What is 4.5 Minors and Cofactors in NCERT Class 12 Mathematics Chapter 4?

4.5 Minors and Cofactors is the topic of Part 2 of Chapter 4 (Ch 4 — Determinants) in NCERT Class 12 Mathematics. This lesson builds intuition through examples, visuals, and graded practice.

The cofactor A_ij of a_ij is the signed minor: A_ij = (u22121)^(i+j) u00b7 M_ij. The (i+j) sign creates the alternating + u2212 + u2212 pattern across the matrix.

What is the formula for det(adj A)?

For an n u00d7 n matrix A, det(adj A) = |A|^(nu22121). For n = 3 this is |A|u00b2. Useful in problems testing knowledge of determinant identities.

The cofactor A_ij of a_ij is the signed minor: A_ij = (−1)^(i+j) · M_ij. The (i+j) sign creates the alternating + − + − pattern across the matrix.

What is the formula for det(adj A)?

For an n × n matrix A, det(adj A) = |A|^(n−1). For n = 3 this is |A|². Useful in problems testing knowledge of determinant identities.

TOPIC 11 OF 16

4.5 Minors and Cofactors

Q: What is a minor of a determinant?

The minor M_ij of element a_ij in an n u00d7 n determinant is the (nu22121) u00d7 (nu22121) determinant obtained by deleting row i and column j from the original.

Q: What is the adjoint of a matrix?

The adjoint adj(A) is the transpose of the matrix of cofactors: (adj A)_ij = A_ji = cofactor of a_ji. It satisfies A u00b7 (adj A) = (adj A) u00b7 A = |A| u00b7 I.

Q: What is the formula for the inverse via adjoint?

For an invertible square matrix A (det A u2260 0): Au207bu00b9 = (1/|A|) u00b7 adj(A). The matrix is invertible iff |A| u2260 0; otherwise no inverse exists.

Q: How do you check if a matrix is invertible?

Compute |A|. If |A| u2260 0 the matrix is invertible (called non-singular). If |A| = 0 the matrix is singular and has no inverse.

🎓 Class 12 Mathematics CBSE Theory Ch 4 — Determinants ⏱ ~15 min

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4.5 Minors and Cofactors

Minor and Cofactor

For an \(n\times n\) determinant \(\Delta=|a_{ij}|\):
The minor \(M_{ij}\) of the element \(a_{ij}\) is the \((n-1)\times(n-1)\) determinant obtained by deleting the \(i\)-th row and \(j\)-th column.

The cofactor \(A_{ij}\) is the signed minor: \[\boxed{\;A_{ij}=(-1)^{i+j}\,M_{ij}\;}\] The \((-1)^{i+j}\) sign produces the chess-board pattern \(\begin{matrix}+ & - & +\\ - & + & -\\ + & - & +\end{matrix}\).

The cofactor expansion (Laplace) along row \(i\):

\[\Delta=\sum_{j=1}^{n}a_{ij}A_{ij}=a_{i1}A_{i1}+a_{i2}A_{i2}+\cdots+a_{in}A_{in}.\]

You may equally expand along column \(j\). The same scalar emerges either way.

Cross-cofactor identity

The expansion of \(\Delta\) along row \(i\) using cofactors of a different row \(k\ (k\ne i)\) gives ZERO: \[\sum_j a_{ij}A_{kj}=0\quad (k\ne i).\] This is the key identity behind \(A\cdot\text{adj}(A)=|A|\cdot I\).

Example of computing minors and cofactors

Example 7. Find minors and cofactors of \(\begin{vmatrix}2 & -4\\ 0 & 3\end{vmatrix}\).

\(M_{11}=3,\ M_{12}=0,\ M_{21}=-4,\ M_{22}=2\). Cofactors: \(A_{11}=+3,\ A_{12}=-0=0,\ A_{21}=-(-4)=4,\ A_{22}=+2\). Det = \(2\cdot 3-(-4)\cdot 0=6\) (expanding along row 1).

Example 8. Find all minors and cofactors of \(\Delta=\begin{vmatrix}1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\end{vmatrix}\) (the identity).

