Mathematics Study Notes
Class XII | Chapter: Matrices & Determinants
Topic: Properties of Determinants and $3 \times 3$ Evaluation
Concept 1: Evaluation of $3 \times 3$ Determinants
A determinant of order 3 is a scalar value associated with a $3 \times 3$ square matrix. It can be evaluated by expanding it along any of its three rows or three columns.
1. Expansion Method
To expand along the first row ($R_1$):
\[ \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} = a_1 \begin{vmatrix} b_2 & c_2 \\ b_3 & c_3 \end{vmatrix} – b_1 \begin{vmatrix} a_2 & c_2 \\ a_3 & c_3 \end{vmatrix} + c_1 \begin{vmatrix} a_2 & b_2 \\ a_3 & b_3 \end{vmatrix} \]
2. Sign Convention (Cofactor Signs)
The sign assigned to each element $a_{ij}$ is $(-1)^{i+j}$. For a $3 \times 3$ matrix, the signs are:
\[ \begin{bmatrix} + & – & + \\ – & + & – \\ + & – & + \end{bmatrix} \]
Solved Examples (Concept 1)
Example 1: Evaluate $\Delta = \begin{vmatrix} 1 & 2 & 4 \\ -1 & 3 & 0 \\ 4 & 1 & 0 \end{vmatrix}$.
Solution:
Expanding along $C_3$ (Column 3) is easiest due to the zeros:
$\Delta = 4 \begin{vmatrix} -1 & 3 \\ 4 & 1 \end{vmatrix} – 0 + 0 = 4[(-1)(1) – (4)(3)] = 4[-1 – 12] = 4(-13) = -52$.
Example 2: Evaluate $\Delta = \begin{vmatrix} 0 & \sin \alpha & -\cos \alpha \\ -\sin \alpha & 0 & \sin \beta \\ \cos \alpha & -\sin \beta & 0 \end{vmatrix}$.
Solution:
Expanding along $R_1$:
$\Delta = 0 – \sin \alpha (0 – \cos \alpha \sin \beta) + (-\cos \alpha)(\sin \alpha \sin \beta – 0)$
$\Delta = \sin \alpha \cos \alpha \sin \beta – \cos \alpha \sin \alpha \sin \beta = 0$.
Example 3: Find values of $x$ if $\begin{vmatrix} x & 3 & 7 \\ 2 & x & -2 \\ 7 & 8 & x \end{vmatrix}$ has a value of 0 when $x=5$ is one root.
Solution:
Substitute $x=5$:
$\begin{vmatrix} 5 & 3 & 7 \\ 2 & 5 & -2 \\ 7 & 8 & 5 \end{vmatrix} = 5(25+16) – 3(10+14) + 7(16-35) = 5(41) – 3(24) + 7(-19) = 205 – 72 – 133 = 0$.
Confirmed.
Example 4: If $A = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{bmatrix}$, show that $|3A| = 27|A|$.
Solution:
$|A| = 1(4-0) = 4$. So $27|A| = 108$.
$|3A| = \begin{vmatrix} 3 & 0 & 3 \\ 0 & 3 & 6 \\ 0 & 0 & 12 \end{vmatrix} = 3(36-0) = 108$. Hence $|3A| = 3^3 |A|$.
Example 5: Prove $\begin{vmatrix} 1 & a & bc \\ 1 & b & ca \\ 1 & c & ab \end{vmatrix} = (a-b)(b-c)(c-a)$.
Solution:
Perform $R_2 \to R_2 – R_1$ and $R_3 \to R_3 – R_1$:
$\begin{vmatrix} 1 & a & bc \\ 0 & b-a & c(a-b) \\ 0 & c-a & b(a-c) \end{vmatrix} = (b-a)(c-a) \begin{vmatrix} 1 & a & bc \\ 0 & 1 & -c \\ 0 & 1 & -b \end{vmatrix}$
$= (b-a)(c-a)[-b – (-c)] = (b-a)(c-a)(c-b) = (a-b)(b-c)(c-a)$.
