Question

Evaluate The Following

{x + \lambda }&x;&x; \\\ x&{x + \lambda }&x; \\\ x&x;&{x + \lambda } \end{array}} \right|$$

Answer 1

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 $A$.

{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.