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If \[I\] is a unit matrix of order \[2 \times 2\] then write down the value of \[\left| I \right|\].

Last updated date: 13th Jul 2024
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Hint: Here the unit matrix is every \[n \times n\] square matrix made of all zeros except for the elements of the main diagonal that are all ones. And the determinant is a scalar value that can be computed from the elements of a square matrix and encodes certain properties of the linear transformation described by the matrix. With these basic concepts we can solve this problem easily.

Complete step-by-step answer:
Given \[I\] is a unit matrix of order \[2 \times 2\]
i.e., \[I = \left[ {\begin{array}{*{20}{c}}
  1&0 \\
\end{array}} \right]\]
The determinant of \[I\] is given by
\[\left| I \right| = \left| {\begin{array}{*{20}{c}}
  1&0 \\
\end{array}} \right|\]
We know that the determinant of matrix \[\left| A \right| = \left| {\begin{array}{*{20}{c}}
  a&b \\
\end{array}} \right|\] is \[ad - bc\].
By using the above formula,
\[\left| I \right| = \left( 1 \right)\left( 1 \right) - \left( 0 \right)\left( 0 \right) = 1 - 0 = 1\]
Thus, the value of \[\left| I \right| = 1\].

Note: In this problem “\[\left| {} \right|\]” denotes the determinant of a matrix. A unit matrix is always a square matrix and the number of rows and number of columns are always equal. The determinant of a unitary matrix is always equal to 1.