By symmetry: \(M_{ii}=1\) for all \(i\), and \(M_{ij}=0\) for \(i\ne j\). Cofactors: \(A_{ii}=+1\), \(A_{ij}=0\) for \(i\ne j\). Hence \(\Delta=1\cdot 1+0+0=1\).

4.5 Adjoint and Inverse of a Matrix

4.5.1 Adjoint of a matrix

Adjoint (or adjugate)

For a square matrix \(A=[a_{ij}]\), the adjoint \(\text{adj}(A)\) is the transpose of the matrix of cofactors: \[\text{adj}(A)=\,[A_{ij}]'=\,[A_{ji}].\] Position \((i,j)\) of \(\text{adj}(A)\) holds the cofactor \(A_{ji}\) (note swap of indices).

Fundamental identity

\[\boxed{\;A\cdot\text{adj}(A)=\text{adj}(A)\cdot A=|A|\cdot I_n\;}\] This is the engine that lets us compute \(A^{-1}\). It comes directly from the cofactor expansion (diagonal entries) and the cross-cofactor identity (off-diagonals are zero).

Worked construction: for \(A=\begin{bmatrix}a & b\\ c & d\end{bmatrix}\) (2×2),

\[\text{adj}(A)=\begin{bmatrix}d & -b\\ -c & a\end{bmatrix}.\]

Verify: \(A\cdot\text{adj}(A)=\begin{bmatrix}ad-bc & 0\\ 0 & ad-bc\end{bmatrix}=(ad-bc)\,I_2=|A|I_2\). ✓

4.5.2 Inverse of a matrix via adjoint

Formula for the inverse

A square matrix \(A\) is invertible iff \(|A|\ne 0\). When invertible: \[\boxed{\;A^{-1}=\dfrac{1}{|A|}\,\text{adj}(A)\;}\] \(A\) is called non-singular when \(|A|\ne 0\) and singular when \(|A|=0\).

Useful identities

For invertible \(A,B\) of order \(n\):

\(\det(A^{-1})=\dfrac{1}{\det A}\).
\(\det(\text{adj}\,A)=|A|^{n-1}\). For \(n=3\): \(\det(\text{adj}\,A)=|A|^2\).
\(\text{adj}(\text{adj}\,A)=|A|^{n-2}\,A\). For \(n=3\): equal to \(|A|\,A\).
\((AB)^{-1}=B^{-1}A^{-1}\).
\(\text{adj}(AB)=\text{adj}(B)\cdot\text{adj}(A)\) (order reversal, like inverse and transpose).

Worked Examples

Example 9. Find the adjoint of \(A=\begin{bmatrix}1 & 4\\ 3 & 2\end{bmatrix}\).

\(\text{adj}(A)=\begin{bmatrix}2 & -4\\ -3 & 1\end{bmatrix}\) (swap diagonal, negate off-diagonal — the 2×2 shortcut).

Example 10. Find the inverse of \(A=\begin{bmatrix}1 & 4\\ 3 & 2\end{bmatrix}\).

\(|A|=2-12=-10\). \(A^{-1}=\dfrac{1}{-10}\begin{bmatrix}2 & -4\\ -3 & 1\end{bmatrix}=\begin{bmatrix}-1/5 & 2/5\\ 3/10 & -1/10\end{bmatrix}\). Verify: \(AA^{-1}=I_2\).

Example 11. Find the inverse of \(A=\begin{bmatrix}2 & 3\\ 1 & 2\end{bmatrix}\) and verify \(A\cdot\text{adj}(A)=|A|I_2\).

\(|A|=4-3=1\). \(\text{adj}(A)=\begin{bmatrix}2 & -3\\ -1 & 2\end{bmatrix}\). \(A^{-1}=\text{adj}(A)\) since \(|A|=1\). Verify: \(A\cdot\text{adj}(A)=\begin{bmatrix}4-3 & -6+6\\ 2-2 & -3+4\end{bmatrix}=\begin{bmatrix}1 & 0\\ 0 & 1\end{bmatrix}=|A|I_2\). ✓

Example 12. Find the inverse of \(A=\begin{bmatrix}2 & 3 & 1\\ 1 & 1 & 2\\ 5 & 1 & 5\end{bmatrix}\).

