Question

Evaluate The Following
\[\left| {\begin{array}{*{20}{c}}
  {x + \lambda }&x&x \\
  x&{x + \lambda }&x \\
  x&x&{x + \lambda }
\end{array}} \right|\]

Answer 1

Hint: First, we should know about the determinants of the matrix $A$ . The determinant of a matrix is a scalar value which is calculated from the entries of that matrix. It is represented by $\left| A \right|$ or $\Delta $ .
To evaluate the determinants of the matrix, we can apply row and column operations.
Formula Section:
Consider a $3 \times 3$ Matrix $A$
$A = \left| {\begin{array}{*{20}{c}}
  a&b&c \\
  d&e&f \\
  g&h&i
\end{array}} \right|$
The formula to find the determinant value of the matrix is
$\Delta = a(ei - fh) - b(di - fg) + c(dh - eg)$ .This is the expansion along the row \[R1\] .
Next, we expand along a column\[\;C1\] , to find the determinant value of the matrix.
$A = \left| {\begin{array}{*{20}{c}}
  a&b&c \\
  d&e&f \\
  g&h&i
\end{array}} \right|$
$\Delta = a(ei - fh) - d(bi - ch) + g(bf - ce)$

Complete step by step answer:
It is given in the problem that the matrix $A$.
\[A = \left| {\begin{array}{*{20}{c}}
  {x + \lambda }&x&x \\
  x&{x + \lambda }&x \\
  x&x&{x + \lambda }
\end{array}} \right|\]
Next, we have to find the determinant value of the given matrix
$\Delta = \det (A) = \left| {\begin{array}{*{20}{c}}
  {x + \lambda }&x&x \\
  x&{x + \lambda }&x \\
  x&x&{x + \lambda }
\end{array}} \right|$
Add all the elements in the Matrix column-wise and place the added value in the column \[\;C1\]of the Matrix.
 i.e Applying the rule $C1 \to C1 + C2 + C3$
$\Delta = \left| {\begin{array}{*{20}{c}}
  {3x + \lambda }&x&x \\
  {3x + \lambda }&{x + \lambda }&x \\
  {3x + \lambda }&x&{x + \lambda }
\end{array}} \right|$
In the above matrix, $3x + \lambda $ is common in Column \[\;C1\], Take $3x + \lambda $ from Column \[\;C1\], we get
 $\Delta = (3x + \lambda )\left| {\begin{array}{*{20}{c}}
  1&x&x \\
  1&{x + \lambda }&x \\
  1&x&{x + \lambda }
\end{array}} \right|$
Next, we are going to Subtract the Elements in Row \[R1\] from the elements in Row \[R2\] and Row \[R3\]
i.e Applying the Rule $R2 \to R2 - R1,R3 \to R3 - R1$
$\Delta = (3x + \lambda )\left| {\begin{array}{*{20}{c}}
  1&x&x \\
  0&\lambda &0 \\
  0&0&\lambda
\end{array}} \right| \xrightarrow{{\begin{array}{*{20}{c}}
  {}&{}
\end{array}}} (1)$
If we expand along the Column \[\;C1\], then the determinant value of the matrix is
$\Delta = a(ei - fh) - d(bi - ch) + g(bf - ce)$
Let us Expand Equation $(1)$ we expand along the Column \[\;C1\]
$\begin{gathered}
  \Delta = (3x + \lambda )[1({\lambda ^2}) - 0 + 0] \\
  \Delta = {\lambda ^2}(3x + \lambda ) \\
\end{gathered} $
This is the Determinant Value of the given matrix.

Note:
To solve this problem easily, we should find the Row or Column with more Zeros for Expansion. In the above expansion, we select a Column \[\;C1\] with two numbers of Zeros. So we can get the answer Easily. If we select the row or column with a lesser number of zeros, the calculation will become more complicated.
We should be clear in the formula for finding the determinants of the matrix.
For Matrix $A$
$A = \left| {\begin{array}{*{20}{c}}
  a&b&c \\
  d&e&f \\
  g&h&i
\end{array}} \right|$
$\Delta = a(ei - fh) - b(di - fg) + c(dh - eg)$
We can expand along any row or column of the matrix.