
If $A = \left[ {\begin{array}{*{20}{c}}
0&1 \\
0&0
\end{array}} \right]$ and $AB = O$ , then $B = $
A. $\left[ {\begin{array}{*{20}{c}}
1&1 \\
1&1
\end{array}} \right]$
B. $\left[ {\begin{array}{*{20}{c}}
0&1 \\
{ - 1}&0
\end{array}} \right]$
C. $\left[ {\begin{array}{*{20}{c}}
0&{ - 1} \\
1&0
\end{array}} \right]$
D. $\left[ {\begin{array}{*{20}{c}}
{ - 1}&0 \\
0&0
\end{array}} \right]$
Answer
161.7k+ views
Hint: Here, we suppose some value for matrix $B$ . Then, use the matrix multiplication rule of the matrices to get the value of a few of the unknowns in matrix $B$. After that, we can say that the product of the matrices $A$ and $B$ equals the zero matrices if all the unknowns of the matrix $B$ are zero, which is not in the options. So, we have to consider values according to that.
Formula Used:
Let $A = \left[ {\begin{array}{*{20}{c}}
{{a_{11}}}&{{a_{12}}} \\
{{a_{21}}}&{{a_{22}}}
\end{array}} \right]$ and $B = \left[ {\begin{array}{*{20}{c}}
{{b_{11}}}&{{b_{12}}} \\
{{b_{21}}}&{{b_{22}}}
\end{array}} \right]$ , then $AB = C = \left[ {\begin{array}{*{20}{c}}
{{c_{11}}}&{{c_{12}}} \\
{{c_{21}}}&{{c_{22}}}
\end{array}} \right]$ .
Here, ${c_{11}} = {a_{11}}{b_{11}} + {a_{12}}{b_{21}}$ ; ${c_{12}} = {a_{11}}{b_{12}} + {a_{12}}{b_{22}}$ ; ${c_{21}} = {a_{21}}{b_{11}} + {a_{22}}{b_{21}}$ ; ${c_{22}} = {a_{21}}{b_{12}} + {a_{22}}{b_{22}}$ .
Complete step by step Solution:
Given, matrix $A = \left[ {\begin{array}{*{20}{c}}
0&1 \\
0&0
\end{array}} \right]$ .
We have here that $AB = O$, where the matrix $B$ is unknown and $O$ is a zero matrix.
Suppose, the matrix $B = \left[ {\begin{array}{*{20}{c}}
a&b \\
c&d
\end{array}} \right]$ .
Therefore, we have $AB = O$ as
$AB = \left[ {\begin{array}{*{20}{c}}
0&0 \\
0&0
\end{array}} \right]$
$ \Rightarrow \left[ {\begin{array}{*{20}{c}}
0&1 \\
0&0
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
a&b \\
c&d
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
0&0 \\
0&0
\end{array}} \right]$ …………………(1.1)
Multiply both the matrices on the left side, and we get
$ \Rightarrow \left[ {\begin{array}{*{20}{c}}
c&d \\
0&0
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
0&0 \\
0&0
\end{array}} \right]$
Equalising both the sides, we get $c = d = 0$ .
If we substitute these values in equation (1.1), we would get
$ \Rightarrow \left[ {\begin{array}{*{20}{c}}
0&1 \\
0&0
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
a&b \\
0&0
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
0&0 \\
0&0
\end{array}} \right]$
Out of the four given options, the best fitted values of $a$ and $b$ will be: $a = - 1$ and $b = 0$ .
$ \Rightarrow \left[ {\begin{array}{*{20}{c}}
0&1 \\
0&0
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
{ - 1}&0 \\
0&0
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
0&0 \\
0&0
\end{array}} \right]$
So, we have our matrix $B = \left[ {\begin{array}{*{20}{c}}
{ - 1}&0 \\
0&0
\end{array}} \right]$ .
Therefore, the correct option is (D).
Note: The product of matrices is only possible if the order of the given matrices is equal or not. If the given matrix is not a square matrix i.e., does not have an equal number of rows and columns, then the product would only be possible if the number of rows of one matrix is equal to the number of columns of the other.
Formula Used:
Let $A = \left[ {\begin{array}{*{20}{c}}
{{a_{11}}}&{{a_{12}}} \\
{{a_{21}}}&{{a_{22}}}
\end{array}} \right]$ and $B = \left[ {\begin{array}{*{20}{c}}
{{b_{11}}}&{{b_{12}}} \\
{{b_{21}}}&{{b_{22}}}
\end{array}} \right]$ , then $AB = C = \left[ {\begin{array}{*{20}{c}}
{{c_{11}}}&{{c_{12}}} \\
{{c_{21}}}&{{c_{22}}}
\end{array}} \right]$ .
Here, ${c_{11}} = {a_{11}}{b_{11}} + {a_{12}}{b_{21}}$ ; ${c_{12}} = {a_{11}}{b_{12}} + {a_{12}}{b_{22}}$ ; ${c_{21}} = {a_{21}}{b_{11}} + {a_{22}}{b_{21}}$ ; ${c_{22}} = {a_{21}}{b_{12}} + {a_{22}}{b_{22}}$ .
Complete step by step Solution:
Given, matrix $A = \left[ {\begin{array}{*{20}{c}}
0&1 \\
0&0
\end{array}} \right]$ .
We have here that $AB = O$, where the matrix $B$ is unknown and $O$ is a zero matrix.
Suppose, the matrix $B = \left[ {\begin{array}{*{20}{c}}
a&b \\
c&d
\end{array}} \right]$ .
Therefore, we have $AB = O$ as
$AB = \left[ {\begin{array}{*{20}{c}}
0&0 \\
0&0
\end{array}} \right]$
$ \Rightarrow \left[ {\begin{array}{*{20}{c}}
0&1 \\
0&0
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
a&b \\
c&d
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
0&0 \\
0&0
\end{array}} \right]$ …………………(1.1)
Multiply both the matrices on the left side, and we get
$ \Rightarrow \left[ {\begin{array}{*{20}{c}}
c&d \\
0&0
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
0&0 \\
0&0
\end{array}} \right]$
Equalising both the sides, we get $c = d = 0$ .
If we substitute these values in equation (1.1), we would get
$ \Rightarrow \left[ {\begin{array}{*{20}{c}}
0&1 \\
0&0
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
a&b \\
0&0
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
0&0 \\
0&0
\end{array}} \right]$
Out of the four given options, the best fitted values of $a$ and $b$ will be: $a = - 1$ and $b = 0$ .
$ \Rightarrow \left[ {\begin{array}{*{20}{c}}
0&1 \\
0&0
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
{ - 1}&0 \\
0&0
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
0&0 \\
0&0
\end{array}} \right]$
So, we have our matrix $B = \left[ {\begin{array}{*{20}{c}}
{ - 1}&0 \\
0&0
\end{array}} \right]$ .
Therefore, the correct option is (D).
Note: The product of matrices is only possible if the order of the given matrices is equal or not. If the given matrix is not a square matrix i.e., does not have an equal number of rows and columns, then the product would only be possible if the number of rows of one matrix is equal to the number of columns of the other.
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
