
If the matrix $\left[ {\begin{array}{*{20}{c}}
0&1&{ - 2} \\
{ - 1}&0&3 \\
\lambda &{ - 3}&0
\end{array}} \right]$is singular, then $\lambda = $
A. $ - 2$
B. $ - 1$
C. $1$
D. $2$
Answer
162k+ views
Hint: In this question, we have to check whether the given matrix is singular or not. To determine that, we must know the properties of a singular matrix, that is, a matrix is said to be singular when its determinant is equal to zero and how to find the determinant of the matrix. To find the answer to the given problem, we will find the determinant of the given matrix.
Formula Used:
1. In order to find the determinant of \[3 \times 3\]matrix, the formula used is given below:
\[A = \left[ {\begin{array}{*{20}{c}}
{{a_{11}}}&{{a_{12}}}&{{a_{13}}} \\
{{a_{21}}}&{{a_{22}}}&{{a_{23}}} \\
{{a_{31}}}&{{a_{32}}}&{{a_{33}}}
\end{array}} \right]\]
\[\left| A \right| = \]\[{a_{11}}{( - 1)^{1 + 1}}\left| {\begin{array}{*{20}{c}}
{{a_{22}}}&{{a_{23}}} \\
{{a_{32}}}&{{a_{33}}}
\end{array}} \right| + {a_{12}}{( - 1)^{1 + 2}}\left| {\begin{array}{*{20}{c}}
{{a_{21}}}&{{a_{23}}} \\
{{a_{31}}}&{{a_{33}}}
\end{array}} \right| + {a_{13}}{( - 1)^{1 + 3}}\left| {\begin{array}{*{20}{c}}
{{a_{21}}}&{{a_{22}}} \\
{{a_{31}}}&{{a_{32}}}
\end{array}} \right|\]
2. In order to find the determinant of \[2 \times 2\]matrix, the formula used is given below:
\[M = \left[ {\begin{array}{*{20}{c}}
{{a_{11}}}&{{a_{12}}} \\
{{a_{21}}}&{{a_{22}}}
\end{array}} \right]\]
\[\left| M \right| = {a_{11}}{a_{22}} - {a_{12}}{a_{21}}\]
Complete step by step Solution:
Assume the given matrix as \[A = \left[ {\begin{array}{*{20}{c}}
0&1&{ - 2} \\
{ - 1}&0&3 \\
\lambda &{ - 3}&0
\end{array}} \right]\]
Now, identify all the elements of the matrix\[A\]as per the given determinant formula.
Here,
\[{a_{11}} = 0,\]\[{a_{12}} = 1,\]\[{a_{13}} = - 2,\]\[{a_{21}} = - 1,\]\[{a_{22}} = 0,\]\[{a_{23}} = 3,\]\[{a_{31}} = \lambda ,\]\[{a_{32}} = - 3,\]\[{a_{33}} = 0\]
Substitute all the values in the determinant formula, as shown below:
\[\left| A \right| = \]\[0{( - 1)^{1 + 1}}\left| {\begin{array}{*{20}{c}}
0&3 \\
{ - 3}&0
\end{array}} \right| + 1{( - 1)^{1 + 2}}\left| {\begin{array}{*{20}{c}}
{ - 1}&3 \\
\lambda &0
\end{array}} \right| + ( - 2){( - 1)^{1 + 3}}\left| {\begin{array}{*{20}{c}}
{ - 1}&0 \\
\lambda &{ - 3}
\end{array}} \right|\]
In order that the given matrix\[A\]to be singular, the determinant of\[A\]should be equal to zero.
\[\left| A \right| = 0\]
\[0{( - 1)^{1 + 1}}\left| {\begin{array}{*{20}{c}}
0&3 \\
{ - 3}&0
\end{array}} \right| + 1{( - 1)^{1 + 2}}\left| {\begin{array}{*{20}{c}}
{ - 1}&3 \\
\lambda &0
\end{array}} \right| + ( - 2){( - 1)^{1 + 3}}\left| {\begin{array}{*{20}{c}}
{ - 1}&0 \\
\lambda &{ - 3}
\end{array}} \right| = 0\]
Consider the above equation as \[eq(1)\].
For solving the above equation, use the 2nd formula, mentioned at the start of the solution, to find the determinant of \[2 \times 2\]matrix.
\[\left| {\begin{array}{*{20}{c}}
0&3 \\
{ - 3}&0
\end{array}} \right| \Rightarrow (0 \times 0) - (3 \times ( - 3)) = 0 - ( - 9) = 9\]
\[\left| {\begin{array}{*{20}{c}}
{ - 1}&3 \\
\lambda &0
\end{array}} \right| \Rightarrow (( - 1) \times 0) - (3 \times \lambda ) = 0 - 3\lambda = - 3\lambda \]
\[\left| {\begin{array}{*{20}{c}}
{ - 1}&0 \\
\lambda &{ - 3}
\end{array}} \right| \Rightarrow (( - 1) \times ( - 3)) - (0 \times \lambda ) = 3 - 0 = 3\]
On substituting the above values in the\[eq(1)\], we get:
\[0{( - 1)^{1 + 1}}\left| {\begin{array}{*{20}{c}}
