Determinant

Introduction

• We studied in the previous classes that the system of two linear equations in two variables namely $\mathrm{a_{1}x\:+\:b_{1}y\:=\:c_{1}\:and\:a_{2}x\:+\:b_{2}y\:=\:c_{2}}$ will have unique solution if $\mathrm{\frac{a_{1}}{b_{1}}\:\neq\:\frac{a_{2}}{b_{2}}\:i.e.\:a_{1}b_{2}\:-\:b_{1}a_{2}\:\neq\:0}$.

• Thus, $\mathrm{(a_{1}b_{2}\:-\:b_{1}a_{2})}$ is the determining factor for the system of two linear equations.

• We define $\mathrm{a_{1}b_{2}\:-\:b_{1}a_{2}}$ to be the determinant of the square matrix of order 2. In this tutorial we define the determinant of the square matrix of order 2 and 3 and understand the properties of determinants along with some solved examples.

Determinant

To every square matrix $\mathrm{A\:=\:[a_{ij}]}$ of order 𝑛, we can associate a real or complex number called the determinant of the square matrix 𝐴, where 𝑎𝑖𝑗 denotes the i, jth element of the matrix 𝐴. Determinant of the matrix is denoted by $\mathrm{det(A)\:or\:\lvert\:A\rvert}$

Determinant of square matrix of order 1:

Note − Determinant of the square matrix of order 1 is defined to the element itself.

Example −

If $\mathrm{A\:=\:[2]}$ then the determinant of the matrix is $\mathrm{\lvert\:A\rvert\:=\:2}$

Determinant of the square matrix of order 2:

Determinant of the square matrix 𝐴 of order 2 is defined as follows −

$\mathrm{A\:=\:\begin{bmatrix}a_{11} & a_{12} \\a_{21} & a_{22} \end{bmatrix}}$

$$\mathrm{\lvert\:A\rvert\:=a_{11}a_{22}\:-\:a_{21}a_{12}}$$

Determinant of the square matrix of order 3

Determinant of the square matrix of order is defined to be the sum of the product of the elements with its corresponding cofactor elements.

If

$\mathrm{A\:=\begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33} \end{bmatrix}}$

The minor element is denoted by $\mathrm{M_{ij}}$ and is the determinant of the reduced matrix. and the cofactor element $\mathrm{A_{ij}}$ is defined by $\mathrm{A_{ij}\:=\:(-1)^{i\:+\:j}M_{ij}}$

Thus, the determinant of the 3 x 3 square matrix is given by

$$\mathrm{\lvert\:A\rvert\:=a_{11}A_{11}\:+\:a_{12}A_{12}\:+\:a_{13}A_{13}}$$

Example

Find the determinant of the following matrix

$\mathrm{A\:=\begin{bmatrix} 1 & -2 & 3 \\ 1 & 0 & 2\\ 0 & 0 & 3 \end{bmatrix}}$

Solution

We will find the determinant by expanding it across the first row. Thus, we calculate the minors and cofactor elements for the elements in the first row.

Minor of 1 is

$\mathrm{\begin{bmatrix} 0 & 2 \\ 0 & 3\\ \end{bmatrix}\:=\:0\:-\:0\:=\:0}$

$\mathrm{cofactor\:of\:1\:=\:(-1)^{1\:+\:1}M_{11}\:=\:M_{11}\:=\:0}$

Minor of -2 is

$\mathrm{\begin{bmatrix} 1 & 2 \\ 0 & 3\\ \end{bmatrix}\:=\:3\:-\:0\:=\:3}$

$\mathrm{cofactor\:of\:-2\:=\:(-1)^{1\:+\:2}M_{12}\:=\:-M_{12}\:=\:-3}$

Minor of 3 is

$\mathrm{\begin{bmatrix} 1 & 2 \\ 0 & 3\\ \end{bmatrix}\:=\:0\:-\:0\:=\:0}$$

$\mathrm{cofactor\:of\:3\:=\:(-1)^{1\:+\:3}M_{13}\:=\:M_{13}\:=\:0}$

Hence, the determinant of the given 3 x 3 matrix is given by,

$$\mathrm{\lvert\:A\rvert\:=a_{11}A_{11}\:+\:a_{12}A_{12}\:+\:a_{13}A_{13}}$$

$$\mathrm{\lvert\:A\rvert\:=1(0)\:+\:2(-3)\:+\:3(0)\:=\:0\:-\:6\:+\:0\:-6}$$

Properties of Determinants

• If each element of a row or a column of a square matrix is zero then the determinant of that matrix becomes zero.

