
If A = $\begin{pmatrix}3 & 2 \\ 0 & 1\end{pmatrix}$ then $(A^{-1})^{3}$ is equal to
A. $\dfrac{1}{27}\begin{pmatrix}1 & -26 \\ 0 & 27\end{pmatrix}$
B. $\dfrac{1}{27}\begin{pmatrix}-1 & 26 \\ 0 & 27\end{pmatrix}$
C. $\dfrac{1}{27}\begin{pmatrix}1 & -26 \\ 0 & -27\end{pmatrix}$
D. $\dfrac{1}{27}\begin{pmatrix}-1 & -26 \\ 0 & -27\end{pmatrix}$
Answer
162k+ views
Hint:
We are given a matrix and need to evaluate $(A^{-1})^{3}$. For this, we need to find the inverse of A and perform matrix multiplication.
Complete Step-by-Step Answer:
Given, A = $\begin{pmatrix}3 & 2 \\ 0 & 1\end{pmatrix}$.
We need to find the inverse of A, for this first we find the adjoint matrix of A.
The adjoint matrix of A is obtained by interchanging the diagonal elements and multiplying the off-diagonal elements by a negative sign.
$\therefore$ $ adj A$ = $\begin{pmatrix}1 & -2 \\ 0& 3\end{pmatrix}$
Determinant of A, $|A| = (3\times1)-(0\times1) = 3$
Inverse of A, $A^{-1} = \dfrac{1}{|A|}adj A$
$A^{-1} = \dfrac{1}{3}\begin{pmatrix}1 & -2 \\ 0 & 3\end{pmatrix}$
Now, we need to find $(A^{-1})^{3}$. For this, we need to multiply $A^{-1}$ three times with itself.
$(A^{-1})^{3} = \dfrac{1}{3}\begin{pmatrix}1 & -2 \\ 0 & 3\end{pmatrix}\times \dfrac{1}{3}\begin{pmatrix}1 & -2 \\ 0 & 3\end{pmatrix}\times \dfrac{1}{3}\begin{pmatrix}1 & -2 \\ 0 & 3\end{pmatrix}$
$(A^{-1})^{3} = \dfrac{1}{27}\begin{pmatrix}1 & -2 \\ 0 & 3\end{pmatrix}\begin{pmatrix}1 & -2 \\ 0 & 3\end{pmatrix}\begin{pmatrix}1 & -2 \\ 0 & 3\end{pmatrix}$
$(A^{-1})^{3} = \dfrac{1}{27}\begin{pmatrix}1\times1+(-2)\times0 & 1\times(-2)+(-2)\times3 \\ 0\times1 + 3\times0 & 0\times(-2)+3\times3\end{pmatrix}\begin{pmatrix}1 & -2 \\ 0 & 3\end{pmatrix}$
$(A^{-1})^{3} = \dfrac{1}{27}\begin{pmatrix}1+ 0 & -2-6 \\ 0 +0 & 0 + 9\end{pmatrix}\begin{pmatrix}1 & -2 \\ 0 & 3\end{pmatrix}$
$(A^{-1})^{3} = \dfrac{1}{27}\begin{pmatrix}1& -8 \\ 0 & 9\end{pmatrix}\begin{pmatrix}1 & -2 \\ 0 & 3\end{pmatrix}$
$(A^{-1})^{3} = \dfrac{1}{27}\begin{pmatrix}1\times1+(-8)\times0 & 1\times(-2)+(-8)\times3 \\ 0\times1+0\times9 & 0\times(-2)+9\times3\end{pmatrix}$
$(A^{-1})^{3} = \dfrac{1}{27}\begin{pmatrix}1+0& -2-24 \\ 0+0 & 0+27\end{pmatrix}$
$(A^{-1})^{3} = \dfrac{1}{27}\begin{pmatrix}1& -26\\ 0 & 27\end{pmatrix}$
Therefore, the correct answer is Option A.
Note:When we have to find $(A^{-1})^{3}$, this means we have to multiply the matrix $A^{-1}$ three times and does not mean to take the cube of the elements of $A^{-1}$. Be careful while multiplying the matrices.
We are given a matrix and need to evaluate $(A^{-1})^{3}$. For this, we need to find the inverse of A and perform matrix multiplication.