0&3 \\
{ - 3}&0
\end{array}} \right| + 1{( - 1)^{1 + 2}}\left| {\begin{array}{*{20}{c}}
{ - 1}&3 \\
\lambda &0
\end{array}} \right| + ( - 2){( - 1)^{1 + 3}}\left| {\begin{array}{*{20}{c}}
{ - 1}&0 \\
\lambda &{ - 3}
\end{array}} \right| = 0\]
\[[0{( - 1)^{1 + 1}} \times 9] + [1{( - 1)^{1 + 2}} \times ( - 3\lambda )] + [( - 2){( - 1)^{1 + 3}} \times 3] = 0\]
Adding the powers,
\[[0{( - 1)^2} \times 9] + [1{( - 1)^3} \times ( - 3\lambda )] + [( - 2){( - 1)^4} \times 3] = 0\]
\[[0(1) \times 9] + [1( - 1) \times ( - 3\lambda )] + [( - 2)(1) \times 3] = 0\]
On solving further, we get:
\[[0 \times 9] + [( - 1) \times ( - 3\lambda )] + [( - 2) \times 3] = 0\]
\[0 + 3\lambda + ( - 6) = 0\]
On simplifying the equation, we get:
\[3\lambda - 6 = 0\]
Taking -6 to the RHS,
\[3\lambda = 6\]
Taking 3 to the RHS and dividing by 6,
\[\lambda = 2\]
Now, it is clear that, in order for the matrix \[A\] to be singular, \[\lambda \] should be equal to 2.
Therefore, the correct option is (D).
Additional Information: In this type of question, where we have to check whether the given matrix is singular or not, we can use another method using the concept of minors and co-factors.
The minor of an element \[({a_{ij}})\] of a square matrix of any order is the determinant of the matrix that is obtained by removing the row (\[{i^{th}}\]row) and the column (\[{j^{th}}\]column) containing the element.
The co-factor of an element \[({a_{ij}})\] of a square matrix of any order is its minor multiplied by \[{( - 1)^{^{i + j}}}\].
Co-factor an element = \[{( - 1)^{i + j}} \times \](minor of the element)
Note: Since, in this question, we have to check whether the given matrix is singular or not, always keep in mind to find the determinant of the given matrix. If the determinant of the matrix is not equal to zero, then the matrix is a non-singular matrix.
Formula Used:
1. In order to find the determinant of \[3 \times 3\]matrix, the formula used is given below:
\[A = \left[ {\begin{array}{*{20}{c}}
{{a_{11}}}&{{a_{12}}}&{{a_{13}}} \\
{{a_{21}}}&{{a_{22}}}&{{a_{23}}} \\
{{a_{31}}}&{{a_{32}}}&{{a_{33}}}
\end{array}} \right]\]
\[\left| A \right| = \]\[{a_{11}}{( - 1)^{1 + 1}}\left| {\begin{array}{*{20}{c}}
{{a_{22}}}&{{a_{23}}} \\
{{a_{32}}}&{{a_{33}}}
\end{array}} \right| + {a_{12}}{( - 1)^{1 + 2}}\left| {\begin{array}{*{20}{c}}
{{a_{21}}}&{{a_{23}}} \\
{{a_{31}}}&{{a_{33}}}
\end{array}} \right| + {a_{13}}{( - 1)^{1 + 3}}\left| {\begin{array}{*{20}{c}}
{{a_{21}}}&{{a_{22}}} \\
{{a_{31}}}&{{a_{32}}}
\end{array}} \right|\]
2. In order to find the determinant of \[2 \times 2\]matrix, the formula used is given below:
\[M = \left[ {\begin{array}{*{20}{c}}
{{a_{11}}}&{{a_{12}}} \\
{{a_{21}}}&{{a_{22}}}
\end{array}} \right]\]
\[\left| M \right| = {a_{11}}{a_{22}} - {a_{12}}{a_{21}}\]
Complete step by step Solution:
Assume the given matrix as \[A = \left[ {\begin{array}{*{20}{c}}
0&1&{ - 2} \\
{ - 1}&0&3 \\
\lambda &{ - 3}&0
\end{array}} \right]\]
Now, identify all the elements of the matrix\[A\]as per the given determinant formula.
Here,
\[{a_{11}} = 0,\]\[{a_{12}} = 1,\]\[{a_{13}} = - 2,\]\[{a_{21}} = - 1,\]\[{a_{22}} = 0,\]\[{a_{23}} = 3,\]\[{a_{31}} = \lambda ,\]\[{a_{32}} = - 3,\]\[{a_{33}} = 0\]
Substitute all the values in the determinant formula, as shown below:
\[\left| A \right| = \]\[0{( - 1)^{1 + 1}}\left| {\begin{array}{*{20}{c}}
0&3 \\
{ - 3}&0
\end{array}} \right| + 1{( - 1)^{1 + 2}}\left| {\begin{array}{*{20}{c}}
{ - 1}&3 \\