• If any two rows or columns of the square matrix are Interchanged then the value of the determinant of the matrix changes by sign.

• If any two rows or columns of a square matrix are same then the value of the determinant is zero

• Determinant of the matrix and its transpose is the same.

• If the ratio of the elements of the one row or column with the corresponding elements of another or column is the same, then the value of the determinant is zero.

• If each element of the row is multiplied by a scalar, 𝑘 then the value of the determinant becomes 𝑘 time the original.

• If 𝑘 is a scalar, then $\mathrm{\lvert\:KA\:\rvert\:=\:K^{3}\lvert\:A\:\rvert}$

• Determinant of the diagonal matrix is the product of the diagonal elements.

• Determinant of the triangular matrices is the product of the diagonal elements.

• $\mathrm{\begin{vmatrix} a_{1}\:+\:x & b_{1} & c_{1} \\ a_{2}\:+\:x & b_{2} & c_{2}\\ a_{3}\:+\:x & b_{3} & c_{3} \end{vmatrix}\:=\:\begin{vmatrix} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2}\\ a_{3} & b_{3} & c_{3} \end{vmatrix}\:+\:\begin{vmatrix} x & b_{1} & c_{1} \\ x & b_{2} & c_{2}\\ x & b_{2} & c_{3} \end{vmatrix}}$

• If value of the determinant of the square matrix containing polynomials in 𝑥 is zero at $\mathrm{x\:=\:0\:then\:(x\:-\:a)}$ is the factor of the determinant.

• Value of the determinant remains the same if we perform elementary operations on rows or columns.

• $ \mathrm{\lvert\:AB\:\rvert\:=\:\lvert\:A\:\rvert\:\lvert\:B\:\rvert\:}$

• $\mathrm{\lvert\:AB\:\rvert\:=\:\lvert\:A\:\rvert\:^{n}}$

Determinants to find the area of the parallelogram

Determinant form in finding the area of a triangle is a formula that is efficient to give the positive magnitude to the area of the triangle. This formula is useful when the vertices of the triangle are known instead of the height and base of the triangle. If (𝑥1, 𝑦1), (𝑥2, 𝑦2) and (𝑥3, 𝑦3) are three vertices of the given triangle then its area is defined in the determinant form as

$ \mathrm{\frac{1}{2}\:\begin{vmatrix} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1\\ x_{3} & y_{3} & 1 \end{vmatrix}}$

area of the parallelogram can be calculated by doubling the area of the triangle formed by any three consecutive vertices of the given triangle.

Solved Examples

1.Without expanding the determinant show that

$\mathrm{\begin{vmatrix} p\:-\:q & q\:-\:r & r\:-\:p \\ q\:-\:r & r\:-\:p & p\:-\:q\\ r\:-\:p & p\:-\:q & q\:-\:r \end{vmatrix}\:=\:0}$

Solution −

Consider the LHS as

$\mathrm{\begin{vmatrix} p\:-\:q & q\:-\:r & r\:-\:p \\ q\:-\:r & r\:-\:p & p\:-\:q\\ r\:-\:p & p\:-\:q & q\:-\:r \end{vmatrix}}$

Now notice that each element of the first row is zero, therefore as per the first property of the determinant discussed above the value of the given determinant is zero.

Thus,

$\mathrm{\begin{vmatrix} p\:-\:q & q\:-\:r & r\:-\:p \\ q\:-\:r & r\:-\:p & p\:-\:q\\ r\:-\:p & p\:-\:q & q\:-\:r \end{vmatrix}\:=\:0}$

2. Show that

$\mathrm{\begin{vmatrix} x& p & p \\ p & x & p\\ p & p & x \end{vmatrix}\:=\:(x\:+\:2p)(x\:-\:p)^{2}}$

Solution −

Consider the LHS as

$\mathrm{\begin{vmatrix} x& p & p \\ p & x & p\\ p & p & x \end{vmatrix}}$

by performing the elementary operation $\mathrm{R_{1}\:\rightarrow\:R_{1}\:+\:R_{2}\:+\:R_{3}}$ we get,

