Determinants

Determinants

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

A determinant is a scalar value assigned to every square matrix. For a 2×2 matrix $\begin{pmatrix}a & b\c & d\end{pmatrix}$, the value is $\mathbf{ad - bc}$; for a 3×3 matrix it is computed by cofactor expansion along any row or column. The determinant of matrix $A$ (written $|A|$ or $\det A$) tells you whether $A$ is invertible: $\det A \neq 0 \Rightarrow A$ is non-singular. The geometric reading: $|\det A|$ equals the area of the parallelogram spanned by the column vectors of a 2×2 matrix, and the volume of the parallelepiped for 3×3. Must-remember identities: $\det(AB)=\det A \cdot \det B$, $\det(A^T)=\det A$, $\det(kA)=k^n\det A$, and $A^{-1}=\dfrac{1}{\det A}\operatorname{adj}(A)$.

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🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months.

Definition and Notation

For a square matrix $A$ of order $n$, the determinant is a number obtained by a specific recursive recipe. It is denoted $|A|$ or $\det A$ and exists only for square matrices. Determinant and matrix are different objects: $A$ is a table of numbers, $|A|$ is one number.

Computing 2×2 and 3×3 Determinants

• 2×2: $\det\begin{pmatrix}a & b\ c & d\end{pmatrix}=ad-bc$.
• 3×3 (expansion along row 1): $\det A = a_{11}(a_{22}a_{33}-a_{23}a_{32}) - a_{12}(a_{21}a_{33}-a_{23}a_{31}) + a_{13}(a_{21}a_{32}-a_{22}a_{31})$.

Cofactor and Minor

The minor $M_{ij}$ is the determinant of the $(n-1)\times(n-1)$ submatrix obtained by deleting row $i$ and column $j$. The cofactor is $C_{ij}=(-1)^{i+j}M_{ij}$, with the sign following the checkerboard pattern $\begin{pmatrix}+&-\ -&+\end{pmatrix}$. Expansion: $\det A = \sum_j a_{ij}C_{ij}$ along any chosen row $i$.

Elementary Operation Properties

Operation	Effect on $|A|$
$R_i \leftrightarrow R_j$ (row swap)	sign flips: $|A|\to -|A|$
$R_i \to kR_i$ (row scaling)	$|A|\to k|A|$
$R_i \to R_i + kR_j$ (add multiple)	no change
Identical/proportional rows or columns	$|A|=0$

Inverse via Adjoint

When $\det A\neq 0$, $A^{-1}=\dfrac{1}{\det A}\operatorname{adj}(A)$, where $\operatorname{adj}(A)$ is the transpose of the cofactor matrix. If $\det A=0$, the matrix is singular and no inverse exists.

Geometric Meaning

For a 2×2 matrix whose columns are vectors $\vec{u},\vec{v}$, $|\det A|$ equals the area of the parallelogram with sides $\vec{u},\vec{v}$. For 3×3, $|\det A|$ equals the volume of the parallelepiped formed by the three column vectors.

CUET-Style Question Patterns

Expect (i) direct 2×2/3×3 evaluation, (ii) MCQs on ”$\det A=0 \iff$ rows linearly dependent”, (iii) using $A^{-1}=\frac{1}{\det A}\operatorname{adj}(A)$ to find the inverse of a 2×2 matrix, and (iv) area/volume questions phrased in vectors.

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🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Product, Transpose and Scalar Rules — Why They Matter

The identity $\det(AB)=\det A \cdot \det B$ is the workhorse behind Cramer’s rule: solving $AX=B$ gives $x_i = \det(A_i)/\det A$. From it follow $\det(A^{-1})=1/\det A$ and $\det(A^n)=(\det A)^n$. The rule $\det(kA)=k^n\det A$ (not $k\det A$) trips up students who forget the dimension: scaling an $n\times n$ matrix by $k$ scales its $n$ rows, so the factor compounds to $k^n$.

Triangular and Block Tricks

For an upper or lower triangular matrix, $\det A$ is simply the product of the diagonal entries. Row-reducing a matrix to triangular form while tracking each swap and row-scaling lets you compute large determinants without cofactor expansion — a standard CUET time-saver.

Common Traps in CUET

• Writing $\det(A+B)=\det A+\det B$ — false in general. Only row/column splits inside one row work.
• Confusing $|A|$ (determinant) with $A$ (matrix) in statements.
• Sign slip in cofactors: $(-1)^{i+j}$ is mandatory; an omitted sign yields a wrong value with a correct-looking expansion.
• Dividing by $\det A$ when computing $A^{-1}$ without verifying $\det A\neq 0$.

Connection to Other CUET Topics

Determinants feed directly into Matrices (inverse, rank), Linear Equations (Cramer’s rule, consistency via $\det A\neq 0$), and Vector Algebra (area of triangle = $\tfrac12|\det(\vec{b}-\vec{a},\vec{c}-\vec{a})|$, scalar triple product = determinant of the $3\times 3$ vector matrix). Mastering the 3×3 expansion here removes friction from JEE-level 3D geometry later.

Worked Micro-Example

For $A=\begin{pmatrix}2 & 1 & 0\ 1 & 3 & 2\ 0 & 1 & 4\end{pmatrix}$, expand along column 3 (two zeros): $\det A = -2\cdot C_{23}+4\cdot C_{33}$. We get $C_{23}=-\det\begin{pmatrix}2&1\0&1\end{pmatrix}=-2$ and $C_{33}=\det\begin{pmatrix}2&1\1&3\end{pmatrix}=5$. So $\det A = -2(-2)+4(5)=4+20=24$. Area of the parallelogram of its first two columns is $24$ square units, and $A$ is invertible with $A^{-1}=\tfrac{1}{24}\operatorname{adj}(A)$.

Practice Prompts

1. Compute $\det\begin{pmatrix}1 & 2 & 3\ 0 & 4 & 5\ 0 & 0 & 6\end{pmatrix}$ without expansion, then verify the diagonal-product rule.
2. If $\det A=5$ for a $3\times 3$ matrix, state $\det(2A)$, $\det(A^2)$, and $\det(A^{-1})$ with reasoning.

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