Complete Step-by-Step Answer:
Given, A = $\begin{pmatrix}3 & 2 \\ 0 & 1\end{pmatrix}$.
We need to find the inverse of A, for this first we find the adjoint matrix of A.
The adjoint matrix of A is obtained by interchanging the diagonal elements and multiplying the off-diagonal elements by a negative sign.
$\therefore$ $ adj A$ = $\begin{pmatrix}1 & -2 \\ 0& 3\end{pmatrix}$
Determinant of A, $|A| = (3\times1)-(0\times1) = 3$
Inverse of A, $A^{-1} = \dfrac{1}{|A|}adj A$
$A^{-1} = \dfrac{1}{3}\begin{pmatrix}1 & -2 \\ 0 & 3\end{pmatrix}$
Now, we need to find $(A^{-1})^{3}$. For this, we need to multiply $A^{-1}$ three times with itself.
$(A^{-1})^{3} = \dfrac{1}{3}\begin{pmatrix}1 & -2 \\ 0 & 3\end{pmatrix}\times \dfrac{1}{3}\begin{pmatrix}1 & -2 \\ 0 & 3\end{pmatrix}\times \dfrac{1}{3}\begin{pmatrix}1 & -2 \\ 0 & 3\end{pmatrix}$
$(A^{-1})^{3} = \dfrac{1}{27}\begin{pmatrix}1 & -2 \\ 0 & 3\end{pmatrix}\begin{pmatrix}1 & -2 \\ 0 & 3\end{pmatrix}\begin{pmatrix}1 & -2 \\ 0 & 3\end{pmatrix}$
$(A^{-1})^{3} = \dfrac{1}{27}\begin{pmatrix}1\times1+(-2)\times0 & 1\times(-2)+(-2)\times3 \\ 0\times1 + 3\times0 & 0\times(-2)+3\times3\end{pmatrix}\begin{pmatrix}1 & -2 \\ 0 & 3\end{pmatrix}$
$(A^{-1})^{3} = \dfrac{1}{27}\begin{pmatrix}1+ 0 & -2-6 \\ 0 +0 & 0 + 9\end{pmatrix}\begin{pmatrix}1 & -2 \\ 0 & 3\end{pmatrix}$
$(A^{-1})^{3} = \dfrac{1}{27}\begin{pmatrix}1& -8 \\ 0 & 9\end{pmatrix}\begin{pmatrix}1 & -2 \\ 0 & 3\end{pmatrix}$
$(A^{-1})^{3} = \dfrac{1}{27}\begin{pmatrix}1\times1+(-8)\times0 & 1\times(-2)+(-8)\times3 \\ 0\times1+0\times9 & 0\times(-2)+9\times3\end{pmatrix}$
$(A^{-1})^{3} = \dfrac{1}{27}\begin{pmatrix}1+0& -2-24 \\ 0+0 & 0+27\end{pmatrix}$
$(A^{-1})^{3} = \dfrac{1}{27}\begin{pmatrix}1& -26\\ 0 & 27\end{pmatrix}$
Therefore, the correct answer is Option A.
Note:When we have to find $(A^{-1})^{3}$, this means we have to multiply the matrix $A^{-1}$ three times and does not mean to take the cube of the elements of $A^{-1}$. Be careful while multiplying the matrices.
Recently Updated Pages
If tan 1y tan 1x + tan 1left frac2x1 x2 right where x frac1sqrt 3 Then the value of y is

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 Electricity and Magnetism Important Concepts and Tips for Exam Preparation

JEE Energetics 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

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

NIT Cutoff Percentile for 2025

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

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

Degree of Dissociation and Its Formula With Solved Example for JEE

Free Radical Substitution Mechanism of Alkanes for JEE Main 2025

JEE Advanced 2025 Notes
