Question

In order that the matrix $\left[ {\begin{array}{*{20}{c}}
  1&2&3 \\
  4&5&6 \\
  3&\lambda &5
\end{array}} \right]$ be non-singular,$\lambda $ should not be equal to
A. $1$
B. $2$
C. $3$
D. $4$

Answer 1

Hint: In this question we have to check whether the given matrix is non-singular or not. To determine that, we must know the properties of a non-singular matrix, that is, a matrix is said to be non-singular when its determinant is not equal to zero.
To solve these types of questions, we must know how to find the determinant of the 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:
Given: Let \[A = \left( {\begin{array}{*{20}{c}}
  1&2&3 \\
  4&5&6 \\
  3&\lambda &5
\end{array}} \right)\]
Step 1: Identify all the elements of matrix A as per the given determinant formula.
Here,
\[{a_{11}} = 1,\]\[{a_{12}} = 2,\]\[{a_{13}} = 3,\]\[{a_{21}} = 4,\]\[{a_{22}} = 5,\]\[{a_{23}} = 6,\]\[{a_{31}} = 3,\]\[{a_{32}} = \lambda ,\]\[{a_{33}} = 5\]
Step 2: Substitute all the values in the determinant formula, as shown below:
\[\left| A \right| = \]\[1{( - 1)^{1 + 1}}\left| {\begin{array}{*{20}{c}}
  5&6 \\
  \lambda &5
\end{array}} \right| + 2{( - 1)^{1 + 2}}\left| {\begin{array}{*{20}{c}}
  4&6 \\
  3&5
\end{array}} \right| + 3{( - 1)^{1 + 3}}\left| {\begin{array}{*{20}{c}}
  4&5 \\
  3&\lambda
\end{array}} \right|\]
Step 3: In order that the given matrix A to be non-singular, the determinant of A should not be equal to zero.
\[\left| A \right| \ne 0\]
\[1{( - 1)^{1 + 1}}\left| {\begin{array}{*{20}{c}}
  5&6 \\
  \lambda &5
\end{array}} \right| + 2{( - 1)^{1 + 2}}\left| {\begin{array}{*{20}{c}}
  4&6 \\
  3&5
\end{array}} \right| + 3{( - 1)^{1 + 3}}\left| {\begin{array}{*{20}{c}}
  4&5 \\
  3&\lambda
\end{array}} \right| \ne 0\]
Step 4: 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.
\[
  \left| {\begin{array}{*{20}{c}}
  5&6 \\
  \lambda &5
\end{array}} \right| \Rightarrow (5 \times 5) - (6 \times \lambda ) = 25 - 6\lambda \\
    \\
 \]
\[\left| {\begin{array}{*{20}{c}}
  4&6 \\
  3&5
\end{array}} \right| \Rightarrow (4 \times 5) - (6 \times 3) = 20 - 18 = 2\]
\[\left| {\begin{array}{*{20}{c}}
  4&5 \\
  3&\lambda
\end{array}} \right| \Rightarrow (4 \times \lambda ) - (5 \times 3) = 4\lambda - 15\]
Step 5: Substituting the above values in the equation obtained in Step 3.
\[1{( - 1)^{1 + 1}}\left| {\begin{array}{*{20}{c}}
  5&6 \\
  \lambda &5
\end{array}} \right| + 2{( - 1)^{1 + 2}}\left| {\begin{array}{*{20}{c}}
  4&6 \\
  3&5
\end{array}} \right| + 3{( - 1)^{1 + 3}}\left| {\begin{array}{*{20}{c}}
  4&5 \\
  3&\lambda
\end{array}} \right| \ne 0\]
\[1{( - 1)^{1 + 1}} \times (25 - 6\lambda ) + 2{( - 1)^{1 + 2}} \times 2 + 3{( - 1)^{1 + 3}} \times (4\lambda - 15) \ne 0\]
Adding the powers,
\[1{( - 1)^2} \times (25 - 6\lambda ) + 2{( - 1)^3} \times 2 + 3{( - 1)^4} \times (4\lambda - 15) \ne 0\]
\[1(1) \times (25 - 6\lambda ) + 2( - 1) \times 2 + 3(1) \times (4\lambda - 15) \ne 0\]
On solving further,
\[1(25 - 6\lambda ) + ( - 2) \times 2 + 3(4\lambda - 15) \ne 0\]
\[(25 - 6\lambda ) + ( - 4) + (12\lambda - 45) \ne 0\]
Opening the brackets and simplifying the equation,
\[25 - 6\lambda - 4 + 12\lambda - 45 \ne 0\]
\[6\lambda - 24 \ne 0\]
Taking -24 to the RHS,
\[6\lambda \ne 24\]
Taking 6 to the RHS and dividing by 24,
\[\lambda \ne 4\]
Step 6: From the above step, it is clear that, in order for the matrix A be non-singular,\[\lambda \] should not be equal to 4.

Therefore, the correct option is (D).

Additional Information: In this type of question, where we have to check whether the given matrix is non-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: In this type of question, where we have to check whether the given matrix is non-singular or not, always keep in mind to find the determinant of the given matrix. If the determinant of the given matrix is not equal to zero, then the given matrix is a non-singular matrix.