First compute \(|A|\). Expand along row 1: \(2(5-2)-3(5-10)+1(1-5)=6+15-4=17\).
Now compute the cofactor matrix:
\(A_{11}=+(5-2)=3\), \(A_{12}=-(5-10)=5\), \(A_{13}=+(1-5)=-4\).
\(A_{21}=-(15-1)=-14\), \(A_{22}=+(10-5)=5\), \(A_{23}=-(2-15)=13\).
\(A_{31}=+(6-1)=5\), \(A_{32}=-(4-1)=-3\), \(A_{33}=+(2-3)=-1\).
Cofactor matrix \(C=\begin{bmatrix}3 & 5 & -4\\ -14 & 5 & 13\\ 5 & -3 & -1\end{bmatrix}\). Adjoint = transpose: \(\text{adj}(A)=\begin{bmatrix}3 & -14 & 5\\ 5 & 5 & -3\\ -4 & 13 & -1\end{bmatrix}\).
\(A^{-1}=\dfrac{1}{17}\,\text{adj}(A)\). Verify a few entries of \(A\cdot A^{-1}\) to confirm.

Example 13. If \(A=\begin{bmatrix}2 & 3\\ 1 & -4\end{bmatrix}\) and \(B=\begin{bmatrix}1 & -2\\ -1 & 3\end{bmatrix}\), verify \((AB)^{-1}=B^{-1}A^{-1}\).

\(AB=\begin{bmatrix}-1 & 5\\ 5 & -14\end{bmatrix}\). \(|AB|=14-25=-11\). \(\text{adj}(AB)=\begin{bmatrix}-14 & -5\\ -5 & -1\end{bmatrix}\). \((AB)^{-1}=\dfrac{1}{-11}\begin{bmatrix}-14 & -5\\ -5 & -1\end{bmatrix}=\begin{bmatrix}14/11 & 5/11\\ 5/11 & 1/11\end{bmatrix}\).

Now \(|A|=-11\), \(A^{-1}=\dfrac{1}{-11}\begin{bmatrix}-4 & -3\\ -1 & 2\end{bmatrix}\). \(|B|=1\), \(B^{-1}=\begin{bmatrix}3 & 2\\ 1 & 1\end{bmatrix}\). Compute \(B^{-1}A^{-1}=\dfrac{1}{-11}\begin{bmatrix}3 & 2\\ 1 & 1\end{bmatrix}\begin{bmatrix}-4 & -3\\ -1 & 2\end{bmatrix}=\dfrac{1}{-11}\begin{bmatrix}-14 & -5\\ -5 & -1\end{bmatrix}=(AB)^{-1}\). ✓

Activity: 2×2 Inverse Recipe Cheat-Sheet

L3 Apply

Materials: Pen, paper.

Predict: What is the inverse of \(\begin{bmatrix}3 & 5\\ 1 & 2\end{bmatrix}\) — without going through cofactors?

Recipe: For 2×2 \(A=\begin{bmatrix}a & b\\ c & d\end{bmatrix}\), compute \(\det A=ad-bc\).
If \(\det A=0\), no inverse.
Otherwise: \(A^{-1}=\dfrac{1}{ad-bc}\begin{bmatrix}d & -b\\ -c & a\end{bmatrix}\). Swap diagonal, negate off-diagonal, divide by \(\det A\).
Apply: \(\det=6-5=1\). Inverse = \(\begin{bmatrix}2 & -5\\ -1 & 3\end{bmatrix}\). Verify by multiplying.
Practice: invert \(\begin{bmatrix}2 & 1\\ 7 & 4\end{bmatrix}\) (det=1, inverse \(\begin{bmatrix}4 & -1\\ -7 & 2\end{bmatrix}\)).

The 2×2 recipe is so quick it should be automatic. For 3×3 the cofactor route is mechanical but lengthy; row-reduction with [A | I] → [I | A⁻¹] is faster in practice. Both methods produce the same inverse — the choice is about speed.

