This MCQ module is based on: 4.1 Introduction
TOPIC 10 OF 16
4.1 Introduction
🎓 Class 12
Mathematics
CBSE
Theory
Ch 4 — Determinants
⏱ ~15 min
🌐 Language: [gtranslate]
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This mathematics assessment will be based on: 4.1 Introduction
Targeting Class 12 level in Matrices, with Advanced difficulty.
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4.1 Introduction
In Chapter 3 we met the matrix \(A\) and the linear system \(AX=B\). This chapter introduces a single number — the determinant? \(\det(A)\) (also written \(|A|\)) — which packages crucial information about \(A\): whether \(A\) is invertible (\(\det A\ne 0\)) or singular? (\(\det A=0\)), and how \(A\) scales areas and volumes geometrically.
Pierre-Simon Laplace
1749 – 1827
French mathematician and astronomer who established systematic methods for evaluating determinants — including the famous cofactor expansion (Laplace expansion) along any row or column. His five-volume Mécanique Céleste applied determinant theory to celestial mechanics, demonstrating the long-term stability of the solar system.
4.2 Determinant
Determinant — definition
For a square matrix \(A\) of order \(n\), the determinant \(\det A\) (or \(|A|\)) is a scalar associated with \(A\). The simplest cases:1 × 1: \(|a|=a\) (just the entry — note this is not the absolute value here, despite the bars).
2 × 2: \(\begin{vmatrix}a & b\\ c & d\end{vmatrix}=ad-bc\).
3 × 3 (expansion along row 1): \[\begin{vmatrix}a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33}\end{vmatrix}=a_{11}(a_{22}a_{33}-a_{23}a_{32})-a_{12}(a_{21}a_{33}-a_{23}a_{31})+a_{13}(a_{21}a_{32}-a_{22}a_{31}).\] The signs alternate \(+,\,-,\,+\). You may expand along any row or any column (Laplace).
Notation
\(|A|\) for a determinant uses bars (vertical lines), e.g. \(\begin{vmatrix}1 & 2\\ 3 & 4\end{vmatrix}=-2\). The matrix itself \(A\) uses brackets: \(\begin{bmatrix}1 & 2\\ 3 & 4\end{bmatrix}\). Different objects — one is a scalar, one is an array.
4.3 Properties of Determinants
The seven properties
For a square matrix \(A\):P1. \(\det(A')=\det(A)\) — transpose preserves determinant.
P2. Swapping two rows (or columns) of \(A\) changes the sign of the determinant.
P3. If two rows (or columns) of \(A\) are identical, \(\det A=0\).
P4. Multiplying any row (or column) by a scalar \(k\) multiplies \(\det A\) by \(k\). Hence for an \(n\times n\) matrix, \(\det(kA)=k^n\det A\).
P5. If a row (or column) is the sum of two row vectors, the determinant splits into a sum: \(\det\begin{pmatrix}\vdots\\ a+b\\ \vdots\end{pmatrix}=\det\begin{pmatrix}\vdots\\ a\\ \vdots\end{pmatrix}+\det\begin{pmatrix}\vdots\\ b\\ \vdots\end{pmatrix}\).
P6. Adding \(k\) times one row to another does NOT change the determinant. (This is the engine of row reduction.)
P7. \(\det(AB)=\det(A)\cdot\det(B)\) (multiplicativity).
Two corollaries used constantly:
- \(\det(I_n)=1\). Hence if \(A\) is invertible, \(\det A\ne 0\) and \(\det(A^{-1})=1/\det A\).
- The determinant of a triangular matrix is the product of its diagonal entries.
Worked Examples
Example 1. Evaluate \(\begin{vmatrix}2 & 4\\ -1 & 2\end{vmatrix}\).
\(2\cdot 2-4\cdot(-1)=4+4=8\).
Example 2. Evaluate \(\Delta=\begin{vmatrix}1 & 2 & 4\\ -1 & 3 & 0\\ 4 & 1 & 0\end{vmatrix}\).
