Question

What is the shortcut to finding the determinant of a $5 \times 5$ matrix?

Answer 1

Hint: A value called the determinant of a matrix $A$ , that we denote by $\det (A)$ or $\left| A \right|$ , is a scalar(number) obtained from the elements of a matrix by applying operations which is characteristic of the matrix. Use Laplace Expansion to get the determinant of any given matrix of order $n \times n$.

Formula Used:
Let $A = {\left[ {{a_{ij}}} \right]_{n \times n}}$ be an $n \times n$ matrix. If $n = 1$ , i.e., $A = \left[ {{a_{11}}} \right]$ , then we define $\det (A) = {a_{11}}$ . If $n > 1$ , we define $\det (A) = \sum\limits_{k = 1}^n {{{( - 1)}^{1 + k}}} {a_{1k}}\det ({A_{1k}})$.

Complete step by step Solution:
The evaluation of the determinant of an $n \times n$ matrix using the definition involves the summation of $n!$ terms, each term is defined as a product of $n$ factors. In this method, the elementary row operations are used to reduce the matrix to a triangular form. Then, the determinant of the upper-triangular or the lower-triangular matrix is evaluated by finding the product of the entries on the main diagonal.
We can use Laplace expansion to find the determinant of a $n \times n$ matrix. It reduces $n \times n$ determinants to that of $(n - 1) \times (n - 1)$ determinants. The formula expanded with respect to the ${i^{th}}$where $A = {\left[ {{a_{ij}}} \right]_{n \times n}}$ is:
$\det (A) = {( - 1)^{i + 1}}{a_{i1}}\det ({A_{i1}}) + \ldots + {( - 1)^{i + n}}{a_{in}}\det ({A_{in}})$
Where ${A_{ij}}$ is the $(n - 1) \times (n - 1)$ matrix obtained by erasing the row $i$ and the column $j$ from $A$ , with respect to $j$ we can write,
$\det (A) = {( - 1)^{j + 1}}{a_{1j}}\det ({A_{1j}}) + \ldots + {( - 1)^{j + n}}{a_{nj}}\det ({A_{nj}})$
The determinant’s linearity implies,
$\det (A) = \sum\limits_{j = 1}^n {{a_{1j}}} \det \left( {\begin{array}{*{20}{c}}
  {{E_j}} \\
  {{A_2}} \\
   \vdots \\
  {{A_n}}
\end{array}} \right)$
Where ${E_j}$ is the canonical basis of the rows i.e., ${E_j}$ is zero except at position $j$ where there is $1$ . Thus, we have to calculate the determinant
$\left| {\begin{array}{*{20}{c}}
  0&0& \cdots &0 \\
  {{a_{21}}}&{{a_{22}}}& \cdots &{{a_{2(j - 1)}}} \\
   \vdots & \vdots & \cdots & \vdots \\
  {{a_{n1}}}&{{a_{n2}}}& \cdots &{{a_{n(j - 1)}}}
\end{array}\begin{array}{*{20}{c}}
  1&0&0& \cdots \\
  {{a_{2j}}}&{{a_{2(j + 1)}}}& \cdots & \cdots \\
   \vdots & \cdots & \vdots & \cdots \\
  {{a_{nj}}}&{{a_{n(j + 1)}}}& \cdots & \cdots
\end{array}\begin{array}{*{20}{c}}
  0 \\
  {{a_{2n}}} \\
   \vdots \\
  {{a_{nn}}}
\end{array}} \right|$
The whole column $j$ does not interfere in the computation of the determinant
$\left| {\begin{array}{*{20}{c}}
  0&0& \cdots &0 \\
  {{a_{21}}}&{{a_{22}}}& \cdots &{{a_{2(j - 1)}}} \\
   \vdots & \vdots & \cdots & \vdots \\
  {{a_{n1}}}&{{a_{n2}}}& \cdots &{{a_{n(j - 1)}}}
\end{array}\begin{array}{*{20}{c}}
  1&0&0& \cdots \\
  0&{{a_{2(j + 1)}}}& \cdots & \cdots \\
   \vdots & \cdots & \vdots & \cdots \\
  0&{{a_{n(j + 1)}}}& \cdots & \cdots
\end{array}\begin{array}{*{20}{c}}
  0 \\
  {{a_{2n}}} \\
   \vdots \\
  {{a_{nn}}}
\end{array}} \right|$
Since, we know that interchanging two rows changes the sign of the determinant,
$\det \left( {\begin{array}{*{20}{c}}
  {{E_j}} \\
  {{A_2}} \\
   \vdots \\
  {{A_n}}
\end{array}} \right) = {( - 1)^{j - 1}}\left| {\begin{array}{*{20}{c}}
  1&0&0& \cdots \\
  0&{{a_{21}}}&{{a_{22}}}& \cdots \\
   \vdots & \vdots & \cdots & \vdots \\
  0&{{a_{n1}}}& \cdots & \cdots
\end{array}\begin{array}{*{20}{c}}
  0&0&0& \cdots \\
  {{a_{2(j - 1)}}}&{{a_{2(j + 1)}}}& \cdots & \cdots \\
   \cdots & \vdots & \cdots & \vdots \\
  {{a_{n(j - 1)}}}&{{a_{n(j + 1)}}}& \cdots & \cdots
\end{array}\begin{array}{*{20}{c}}
  0 \\
  {{a_{2n}}} \\
   \vdots \\
  {{a_{nn}}}
\end{array}} \right| = {( - 1)^{1 + j}}\det ({A_{ij}})$
Altogether, we have the expansion along the first row: $\det (A) = {a_{11}}\det ({A_{i1}}) \pm \ldots \pm {a_{1n}}\det ({A_{1n}})$ .

Note: Note that the determinant $\left| {\begin{array}{*{20}{c}}
  a&b \\
  c&d
\end{array}} \right| = ad - bc$ is a special case of Laplace’s formula. The sign of the determinant changes if any two rows (or columns) are switched. If any two rows (or columns) in a determinant are identical or proportional then, the value of the determinant is zero.