Competency-Based Questions

Scenario: A 3×3 matrix \(A\) represents a transformation in 3-D. Its determinant is 5.

Q1. Is \(A\) invertible?

L3 Apply

Yes. \(|A|=5\ne 0\), so \(A\) is non-singular and \(A^{-1}\) exists.

Q2. Compute \(\det(\text{adj}\,A)\).

L3 Apply

Answer: For \(n=3\): \(\det(\text{adj}\,A)=|A|^{n-1}=5^2=25\).

Q3. (T/F) "If \(A\) is a non-singular square matrix, then \(\text{adj}(\text{adj}\,A)\) is always equal to \(A\)." Justify.

L5 Evaluate

False (in general). The identity is \(\text{adj}(\text{adj}\,A)=|A|^{n-2}\,A\). For \(n=2\) this gives \(|A|^0\,A=A\) ✓. For \(n=3\): \(|A|\,A\), which equals \(A\) only if \(|A|=1\). So true only in special cases.

Q4. Compute \(\det A^{-1}\) given \(\det A=2\) for some 4×4 matrix.

L3 Apply

Answer: \(\det(A^{-1})=1/\det A=1/2\).

Q5. Design: build a 2×2 matrix that has determinant exactly 2 AND has integer entries AND has integer inverse. Justify your construction.

L6 Create

Solution: For \(A^{-1}\) to be integer, \(\det A=\pm 1\) (since \(A^{-1}=\text{adj}(A)/|A|\)). With \(|A|=2\) the inverse will have half-integers — not integer. So no such matrix exists; the question reveals the role of det = ±1 (unimodular matrices) in producing integer inverses. To salvage: ask for \(|A|=1\), e.g. \(\begin{bmatrix}3 & 2\\ 1 & 1\end{bmatrix}\) with inverse \(\begin{bmatrix}1 & -2\\ -1 & 3\end{bmatrix}\). Both integer ✓.

Assertion–Reason Questions

Assertion (A): Every square matrix has an inverse.
Reason (R): The formula \(A^{-1}=\text{adj}(A)/|A|\) requires \(|A|\ne 0\).

(a) Both true, R explains A.

(b) Both true, R doesn't explain A.

(c) A true, R false.

(d) A false, R true.

Answer: (d). A is false (singular matrices have no inverse). R is true and is the precise reason A is false.

Assertion (A): \(A\cdot\text{adj}(A)=|A|\cdot I\).
Reason (R): The diagonal entries of \(A\cdot\text{adj}(A)\) are cofactor expansions of \(\det A\) along that row; the off-diagonals are cross-row cofactor expansions, which vanish.

(a) Both true, R explains A.

(b) Both true, R doesn't explain A.

(c) A true, R false.

(d) A false, R true.

Answer: (a). R is the precise proof outline of A.

Assertion (A): \((AB)^{-1}=A^{-1}B^{-1}\) for invertible \(A,B\).
Reason (R): Inverse of a product is product of inverses in original order.

(a) Both true, R explains A.

(b) Both true, R doesn't explain A.

(c) A true, R false.

(d) A false, R true.

Answer: (d). Both A and R are FALSE. The correct identity is \((AB)^{-1}=B^{-1}A^{-1}\) — order REVERSES. (Same as \((AB)'\).)

Frequently Asked Questions

What is a minor of a determinant?

The minor M_ij of element a_ij is the (n−1) × (n−1) determinant obtained by deleting row i and column j.

What is a cofactor?

The cofactor A_ij = (−1)^(i+j) · M_ij. The (i+j) sign creates the alternating + − + − pattern.

What is the adjoint of a matrix?

adj(A) is the transpose of the matrix of cofactors: (adj A)_ij = A_ji.

What is the formula for the inverse via adjoint?

A⁻¹ = (1/|A|) · adj(A), valid when |A| ≠ 0.

How do you check if a matrix is invertible?

Compute |A|. Invertible iff |A| ≠ 0.

What is det(adj A)?

det(adj A) = |A|^(n−1). For n = 3 this is |A|².

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