Expand along column 3 (two zeros): \(\Delta=4\cdot\begin{vmatrix}-1 & 3\\ 4 & 1\end{vmatrix}-0+0=4\cdot(-1-12)=-52\).
Example 3. Find x if \(\begin{vmatrix}x & 2\\ 18 & x\end{vmatrix}=\begin{vmatrix}6 & 2\\ 18 & 6\end{vmatrix}\).
RHS = 36 − 36 = 0. LHS = x² − 36. So x² = 36 ⇒ x = ±6.
Example 4. Show that \(\begin{vmatrix}1 & 1 & 1\\ a & b & c\\ a^2 & b^2 & c^2\end{vmatrix}=(a-b)(b-c)(c-a)\) (Vandermonde 3×3).
Apply C₂ → C₂ − C₁ and C₃ → C₃ − C₁ (P6, doesn't change determinant):
\(\begin{vmatrix}1 & 0 & 0\\ a & b-a & c-a\\ a^2 & b^2-a^2 & c^2-a^2\end{vmatrix}=\begin{vmatrix}b-a & c-a\\ (b-a)(b+a) & (c-a)(c+a)\end{vmatrix}\)
\(=(b-a)(c-a)\begin{vmatrix}1 & 1\\ b+a & c+a\end{vmatrix}=(b-a)(c-a)(c+a-b-a)=(b-a)(c-a)(c-b)\).
With sign rearrangement, this equals \((a-b)(b-c)(c-a)\). \(\square\)
4.4 Area of a Triangle
Area formula via determinant
The area of the triangle with vertices \(A(x_1,y_1)\), \(B(x_2,y_2)\), \(C(x_3,y_3)\) is
\[\boxed{\;\text{Area}=\dfrac{1}{2}\,\left|\,\begin{vmatrix}x_1 & y_1 & 1\\ x_2 & y_2 & 1\\ x_3 & y_3 & 1\end{vmatrix}\,\right|\;}\]
The outer absolute value bars handle the sign. The three points are collinear iff this determinant is zero.
Example 5. Find the area of the triangle with vertices A(3, 8), B(−4, 2), C(5, −1).
\(\Delta=\begin{vmatrix}3 & 8 & 1\\ -4 & 2 & 1\\ 5 & -1 & 1\end{vmatrix}=3(2+1)-8(-4-5)+1(4-10)=9+72-6=75\). Area = (1/2)·|75| = 37.5 sq. units.
Example 6. Show that points (1, 2), (2, 4), (3, 6) are collinear.
\(\begin{vmatrix}1 & 2 & 1\\ 2 & 4 & 1\\ 3 & 6 & 1\end{vmatrix}=1(4-6)-2(2-3)+1(12-12)=-2+2+0=0\). Determinant is 0, so the three points are collinear. (Indeed they all lie on \(y=2x\).)
Activity: Use Properties to Speed Up Computation
L3 ApplyMaterials: Pen, paper.
Predict: What is \(\begin{vmatrix}102 & 18 & 36\\ 1 & 3 & 4\\ 17 & 3 & 6\end{vmatrix}\)? Try without expanding directly.
- Notice row 1 = 6 · row 3 (since 102=6·17, 18=6·3, 36=6·6).
- By P3 (two proportional rows force det = 0 — actually we need P4 + P3 combined: factor 6 out of R₁, then R₁ = R₃, so det = 0).
- Verify: factor 6 out: \(6\cdot\begin{vmatrix}17 & 3 & 6\\ 1 & 3 & 4\\ 17 & 3 & 6\end{vmatrix}\). Rows 1 and 3 identical ⇒ det = 0. So original det = 6·0 = 0.
- Lesson: spot row/column relations BEFORE expanding. Saves enormous time.
