Question
Answers

Let $\left( {\begin{array}{*{20}{c}}
  {x + \lambda }&x&x \\
  x&{x + \lambda }&x \\
  x&x&{x + \lambda }
\end{array}} \right)$ then ${A^{ - 1}}$ exists if
$(A)x \ne 0$
$(B)\lambda \ne 0$
$(C)3x + \lambda \ne 0$
$(D)x \ne 0,\lambda \ne 0$

Answer
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Hint: In these types of questions we need to find the inverse of the given term by using the formula and apply elementary transformations. Then expand the matrix along the ${R_3}$ row and by simplification we get the required answer.

Formula used: ${A^{ - 1}} = \dfrac{1}{{\left| A \right|}}adj{\text{ }}A$

Complete step-by-step solution:
It is given that the question stated as a matrix as below
 Let us consider, $A = $$\left( {\begin{array}{*{20}{c}}
  {x + \lambda }&x&x \\
  x&{x + \lambda }&x \\
  x&x&{x + \lambda }
\end{array}} \right)$
We can find the inverse of the given matrix by using the formula, ${A^{ - 1}} = \dfrac{1}{{\left| A \right|}}adj{\text{ }}A$
But we can find it as, when the condition exists $\left| A \right| \ne 0$
Hence we need to show the determinant for the given matrix is non-zero.
To show: $\left| {\left( {\begin{array}{*{20}{c}}
  {x + \lambda }&x&x \\
  x&{x + \lambda }&x \\
  x&x&{x + \lambda }
\end{array}} \right)} \right| \ne 0$ under which conditions
Firstly we have to apply the elementary transformations in the rows
${R_1} \to {R_1} - {R_2}$ and ${R_3} \to {R_3} - {R_2}$
Now to make changes in the row $1$ and row $3$ we are applied elementary operations, which means entries of the row $2$ is subtracting from the entries of the row $1$ respectively.
Similarly entries of row $2$ is subtracting from the entries of row $3$ to make changes in the row $3$respectively,
We get, $\left| {\left( {\begin{array}{*{20}{c}}
  \lambda &{ - \lambda }&0 \\
  x&{x + \lambda }&x \\
  0&{ - \lambda }&\lambda
\end{array}} \right)} \right| \ne 0$
Expanding the matrix along ${R_1}$,
$ \Rightarrow \lambda (\lambda (x + \lambda ) + x\lambda ) - x( - {\lambda ^2} + 0) \ne 0$(For inverse)
On open brackets we multiply the terms we get,
$ \Rightarrow \lambda (\lambda x + {\lambda ^2} + x\lambda ) + x{\lambda ^2} + 0 \ne 0$
Open the brackets we multiply the terms and we get,
$ \Rightarrow x{\lambda ^2} + {\lambda ^3} + x{\lambda ^2} + x{\lambda ^2} + 0 \ne 0$
Adding the same terms we get,
$ \Rightarrow 3x{\lambda ^2} + {\lambda ^3} \ne 0$
Taking ${\lambda ^2}$ as common , we get,
$ \Rightarrow {\lambda ^2}(3x + \lambda ) \ne 0$
These two individually are not equal to zero
Separately, we put both are not equal to zero
${\lambda ^2} \ne 0$ And $3x + \lambda \ne 0$
We know when the square of the number is not equal to zero hence the number cannot be zero it is not possible.
So we can say, $\lambda \ne 0$ And $3x + \lambda \ne 0$

Hence the correct options are $(B)$ and $(C)$ that is $\lambda \ne 0$ and $3x + \lambda \ne 0$.

Note: Whenever we find the inverse of any matrix, the matrix should be a square matrix means number of rows equal to number of columns in the matrix.
Make sure to take the signs right. The inverse of a matrix can be found by using row or column operations not both in a particular answer.