
What is the determinant of a \[1 \times 1\] matrix?
Answer
162.9k+ views
Hint: The determinant is defined as a scalar value related to the square matrix. If $X$ is a matrix, then the matrix's determinant is represented by $\left| X \right|$ or $\det \left( X \right)$. Let, \[1 \times 1\] as an example to answer the given question. Determinants are regarded as matrix scaling factors. They can be regarded as matrices' expanding out and shrinking in functions. Determinants accept a square matrix as input and output a single number.
Complete step by step Solution:
Let, $A = \left[ 3 \right]$ be the \[1 \times 1\] matrix
As there is only one element in the matrix. There is no chance of using any formula or calculation. So, directly write the element as the determinant of the matrix.
Determinant of matrix $A = \left[ 3 \right]$will be,
det$A = \left| 3 \right| = 3$
This implies that the determinant of \[1 \times 1\] matrix is the element of the matrix itself.
Note: The key concept involved in solving this problem is the good knowledge of Determinant of a matrix. Students must remember that to find a determinant of $2 \times 2$ matrix there is determinant formula i.e., det $A = ad - bc$ where $A = \left( {\begin{array}{*{20}{c}}
a&b \\
c&d
\end{array}} \right)$. Similarly, for $3 \times 3$matrix det $B = a\left( {\det \left( {\begin{array}{*{20}{c}}
e&f \\
h&i
\end{array}} \right)} \right) - b\left( {\det \left( {\begin{array}{*{20}{c}}
d&f \\
g&i
\end{array}} \right)} \right) + c\left( {\det \left( {\begin{array}{*{20}{c}}
d&e \\
g&h
\end{array}} \right)} \right)$ where matrix $B = \left( {\begin{array}{*{20}{c}}
a&b&c \\
d&e&f \\
g&h&i
\end{array}} \right)$. Also, in linear algebra determinants and matrices are used to solve linear equations by applying Cramer's rule to a collection of non-homogeneous linear equations. Determinants are only computed for square matrices. If the determinant of a matrix is zero, it is referred to as singular, and if it is one, it is referred to as unimodular. The determinant of the matrix must be non-singular, that is, its value must be nonzero, for the system of equations to have a unique solution.
Complete step by step Solution:
Let, $A = \left[ 3 \right]$ be the \[1 \times 1\] matrix
As there is only one element in the matrix. There is no chance of using any formula or calculation. So, directly write the element as the determinant of the matrix.
Determinant of matrix $A = \left[ 3 \right]$will be,
det$A = \left| 3 \right| = 3$
This implies that the determinant of \[1 \times 1\] matrix is the element of the matrix itself.
Note: The key concept involved in solving this problem is the good knowledge of Determinant of a matrix. Students must remember that to find a determinant of $2 \times 2$ matrix there is determinant formula i.e., det $A = ad - bc$ where $A = \left( {\begin{array}{*{20}{c}}
a&b \\
c&d
\end{array}} \right)$. Similarly, for $3 \times 3$matrix det $B = a\left( {\det \left( {\begin{array}{*{20}{c}}
e&f \\
h&i
\end{array}} \right)} \right) - b\left( {\det \left( {\begin{array}{*{20}{c}}
d&f \\
g&i
\end{array}} \right)} \right) + c\left( {\det \left( {\begin{array}{*{20}{c}}
d&e \\
g&h
\end{array}} \right)} \right)$ where matrix $B = \left( {\begin{array}{*{20}{c}}
a&b&c \\
d&e&f \\
g&h&i
\end{array}} \right)$. Also, in linear algebra determinants and matrices are used to solve linear equations by applying Cramer's rule to a collection of non-homogeneous linear equations. Determinants are only computed for square matrices. If the determinant of a matrix is zero, it is referred to as singular, and if it is one, it is referred to as unimodular. The determinant of the matrix must be non-singular, that is, its value must be nonzero, for the system of equations to have a unique solution.
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