$\mathrm{=\:\begin{vmatrix} x\:+\:2p& x\:+\:2p & x\:+\:2p \\ p & x & p\\ p & p & x \end{vmatrix}}$

Notice that $\mathrm{x\:+\:2p}$ is factor of the above determinant,

Thus,

$\mathrm{(x\:+\:2p)\:\begin{vmatrix} 1& 1 & 1 \\ p & x & p\\ p & p & x \end{vmatrix}}$

Now performing the elementary operation $\mathrm{R_{2}\:\rightarrow\:R_{2}\:+\:R_{3}}$ we get,

$\mathrm{(x\:+\:2p)\:\begin{vmatrix}1& 1 & 1 \\ 0 & x\:-\:p & \:-\:x\\ p & p & x\end{vmatrix}}$

Notice that (𝑥 − 𝑝) is the factor of the above determinant. Thus,

Now, expanding the remaining determinant value we get,

$$\mathrm{=\:(x\:+\:2p)(x\:-\:p)\:[1(x\:+\:p)\:-\:1(0\:+\:p)\:+\:1(0\:-\:p)]}$$

$$\mathrm{=\:(x\:+\:2p)(x\:-\:p)(x\:-\:p)}$$

$$\mathrm{=\:(x\:+\:2p)(x\:-\:p)^{2}}$$

Hence,

$$\mathrm{\begin{vmatrix} x & p & p \\ p & x & p\\ p & p & x \end{vmatrix}\:=\:(x\:+\:2p)(x\:-\:p)^{2}}$$

Conclusion

• To every square matrix $\mathrm{A\:=\:[a_{ij}]}$of order $\mathrm{n}$, we can associate a real or complex number called the determinant of the square matrix 𝐴, where $\mathrm{a_{ij}}$ denotes the i, jth element of the matrix 𝐴. Determinant of the matrix is denoted by $\mathrm{det(A)\:or\:\lvert\:A\:\rvert}$

• The area of the triangle in the determinant form is a formula defined by $\mathrm{\frac{1}{2}\:\lvert\:x_{1}(y_{2}\:-\:y_{3})\:+\:x_{2}(y_{3}\:-\:y_{1})\:+\:x_{3}(y_{1}\:-\:y_{2})\:\rvert\:Where\:(x_{1},y_{1})\:(x_{2},y_{2})\:and\:(x_{3},y_{3})}$ are three vertices of the given triangle.

• Since the value of the determinant can be negative or positive, the absolute value of the determinant is taken to mention the calculated value of the determinant as the area of the triangle.

• The area of any quadrilateral can be found by adding areas of two triangles which are formed by dividing a quadrilateral by a diagonal.

• The area of the Parallelogram formula in determinant form is given by $\mathrm{\lvert\:x_{1}(y_{2}\:-\:y_{3})\:+\:x_{2}(y_{3}\:-\:y_{1})\:+\:x_{3}(y_{1}\:-\:y_{2})\:\rvert}$

• If the area of the triangle formed by three points is zero, then the three points are collinear is the application of determinant in the geometry.

FAQs

1. Can a determinant be negative?

The determinant is not a matrix; it is a real number. It's possible for the determinant to be negative. Except for the fact that they both employ vertical lines, it has nothing to do with absolute value.

2. Can we add two determinants?

We can combine two determinants together by just adding these two columns if they only differ by one column.

3. What if the determinant is 0?

When a matrix's determinant is 0, the matrix cannot be inverted. This is the main need to determine whether or not the matrix is invertible. The matrix in question is referred to as a singular matrix. The matrix-related system of equations is linearly dependent when the determinant is zero.

4. What is the formula to find the area of the parallelogram?

The area of the Parallelogram formula in determinant form is given by $\mathrm{\lvert\:x_{1}(y_{2}\:-\:y_{3})\:+\:x_{2}(y_{3}\:-\:y_{1})\:+\:x_{3}(y_{1}\:-\:y_{2})\:\rvert}$

5. Is the determinant a scalar?

The determinant in mathematics is a scalar quantity that is a function of the rows and columns of a square matrix. It enables characterising a few aspects of the matrix and the linear map that the matrix represents.