4 \\ \end{matrix} \right]$
So, option is A correct.
Note:
${{A}^{-1}}$ exists only when $|A|\ne 0$. We have to remember the formula for ${{A}^{-1}}$. Sometimes students make mistakes while solving $adjA$ and $|A|$ for a matrix. If the inverse of a matrix exists, we can find the adjoint of the given matrix and divide it by the determinant of the matrix.
Recently Updated Pages
Geometry of Complex Numbers – Topics, Reception, Audience and Related Readings

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

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

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

Atomic Structure - Electrons, Protons, Neutrons and Atomic Models

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

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

Degree of Dissociation and Its Formula With Solved Example for JEE

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

Instantaneous Velocity - Formula based Examples for JEE

JEE Advanced 2025 Notes

JEE Main Chemistry Question Paper with Answer Keys and Solutions

Total MBBS Seats in India 2025: Government and Private Medical Colleges
