Let \[A=\left[ \begin{matrix}
   x+\lambda & x & x \\
   x & x+\lambda & x \\
   x & x & x+\lambda \\
\end{matrix} \right]\], then \[{{A}^{-1}}\] exist if

Hint: This question is from the topic of matrix and determinant. In this question, we will first know what should be conditions for the existence of the inverse of the matrix that is \[{{A}^{-1}}\]. After that, we will find out the condition in which \[{{A}^{-1}}\] exist if \[A=\left[ \begin{matrix}
   x+\lambda & x & x \\
   x & x+\lambda & x \\
   x & x & x+\lambda \\
\end{matrix} \right]\].

Complete step by step solution:
Let us solve this question.
In this question, we have given the matrix A that is \[A=\left[ \begin{matrix}
   x+\lambda & x & x \\
   x & x+\lambda & x \\
   x & x & x+\lambda \\
\end{matrix} \right]\] and we have to check the condition for existence of \[{{A}^{-1}}\].
Let us first know the conditions for the existence of \[{{A}^{-1}}\].
The conditions are:
1) The matrix A should be a square matrix that is the number of rows and number of columns should be the same.
2) The determinant of matrix A should not be equal to zero.
So, in the given matrix \[A=\left[ \begin{matrix}
   x+\lambda & x & x \\
   x & x+\lambda & x \\
   x & x & x+\lambda \\
\end{matrix} \right]\], the number of rows and number of columns are the same, that is 3.
Now, let us first find out the determinant of matrix A.
\[\left| A \right|=\left( x+\lambda \right)\left( \left( x+\lambda \right)\left( x+\lambda \right)-\left( x \right)\left( x \right) \right)-x\left( x\left( x+\lambda \right)-\left( x \right)\left( x \right) \right)+x\left( \left( x \right)\left( x \right)-x\left( x+\lambda \right) \right)\]
The above equation can also be written by using the formula \[{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab\] as
\[\Rightarrow \left| A \right|=\left( x+\lambda \right)\left( {{x}^{2}}+{{\lambda }^{2}}+2x\lambda -{{x}^{2}} \right)-x\left( {{x}^{2}}+\lambda x-{{x}^{2}} \right)+x\left( {{x}^{2}}-{{x}^{2}}-\lambda x \right)\]
The above equation can also be written as
\[\Rightarrow \left| A \right|=\left( x+\lambda \right)\left( {{x}^{2}}+{{\lambda }^{2}}+2x\lambda -{{x}^{2}} \right)-x\left( \lambda x \right)+x\left( -\lambda x \right)\]
The above equation can also be written as
\[\Rightarrow \left| A \right|=\left( x+\lambda \right)\left( {{\lambda }^{2}}+2x\lambda \right)-2{{x}^{2}}\lambda \]
The above equation can also be written as
\[\Rightarrow \left| A \right|=\lambda \left[ \left( x+\lambda \right)\left( \lambda +2x \right)-2{{x}^{2}} \right]\]
Using the formula of foil method that is (a+b)(c+d)=ac+ad+bc+bd, we can write
\[\Rightarrow \left| A \right|=\lambda \left[ x\lambda +2{{x}^{2}}+{{\lambda }^{2}}+2x\lambda -2{{x}^{2}} \right]\]
The above can also be written as
\[\Rightarrow \left| A \right|=\lambda \left[ x\lambda +{{\lambda }^{2}}+2x\lambda \right]\]
The above can also be written as
\[\Rightarrow \left| A \right|=\lambda \left[ 3x\lambda +{{\lambda }^{2}} \right]={{\lambda }^{2}}\left[ 3x+\lambda \right]\]
\[\Rightarrow \left| A \right|={{\lambda }^{2}}\left[ 3x+\lambda \right]\]
The condition is determinant of A cannot be zero or \[\left| A \right|\ne 0\]
So, we can say that
\[\lambda \ne 0\]
And
\[3x+\lambda \ne 0\]
Or, \[\lambda \ne -3x\]
So, we can say that if \[A=\left[ \begin{matrix}
   x+\lambda & x & x \\
   x & x+\lambda & x \\
   x & x & x+\lambda \\
\end{matrix} \right]\], then \[{{A}^{-1}}\] will exist if \[\lambda \ne 0\] and \[\lambda \ne -3x\]

Note: As we can see that this question is from the topic of matrix and determinant, so we should have a better knowledge in that topic. We should remember the following formulas to solve this type of question easily:
Foil method: (a+b)(c+d)=ac+ad+bc+bd
\[{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab\]
We should know how to find the determinant of any matrix.
Suppose, we have given a matrix \[A=\left[ \begin{matrix}
   a & b & c \\
   d & e & f \\
   g & h & i \\
\end{matrix} \right]\], then determinant of A will be
\[\left| A \right|=a\left( e\times i-f\times h \right)-b\left( d\times i-f\times g \right)+c\left( d\times h-e\times g \right)\]