Property recognition is more powerful than brute computation. Many JEE/competitive determinant problems collapse to 0 after one observation. Train your eye to spot proportional rows/columns, identical rows, or rows that combine to make zeros.
Competency-Based Questions
Scenario: A linear transformation in 2-D defined by \(A=\begin{bmatrix}3 & 1\\ 2 & 4\end{bmatrix}\) sends a unit square to a parallelogram.
Q1. The area of the parallelogram is:
L3 ApplyAnswer: \(|\det A|=|3\cdot 4-1\cdot 2|=10\) square units. The unit square (area 1) is scaled by \(|\det A|\).
Q2. (T/F) "If \(\det A=0\), the linear map collapses 2-D to a line or point." Justify.
L5 EvaluateTrue. Zero determinant means zero area for the image of the unit square — so the map's image is degenerate (lower dimension). This is the geometric meaning of "singular".
Q3. Compute \(\begin{vmatrix}\cos\theta & -\sin\theta\\ \sin\theta & \cos\theta\end{vmatrix}\). What's special about it?
L3 ApplyAnswer: \(\cos^2\theta+\sin^2\theta=1\). The 2-D rotation matrix has determinant 1 — rotations preserve area (and orientation).
Q4. Apply: a 3×3 matrix has \(\det A=5\). What is \(\det(2A)\)?
L3 ApplyAnswer: \(\det(2A)=2^3\det A=8\cdot 5=40\). Each of the 3 rows picks up a factor of 2.
Q5. Design: find k so that the points (1, k), (4, −2), (−3, 6) are collinear.
L6 CreateSolution: Set \(\begin{vmatrix}1 & k & 1\\ 4 & -2 & 1\\ -3 & 6 & 1\end{vmatrix}=0\). Expand: \(1(-2-6)-k(4+3)+1(24-6)=-8-7k+18=10-7k=0\). So \(k=10/7\).
Assertion–Reason Questions
Assertion (A): If two rows of a square matrix are identical, its determinant is zero.
Reason (R): Swapping the two equal rows changes the sign yet leaves the matrix unchanged, so \(\det A=-\det A\), forcing \(\det A=0\).
Reason (R): Swapping the two equal rows changes the sign yet leaves the matrix unchanged, so \(\det A=-\det A\), forcing \(\det A=0\).
Answer: (a). R is the elegant slick proof of A using P2.
Assertion (A): \(\det(AB)=\det(A)\cdot\det(B)\).
Reason (R): Determinants measure scaling; composing two scalings multiplies the scale factors.
Reason (R): Determinants measure scaling; composing two scalings multiplies the scale factors.
Answer: (a). R is the geometric reason for the algebraic statement A.
Assertion (A): Three collinear points form a triangle of zero area.
Reason (R): The "triangle" with collinear vertices is degenerate (a line segment), and the determinant formula gives 0.
Reason (R): The "triangle" with collinear vertices is degenerate (a line segment), and the determinant formula gives 0.
Answer: (a). R is the why; A is the what.
Frequently Asked Questions
What is a determinant?
For a square matrix A, the determinant det(A) (or |A|) is a scalar that captures whether A is invertible (det ≠ 0) or singular (det = 0). It generalises the area/volume scaling factor.
How do you compute a 2×2 determinant?
|a b; c d| = ad − bc.
How do you compute a 3×3 determinant?
Expand along any row or column with alternating signs +, −, +.
What are the properties of determinants?
det(A')=det(A); swapping rows/columns flips sign; equal rows ⇒ det = 0; scalar multiple of a row scales det; row sums split; adding a multiple of one row to another preserves det; det(AB) = det(A)·det(B).
How do you find the area of a triangle using determinants?
Area = (1/2)·|det of the 3×3 matrix [[x₁,y₁,1],[x₂,y₂,1],[x₃,y₃,1]]|.
What does it mean for a matrix to be singular?
det = 0. Singular matrices have no inverse and the linear map collapses some non-zero vector to zero.
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