Chapter 4: Determinants

To every square matrix $A = [a_{ij}]$ of order $n$, we can associate a number (real or complex) called determinant of the square matrix $A$, denoted by det $A$ or $|A|$ or $\Delta$.

Properties of Determinants

1. The value of the determinant remains unchanged if its rows and columns are interchanged. $|A| = |A^T|$.
2. If any two rows (or columns) of a determinant are interchanged, then sign of determinant changes.
3. If any two rows (or columns) of a determinant are identical (all corresponding elements are same), then value of determinant is zero.
4. If each element of a row (or a column) of a determinant is multiplied by a constant $k$, then its value gets multiplied by $k$.
5. If some or all elements of a row or column of a determinant are expressed as sum of two (or more) terms, then the determinant can be expressed as sum of two (or more) determinants.

Adjoint and Inverse of a Matrix

• Adjoint: The adjoint of a square matrix $A = [a_{ij}]{n \times n}$ is defined as the transpose of the matrix $[A{ij}]{n \times n}$, where $A{ij}$ is the cofactor of the element $a_{ij}$. Adjoint of the matrix $A$ is denoted by $\text{adj} A$.
• Inverse: A square matrix of order $n$ is invertible if and only if $A$ is nonsingular matrix (i.e., $|A| \neq 0$). The inverse is given by: $$A^{-1} = \frac{1}{|A|}\text{adj}(A)$$

Applications of Determinants and Matrices

Solution of system of linear equations using matrix method: If $AX = B$, then $X = A^{-1}B$.

• If $|A| \neq 0$, the system is consistent and has a unique solution.
• If $|A| = 0$ and $(\text{adj} A) B \neq 0$, the system is inconsistent (no solution).
• If $|A| = 0$ and $(\text{adj} A) B = 0$, the system may be either consistent (infinitely many solutions) or inconsistent.

Competency-Based Questions

1. [CBSE 2023] Find the area of the triangle whose vertices are $(-2, -3), (3, 2)$ and $(-1, -8)$ by using determinants.

Solution:

The area of a triangle with vertices $(x_1, y_1), (x_2, y_2)$ and $(x_3, y_3)$ is given by the absolute value of:

$$ \Delta = \frac{1}{2} \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} $$

Substituting the coordinates:

$$ \Delta = \frac{1}{2} \begin{vmatrix} -2 & -3 & 1 \\ 3 & 2 & 1 \\ -1 & -8 & 1 \end{vmatrix} $$

Evaluating the determinant:

$$ \Delta = \frac{1}{2} [-2(2 - (-8)) - (-3)(3 - (-1)) + 1(-24 - (-2))] $$

$$ \Delta = \frac{1}{2} [-2(10) + 3(4) + 1(-22)] $$

$$ \Delta = \frac{1}{2} [-20 + 12 - 22] = \frac{1}{2} [-30] = -15 $$

Area cannot be negative, so taking the absolute value, Area $= 15$ sq. units.

2. [Sample Paper 2022] Solve the system of equations using matrix method: $3x - 2y + 3z = 8$ $2x + y - z = 1$ $4x - 3y + 2z = 4$

Solution:

The system can be written as $AX = B$:

$A = \begin{bmatrix} 3 & -2 & 3 \\ 2 & 1 & -1 \\ 4 & -3 & 2 \end{bmatrix}, X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, B = \begin{bmatrix} 8 \\ 1 \\ 4 \end{bmatrix}$

First, find $|A|$:

$|A| = 3(2 - 3) - (-2)(4 - (-4)) + 3(-6 - 4) = 3(-1) + 2(8) + 3(-10) = -3 + 16 - 30 = -17 \neq 0$.

So, $A^{-1}$ exists.

Find Cofactors $A_{ij}$:

$A_{11} = -1, A_{12} = -8, A_{13} = -10$

$A_{21} = -5, A_{22} = -6, A_{23} = 1$

$A_{31} = -1, A_{32} = 9, A_{33} = 7$

$\text{adj}(A) = \begin{bmatrix} -1 & -5 & -1 \\ -8 & -6 & 9 \\ -10 & 1 & 7 \end{bmatrix}$

$X = A^{-1}B = \frac{1}{-17} \begin{bmatrix} -1 & -5 & -1 \\ -8 & -6 & 9 \\ -10 & 1 & 7 \end{bmatrix} \begin{bmatrix} 8 \\ 1 \\ 4 \end{bmatrix}$

$X = \frac{1}{-17} \begin{bmatrix} -8 - 5 - 4 \\ -64 - 6 + 36 \\ -80 + 1 + 28 \end{bmatrix} = \frac{1}{-17} \begin{bmatrix} -17 \\ -34 \\ -51 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$

So, $x = 1, y = 2, z = 3$.

3. [CBSE 2020] If $A = \begin{bmatrix} 2 & 3 \\ 1 & -4 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & -2 \\ -1 & 3 \end{bmatrix}$, then verify that $(AB)^{-1} = B^{-1} A^{-1}$.

Solution:

This is left as an exercise for the student for verification purpose. Calculate $AB$, find its inverse, then separately find inverses of $A$ and $B$, their product, and compare.