Concept 2: Properties of Determinants
Properties help in simplifying determinants without direct expansion. These are extremely important for solving problems quickly in CBSE and ISC board examinations.
Key Properties
-
1. Reflection Property: The value of a determinant remains unchanged if its rows and columns are interchanged:
\[ |A| = |A^T| \]
Example (2×2):\[ A = \begin{bmatrix} 2 & 3 \\ 5 & 7 \end{bmatrix}, \quad |A| = (2)(7) – (3)(5) = 14 – 15 = -1 \]
\[ A^T = \begin{bmatrix} 2 & 5 \\ 3 & 7 \end{bmatrix}, \quad |A^T| = (2)(7) – (5)(3) = 14 – 15 = -1 \]
-
2. Switching Property: If any two rows (or columns) are interchanged, the sign of the determinant changes.
Example (2×2):
\[ C = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}, \quad |C| = 1\cdot4 – 2\cdot3 = 4 – 6 = -2 \]
Interchange \(R_1\) and \(R_2\):
\[ C’ = \begin{bmatrix} 3 & 4 \\ 1 & 2 \end{bmatrix}, \quad |C’| = 3\cdot2 – 4\cdot1 = 6 – 4 = 2 = -(-2) = -|C| \]
-
3. Repetition Property: If any two rows (or columns) are identical or proportional, the determinant is zero.
Example (3×3):
\[ F = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 1 & 2 & 3 \end{bmatrix} \quad (R_1 = R_3), \quad |F| = 0 \]
-
4. Scalar Multiple Property: If each element of a row (or column) is multiplied by a constant \(k\), then the determinant is multiplied by \(k\).
Example (2×2):
\[ G = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}, \quad |G| = -2 \]
Multiply \(R_1\) by \(k=3\):
\[ G’ = \begin{bmatrix} 3 & 6 \\ 3 & 4 \end{bmatrix}, \quad |G’| = 12 – 18 = -6 = 3 \times (-2) \]
-
5. Common Factor Property: If a common factor exists in any row or column, it can be taken out of the determinant.
Example (3×3):
\[ K = \begin{bmatrix} 3 & 6 & 9 \\ 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix} = 3 \begin{bmatrix} 1 & 2 & 3 \\ 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix} \]
\[ |K| = 3 \times 0 = 0 \]
-
6. Invariance Property (Row/Column Operation): The value of the determinant remains unchanged if we apply:
\[ R_i \rightarrow R_i + kR_j \quad \text{or} \quad C_i \rightarrow C_i + kC_j \]
Example (2×2):\[ L = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}, \quad |L| = -2 \]
\(R_2 \rightarrow R_2 – 3R_1\):
\[ L’ = \begin{bmatrix} 1 & 2 \\ 0 & -2 \end{bmatrix}, \quad |L’| = -2 \]
-
7. Zero Row/Column Property: If all elements of any row or column are zero, then the determinant is zero.
Example (3×3):
\[ P = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 0 & 0 \\ 4 & 5 & 6 \end{bmatrix}, \quad |P| = 0 \]
-
8. Proportional Rows/Columns Property: If two rows (or columns) are proportional, the determinant is zero.
Example (3×3):
\[ R = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 7 & 8 & 9 \end{bmatrix}, \quad R_2 = 2 \times R_1 \Rightarrow |R| = 0 \]
-
9. Factor Multiplication Property: If all elements of a determinant are multiplied by a constant \(k\), then the determinant is multiplied by \(k^n\) (for an \(n \times n\) determinant).
Example (2×2):
\[ S = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}, \quad |S| = -2 \]
Multiply all elements by \(k=2\):
\[ S’ = \begin{bmatrix} 2 & 4 \\ 6 & 8 \end{bmatrix}, \quad |S’| = 2\cdot8 – 4\cdot6 = 16 – 24 = -8 = 2^2 \times (-2) = k^2|S| \]
Example (3×3):
\[ T = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{bmatrix}, \quad |T| = 6 \]
Multiply all elements by \(k=2\):
\[ T’ = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 6 \end{bmatrix}, \quad |T’| = 2\cdot4\cdot6 = 48 = 2^3 \times 6 = k^3|T| \]
-
10. Sum Property (Linearity): If elements of a row (or column) are expressed as the sum of two terms, the determinant can be split into two determinants.
