
Let $A = \left[ {\begin{array}{*{20}{c}}
0&0&{ - 1} \\
0&{ - 1}&0 \\
{ - 1}&0&0
\end{array}} \right]$ , the only correct statement about the matrix $A$ is
A. ${A^2} = I$
B. $A = ( - 1)I$ , where $I$ is a unit matrix
C. ${A^{ - 1}}$ does not exist
D. $A$ is a zero matrix
Answer
162k+ views
Hint: The matrix is referred to as singular if the determinant is zero, in which case there would not be an inverse. Inverses are only present in non-singular matrices. Further, check if all of the main diagonal elements are $1$ and the remaining elements are $0$ in the given matrix. The matrix is an identity matrix if the criterion is met.
Formula Used: If the determinant of the matrix $A$ i.e., $\left| A \right| \ne 0$ , then the inverse of matrix $A$ exists or if $\left| A \right| = 0$ , then the inverse does not exist.
Complete step-by-step solution:
We have been provided with the matrix $A = \left[ {\begin{array}{*{20}{c}}
0&0&{ - 1} \\
0&{ - 1}&0 \\
{ - 1}&0&0
\end{array}} \right]$
If we multiply the given matrix $A$ with itself, then we will have
$A \cdot A = \left[ {\begin{array}{*{20}{c}}
0&0&{ - 1} \\
0&{ - 1}&0 \\
{ - 1}&0&0
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
0&0&{ - 1} \\
0&{ - 1}&0 \\
{ - 1}&0&0
\end{array}} \right]$
$ \Rightarrow {A^2} = \left[ {\begin{array}{*{20}{c}}
1&0&0 \\
0&1&0 \\
0&0&1
\end{array}} \right]$
$ \Rightarrow {A^2} = I$
Suppose the identity matrix is multiplied by $( - 1)$ , we get
$( - 1)I = \left[ {\begin{array}{*{20}{c}}
{ - 1}&0&0 \\
0&{ - 1}&0 \\
0&0&{ - 1}
\end{array}} \right]$
But matrix $A = \left[ {\begin{array}{*{20}{c}}
0&0&{ - 1} \\
0&{ - 1}&0 \\
{ - 1}&0&0
\end{array}} \right]$
$ \Rightarrow A \ne ( - 1)I$
Now, we determine the determinant of the matrix $A$
$\left| A \right| = \left| {\begin{array}{*{20}{c}}
0&0&{ - 1} \\
0&{ - 1}&0 \\
{ - 1}&0&0
\end{array}} \right|$
$\left| A \right| = 0(0 - 0) - 0(0 + 0) + ( - 1)(0 - 1)$
$ \Rightarrow \left| A \right| = 1$
So, we have $\left| A \right| = 1 \ne 0$ , which implies that ${A^{ - 1}}$ of the matrix exists.
And as we can see clearly, the given matrix $A$ does not have all of its elements as zero.
$ \Rightarrow A$ is not a zero matrix.
Hence, the correct option would be (A) as ${A^2} = I$ .
Note: Another way of finding inverse of matrix $A = \left[ {\begin{array}{*{20}{c}}
a&b \\
c&d
\end{array}} \right]$ . Use the formula ${A^{ - 1}} = \dfrac{1}{{ad - bc}}\left[ {\begin{array}{*{20}{c}}
d&{ - b} \\
{ - c}&a
\end{array}} \right]$ , where $(ad - bc)$ is the determinant and $(ad - bc) \ne 0$ . If this is the condition, then the inverse of the matrix will exist. Otherwise, if $(ad - bc) = 0$ then the inverse would not exist.
Formula Used: If the determinant of the matrix $A$ i.e., $\left| A \right| \ne 0$ , then the inverse of matrix $A$ exists or if $\left| A \right| = 0$ , then the inverse does not exist.
Complete step-by-step solution:
We have been provided with the matrix $A = \left[ {\begin{array}{*{20}{c}}
0&0&{ - 1} \\
0&{ - 1}&0 \\
{ - 1}&0&0
\end{array}} \right]$
If we multiply the given matrix $A$ with itself, then we will have
$A \cdot A = \left[ {\begin{array}{*{20}{c}}
0&0&{ - 1} \\
0&{ - 1}&0 \\
{ - 1}&0&0
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
0&0&{ - 1} \\
0&{ - 1}&0 \\
{ - 1}&0&0
\end{array}} \right]$
$ \Rightarrow {A^2} = \left[ {\begin{array}{*{20}{c}}
1&0&0 \\
0&1&0 \\
0&0&1
\end{array}} \right]$
$ \Rightarrow {A^2} = I$
Suppose the identity matrix is multiplied by $( - 1)$ , we get
$( - 1)I = \left[ {\begin{array}{*{20}{c}}
{ - 1}&0&0 \\
0&{ - 1}&0 \\
0&0&{ - 1}
\end{array}} \right]$
But matrix $A = \left[ {\begin{array}{*{20}{c}}
0&0&{ - 1} \\
0&{ - 1}&0 \\
{ - 1}&0&0
\end{array}} \right]$
$ \Rightarrow A \ne ( - 1)I$
Now, we determine the determinant of the matrix $A$
$\left| A \right| = \left| {\begin{array}{*{20}{c}}
0&0&{ - 1} \\
0&{ - 1}&0 \\
{ - 1}&0&0
\end{array}} \right|$
$\left| A \right| = 0(0 - 0) - 0(0 + 0) + ( - 1)(0 - 1)$
$ \Rightarrow \left| A \right| = 1$
So, we have $\left| A \right| = 1 \ne 0$ , which implies that ${A^{ - 1}}$ of the matrix exists.
And as we can see clearly, the given matrix $A$ does not have all of its elements as zero.
$ \Rightarrow A$ is not a zero matrix.
Hence, the correct option would be (A) as ${A^2} = I$ .
Note: Another way of finding inverse of matrix $A = \left[ {\begin{array}{*{20}{c}}
a&b \\
c&d
\end{array}} \right]$ . Use the formula ${A^{ - 1}} = \dfrac{1}{{ad - bc}}\left[ {\begin{array}{*{20}{c}}
d&{ - b} \\
{ - c}&a
\end{array}} \right]$ , where $(ad - bc)$ is the determinant and $(ad - bc) \ne 0$ . If this is the condition, then the inverse of the matrix will exist. Otherwise, if $(ad - bc) = 0$ then the inverse would not exist.
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

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

Displacement-Time Graph and Velocity-Time Graph for JEE

JEE Main 2026 Syllabus PDF - Download Paper 1 and 2 Syllabus by NTA

JEE Main Eligibility Criteria 2025

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

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

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

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