\lambda &0
\end{array}} \right| + ( - 2){( - 1)^{1 + 3}}\left| {\begin{array}{*{20}{c}}
{ - 1}&0 \\
\lambda &{ - 3}
\end{array}} \right|\]
In order that the given matrix\[A\]to be singular, the determinant of\[A\]should be equal to zero.
\[\left| A \right| = 0\]
\[0{( - 1)^{1 + 1}}\left| {\begin{array}{*{20}{c}}
0&3 \\
{ - 3}&0
\end{array}} \right| + 1{( - 1)^{1 + 2}}\left| {\begin{array}{*{20}{c}}
{ - 1}&3 \\
\lambda &0
\end{array}} \right| + ( - 2){( - 1)^{1 + 3}}\left| {\begin{array}{*{20}{c}}
{ - 1}&0 \\
\lambda &{ - 3}
\end{array}} \right| = 0\]
Consider the above equation as \[eq(1)\].
For solving the above equation, use the 2nd formula, mentioned at the start of the solution, to find the determinant of \[2 \times 2\]matrix.
\[\left| {\begin{array}{*{20}{c}}
0&3 \\
{ - 3}&0
\end{array}} \right| \Rightarrow (0 \times 0) - (3 \times ( - 3)) = 0 - ( - 9) = 9\]
\[\left| {\begin{array}{*{20}{c}}
{ - 1}&3 \\
\lambda &0
\end{array}} \right| \Rightarrow (( - 1) \times 0) - (3 \times \lambda ) = 0 - 3\lambda = - 3\lambda \]
\[\left| {\begin{array}{*{20}{c}}
{ - 1}&0 \\
\lambda &{ - 3}
\end{array}} \right| \Rightarrow (( - 1) \times ( - 3)) - (0 \times \lambda ) = 3 - 0 = 3\]
On substituting the above values in the\[eq(1)\], we get:
\[0{( - 1)^{1 + 1}}\left| {\begin{array}{*{20}{c}}
0&3 \\
{ - 3}&0
\end{array}} \right| + 1{( - 1)^{1 + 2}}\left| {\begin{array}{*{20}{c}}
{ - 1}&3 \\
\lambda &0
\end{array}} \right| + ( - 2){( - 1)^{1 + 3}}\left| {\begin{array}{*{20}{c}}
{ - 1}&0 \\
\lambda &{ - 3}
\end{array}} \right| = 0\]
\[[0{( - 1)^{1 + 1}} \times 9] + [1{( - 1)^{1 + 2}} \times ( - 3\lambda )] + [( - 2){( - 1)^{1 + 3}} \times 3] = 0\]
Adding the powers,
\[[0{( - 1)^2} \times 9] + [1{( - 1)^3} \times ( - 3\lambda )] + [( - 2){( - 1)^4} \times 3] = 0\]
\[[0(1) \times 9] + [1( - 1) \times ( - 3\lambda )] + [( - 2)(1) \times 3] = 0\]
On solving further, we get:
\[[0 \times 9] + [( - 1) \times ( - 3\lambda )] + [( - 2) \times 3] = 0\]
\[0 + 3\lambda + ( - 6) = 0\]
On simplifying the equation, we get:
\[3\lambda - 6 = 0\]
Taking -6 to the RHS,
\[3\lambda = 6\]
Taking 3 to the RHS and dividing by 6,
\[\lambda = 2\]
Now, it is clear that, in order for the matrix \[A\] to be singular, \[\lambda \] should be equal to 2.
Therefore, the correct option is (D).
Additional Information: In this type of question, where we have to check whether the given matrix is singular or not, we can use another method using the concept of minors and co-factors.
The minor of an element \[({a_{ij}})\] of a square matrix of any order is the determinant of the matrix that is obtained by removing the row (\[{i^{th}}\]row) and the column (\[{j^{th}}\]column) containing the element.
The co-factor of an element \[({a_{ij}})\] of a square matrix of any order is its minor multiplied by \[{( - 1)^{^{i + j}}}\].
Co-factor an element = \[{( - 1)^{i + j}} \times \](minor of the element)
Note: Since, in this question, we have to check whether the given matrix is singular or not, always keep in mind to find the determinant of the given matrix. If the determinant of the matrix is not equal to zero, then the matrix is a non-singular matrix.
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

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

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

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

1 Billion in Rupees

Which one is a true fish A Jellyfish B Starfish C Dogfish class 11 biology CBSE