Example (2×2):
\[ U = \begin{bmatrix} a+b & c+d \\ e & f \end{bmatrix} = \begin{bmatrix} a & c \\ e & f \end{bmatrix} + \begin{bmatrix} b & d \\ e & f \end{bmatrix} \]
Check numerically:
\[ U = \begin{bmatrix} 2+3 & 4+5 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 5 & 9 \\ 1 & 2 \end{bmatrix}, \quad |U| = 5\cdot2 – 9\cdot1 = 10 – 9 = 1 \]
\[ \begin{bmatrix} 2 & 4 \\ 1 & 2 \end{bmatrix} = 0, \quad \begin{bmatrix} 3 & 5 \\ 1 & 2 \end{bmatrix} = 1, \quad \text{Sum} = 0 + 1 = 1 = |U| \]
-
11. Triangular Matrix Property: If the matrix is upper or lower triangular, then the determinant is equal to the product of its diagonal elements.
Example (2×2):
\[ W = \begin{bmatrix} 3 & 0 \\ 5 & -2 \end{bmatrix} \ (\text{lower triangular}), \quad |W| = 3 \times (-2) = -6 \]
Example (3×3):
\[ X = \begin{bmatrix} 2 & 1 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{bmatrix} \ (\text{upper triangular}), \quad |X| = 2\cdot4\cdot6 = 48 \]
-
12. Identity Matrix Property: The determinant of an identity matrix is always 1.
Example (2×2): \[ |I_2| = 1 \]
Example (3×3): \[ |I_3| = 1 \]
-
13. Determinant of Product: For square matrices \(A\) and \(B\) of same order:
\[ |AB| = |A||B| \]
Example (2×2):
\[ A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}, \quad B = \begin{bmatrix} 2 & 0 \\ 1 & 2 \end{bmatrix} \]
\[ |A| = -2, \quad |B| = 4, \quad |A||B| = -8 \]
\[ AB = \begin{bmatrix} 4 & 4 \\ 10 & 8 \end{bmatrix}, \quad |AB| = 32 – 40 = -8 \]
-
14. Determinant of Inverse: If \(A\) is invertible, then:
\[ |A^{-1}| = \frac{1}{|A|} \]
Example (2×2):
\[ A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}, \quad |A| = -2 \]
\[ |A^{-1}| = -0.5 = \frac{1}{-2} \]
Example (3×3):
\[ B = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{bmatrix}, \quad |B| = 6 \]
\[ |B^{-1}| = \frac16 = \frac{1}{|B|} \]
Solved Examples (Concept 2)
Example 1: Without expanding, show that $\begin{vmatrix} 102 & 18 & 36 \\ 1 & 3 & 4 \\ 17 & 3 & 6 \end{vmatrix} = 0$.
Solution:
In $R_1$, take 6 common:
$6 \begin{vmatrix} 17 & 3 & 6 \\ 1 & 3 & 4 \\ 17 & 3 & 6 \end{vmatrix}$.
Since $R_1 = R_3$, the determinant is $6(0) = 0$.
Example 2: Evaluate $\begin{vmatrix} 2 & 7 & 65 \\ 3 & 8 & 75 \\ 5 & 9 & 86 \end{vmatrix}$ using properties.
Solution:
Apply $C_3 \to C_3 – 9C_2$:
$\begin{vmatrix} 2 & 7 & 65-63 \\ 3 & 8 & 75-72 \\ 5 & 9 & 86-81 \end{vmatrix} = \begin{vmatrix} 2 & 7 & 2 \\ 3 & 8 & 3 \\ 5 & 9 & 5 \end{vmatrix}$.
Since $C_1 = C_3$, the value is 0.
