Math Doubts

Determinant of a 2 × 2 matrix


$\begin{vmatrix} a & b \\ c & d \\ \end{vmatrix}$ $\,=\,$ $ad-bc$


Let $M$ be a matrix. It has four elements $a$, $b$, $c$ and $d$. The four elements are arranged in a matrix form as follows.

$M$ $\,=\,$ $\begin{bmatrix} a & b \\ c & d \\ \end{bmatrix}$

The matrix $M$ has two rows and two columns. Hence, it is called a $2 \times 2$ matrix. It is also called a square matrix.


According to the determinant of a matrix, the determinant of matrix $M$ is written in mathematical form as $det(M)$ or $|M|$

$(1).\,\,\,$ $det(M)$ $\,=\,$ $\begin{vmatrix} a & b \\ c & d \\ \end{vmatrix}$

$(2).\,\,\,$ $|M|$ $\,=\,$ $\begin{vmatrix} a & b \\ c & d \\ \end{vmatrix}$

You can follow any one of them for expressing the determinant of any $2$ by $2$ matrix in mathematical form.


There are three mathematical steps for finding the determinant of any two by two matrix.

  1. Find the product of diagonal elements. In this case, $a$ and $d$ are diagonal elements.
  2. Evaluate the product of anti-diagonal elements. In this case, $b$ and $c$ are anti-diagonal elements.
  3. Find the subtraction of the product of anti-diagonal elements from the product of diagonal elements for evaluating the determinant of any matrix of order $2$.

Therefore, the determinant of a square matrix of order two can be expressed in algebraic form as follows.

$\,\,\,\therefore\,\,\,\,\,\,$ $|M|$ $\,=\,$ $ad-bc$


Find the determinant of matrix $P$ $\,=\,$ $\begin{bmatrix} 2 & -6 \\ 3 & 7 \\ \end{bmatrix}$

$\implies$ $|P|$ $\,=\,$ $\begin{vmatrix} 2 & -6 \\ 3 & 7 \\ \end{vmatrix}$

$\implies$ $|P|$ $\,=\,$ $2 \times 7 \,-\, (-6) \times 3$

$\implies$ $|P|$ $\,=\,$ $14 \,-\, (-18)$

$\implies$ $|P|$ $\,=\,$ $14+18$

$\,\,\,\therefore\,\,\,\,\,\,$ $|P|$ $\,=\,$ $32$

Thus, we can determine the determinant of any matrix of order two in matrices.

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