Question

The matrix\[\left[ {\begin{array}{*{20}{c}}
  2&\lambda &{ - 4} \\
  { - 1}&3&4 \\
  1&{ - 2}&{ - 3}
\end{array}} \right]\]is non-singular, if
A. \[\lambda \ne - 2\]
B. \[\lambda \ne 2\]
C. \[\lambda \ne 3\]
D. \[\lambda \ne - 3\]

Answer 1

Hint: In the given problem we have to find the values of \[\lambda \] for which the given matrix \[\left[ {\begin{array}{*{20}{c}}
  2&\lambda &{ - 4} \\
  { - 1}&3&4 \\
  1&{ - 2}&{ - 3}
\end{array}} \right]\]will be non-singular. Non-singular matrix is a matrix whose determinant is non-zero. So, we have to find the values of \[\lambda \] for which determinant of the above matrix will be a non-zero value.

Formula used:
The determinant of a \[3 \times 3\] matrix \[A = \left[ {\begin{array}{*{20}{c}}
  a&b&c \\
  d&e&f \\
  g&h&i
\end{array}} \right]\] is \[\left| A \right| = a \times \left| {\begin{array}{*{20}{c}}
  e&f \\
  h&i
\end{array}} \right| - b \times \left| {\begin{array}{*{20}{c}}
  d&f \\
  g&i
\end{array}} \right| + c \times \left| {\begin{array}{*{20}{c}}
  d&e \\
  g&h
\end{array}} \right|\]

Complete step by step solution:
We are given that matrix \[\left[ {\begin{array}{*{20}{c}}
  2&\lambda &{ - 4} \\
  { - 1}&3&4 \\
  1&{ - 2}&{ - 3}
\end{array}} \right]\] is non-singular.
Let A=\[\left[ {\begin{array}{*{20}{c}}
  2&\lambda &{ - 4} \\
  { - 1}&3&4 \\
  1&{ - 2}&{ - 3}
\end{array}} \right]\]
 Non-singular matrix is a matrix whose determinant is non-zero. Therefore \[\left| A \right| \ne 0\]
We have to check for values of \[\lambda \] for which the determinant of the above matrix will be a non-zero value.
Let us calculate the determinant of the above matrix
\[\left| {\begin{array}{*{20}{c}}
  2&\lambda &{ - 4} \\
  { - 1}&3&4 \\
  1&{ - 2}&{ - 3}
\end{array}} \right|\]
On performing row operations, we get,
\[\left| A \right| = \left| {\begin{array}{*{20}{c}}
  2&\lambda &{ - 4} \\
  { - 1}&3&4 \\
  1&{ - 2}&{ - 3}
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
  1&{\lambda + 3}&0 \\
  { - 1}&3&4 \\
  1&{ - 2}&{ - 3}
\end{array}} \right|\]\[\left[ {{R_1} \to {R_2} + {R_1}} \right]\]
\[ = \left| {\begin{array}{*{20}{c}}
  1&{\lambda + 3}&0 \\
  0&1&1 \\
  0&{ - \lambda - 5}&{ - 3}
\end{array}} \right|\left[
  {R_2} \to {R_2} + {R_3},
  {R_3} \to {R_3} - {R_1}
  \right]\]
On applying the formula for the determinant, we get
\[ = 1[(1)( - 3) - (1)( - \lambda - 5)]\]
\[ = 1[ - 3 + \lambda + 5]\]
\[ = 2 + \lambda \]
Therefore \[\left| A \right| = 2 + \lambda \]
The condition for the matrix to be non-singular is, the determinant of the matrix should have non-zero value.
\[ \Rightarrow \left| A \right| \ne 0\]
\[ \Rightarrow 2 + \lambda \ne 0\]
\[ \Rightarrow \lambda \ne - 2\]
This is a required condition for which the matrix will be non-singular.

Option A. is the correct answer.

Note: One must know that for every square matrix, we can associate a number which is called as the determinant of the matrix. Singular matrix is a matrix whose determinant is zero and non-singular matrix is a matrix whose determinant is non-zero. In order to reduce a matrix for easier calculation of determinant we can use row operations.