Example 3: Using properties, prove $\begin{vmatrix} a-b-c & 2a & 2a \\ 2b & b-c-a & 2b \\ 2c & 2c & c-a-b \end{vmatrix} = (a+b+c)^3$.
Solution:
Apply $R_1 \to R_1 + R_2 + R_3$:
$\begin{vmatrix} a+b+c & a+b+c & a+b+c \\ 2b & b-c-a & 2b \\ 2c & 2c & c-a-b \end{vmatrix} = (a+b+c) \begin{vmatrix} 1 & 1 & 1 \\ 2b & b-c-a & 2b \\ 2c & 2c & c-a-b \end{vmatrix}$.
Applying $C_2 \to C_2 – C_1$ and $C_3 \to C_3 – C_1$ results in a lower triangular determinant whose product is $(a+b+c)^2$. Total: $(a+b+c)^3$.
Example 4: If $A$ is a skew-symmetric matrix of order 3, prove $|A| = 0$.
Solution:
$A^T = -A \implies |A^T| = |-A|$.
Since order is 3, $|-A| = (-1)^3 |A| = -|A|$.
We know $|A^T| = |A|$, so $|A| = -|A| \implies 2|A| = 0 \implies |A| = 0$.
Example 5: Find $|A|$ if $A = \begin{bmatrix} x+y & y+z & z+x \\ z & x & y \\ 1 & 1 & 1 \end{bmatrix}$.
Solution:
Apply $R_1 \to R_1 + R_2$:
$\begin{vmatrix} x+y+z & x+y+z & x+y+z \\ z & x & y \\ 1 & 1 & 1 \end{vmatrix} = (x+y+z) \begin{vmatrix} 1 & 1 & 1 \\ z & x & y \\ 1 & 1 & 1 \end{vmatrix}$.
Since $R_1 = R_3$, $|A| = 0$.
Self Exercise & HOTS
If $a, b, c$ are in A.P., find the value of $\begin{vmatrix} x+2 & x+3 & x+2a \\ x+3 & x+4 & x+2b \\ x+4 & x+5 & x+2c \end{vmatrix}$.
Prove that $\begin{vmatrix} y+z & z & y \\ z & z+x & x \\ y & x & x+y \end{vmatrix} = 4xyz$.
Show that $\begin{vmatrix} (b+c)^2 & a^2 & a^2 \\ b^2 & (c+a)^2 & b^2 \\ c^2 & c^2 & (a+b)^2 \end{vmatrix} = 2abc(a+b+c)^3$.
Without expanding, show that $\begin{vmatrix} \sin^2 A & \cos^2 A & 1 \\ \sin^2 B & \cos^2 B & 1 \\ \sin^2 C & \cos^2 C & 1 \end{vmatrix} = 0$.
Solve for $x$: $\begin{vmatrix} x+a & b & c \\ a & x+b & c \\ a & b & x+c \end{vmatrix} = 0$.
If $a, b, c$ are real numbers and $\Delta = \begin{vmatrix} b+c & c+a & a+b \\ c+a & a+b & b+c \\ a+b & b+c & c+a \end{vmatrix} = 0$, show that either $a+b+c=0$ or $a=b=c$.
Prove that the following determinant results in the cyclic product: \[ \begin{vmatrix} 1 & 1 & 1 \\ \alpha & \beta & \gamma \\ \alpha^3 & \beta^3 & \gamma^3 \end{vmatrix} = (\alpha – \beta)(\beta – \gamma)(\gamma – \alpha)(\alpha + \beta + \gamma) \]
If $\alpha, \beta, \gamma$ are the angles of a triangle ($ \alpha + \beta + \gamma = \pi $), prove that: \[ \begin{vmatrix} \sin^2 \alpha & \sin \alpha \cos \alpha & \cos^2 \alpha \\ \sin^2 \beta & \sin \beta \cos \beta & \cos^2 \beta \\ \sin^2 \gamma & \sin \gamma \cos \gamma & \cos^2 \gamma \end{vmatrix} = -\sin(\alpha – \beta)\sin(\beta – \gamma)\sin(\gamma – \alpha) \]
Show that the following determinant is independent of $\alpha, \beta, \gamma$ and find its constant value: \[ \Delta = \begin{vmatrix} \cos(\alpha+\beta) & -\sin(\alpha+\beta) & \cos 2\beta \\ \sin \alpha & \cos \alpha & \sin \beta \\ -\cos \alpha & \sin \alpha & \cos \beta \end{vmatrix} \]
If $\alpha, \beta, \gamma$ are the roots of the cubic equation $x^3 + px + q = 0$, evaluate: \[ \begin{vmatrix} \alpha & \beta & \gamma \\ \beta & \gamma & \alpha \\ \gamma & \alpha & \beta \end{vmatrix} \]
Prove that: \[ \begin{vmatrix} (\beta + \gamma)^2 & \alpha^2 & \alpha^2 \\ \beta^2 & (\gamma + \alpha)^2 & \beta^2 \\ \gamma^2 & \gamma^2 & (\alpha + \beta)^2 \end{vmatrix} = 2\alpha\beta\gamma(\alpha + \beta + \gamma)^3 \]
Without expanding, prove that: $\begin{vmatrix} a-b & b-c & c-a \\ b-c & c-a & a-b \\ c-a & a-b & b-c \end{vmatrix} = 0$
If $x, y, z$ are in Arithmetic Progression (A.P.), evaluate: $\begin{vmatrix} p+2 & p+3 & p+x \\ p+3 & p+4 & p+y \\ p+4 & p+5 & p+z \end{vmatrix}$
Prove that: $\begin{vmatrix} 1 & a & a^2-bc \\ 1 & b & b^2-ca \\ 1 & c & c^2-ab \end{vmatrix} = 0$
Using properties of determinants, show that: $\begin{vmatrix} x+k & y & z \\ x & y+k & z \\ x & y & z+k \end{vmatrix} = k^2(x+y+z+k)$
If $a, b, c$ are all distinct and $\begin{vmatrix} a & a^2 & 1+a^3 \\ b & b^2 & 1+b^3 \\ c & c^2 & 1+c^3 \end{vmatrix} = 0$, then prove that $abc = -1$.
Evaluate: $\begin{vmatrix} \sin^2 \theta & \cos^2 \theta & 1 \\ \cos^2 \theta & \sin^2 \theta & 1 \\ -10 & 12 & 2 \end{vmatrix}$
Show that: $\begin{vmatrix} y+z & x & x \\ y & z+x & y \\ z & z & x+y \end{vmatrix} = 4xyz$
Prove that: $\begin{vmatrix} a+b+2c & a & b \\ c & b+c+2a & b \\ c & a & c+a+2b \end{vmatrix} = 2(a+b+c)^3$
If $\Delta = \begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ a^3 & b^3 & c^3 \end{vmatrix}$, prove that $\Delta = (a-b)(b-c)(c-a)(a+b+c)$.
Without expanding, show that $\begin{vmatrix} 0 & p-q & p-r \\ q-p & 0 & q-r \\ r-p & r-q & 0 \end{vmatrix} = 0$.
Find the value of $k$ if $\begin{vmatrix} a-b-c & 2a & 2a \\ 2b & b-c-a & 2b \\ 2c & 2c & c-a-b \end{vmatrix} = (a+b+c)^k$.
Prove that: $\begin{vmatrix} 1+a & 1 & 1 \\ 1 & 1+b & 1 \\ 1 & 1 & 1+c \end{vmatrix} = abc\left(1 + \frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right)$.
Evaluate: $\begin{vmatrix} \omega & \omega^2 & 1 \\ \omega^2 & 1 & \omega \\ 1 & \omega & \omega^2 \end{vmatrix}$ where $\omega$ is a complex cube root of unity.
Prove that: $\begin{vmatrix} a^2+1 & ab & ac \\ ab & b^2+1 & bc \\ ac & bc & c^2+1 \end{vmatrix} = 1 + a^2 + b^2 + c^2$