Quick answer: For a 2 × 2 2 \times 2 2 × 2 ∣ a b c d ∣ = a d − b c \begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc a c b d = a d − b c 3 × 3 3 \times 3 3 × 3 cofactors C i j = ( − 1 ) i + j M i j C_{ij} = (-1)^{i + j}M_{ij} C ij = ( − 1 ) i + j M ij inverse of a square matrix exists only when ∣ A ∣ ≠ 0 \lvert A\rvert \ne 0 ∣ A ∣ = 0 A − 1 = 1 ∣ A ∣ adj A A^{-1} = \frac{1}{\lvert A\rvert}\text{adj}A A − 1 = ∣ A ∣ 1 adj A ∣ A B ∣ = ∣ A ∣ ∣ B ∣ \lvert AB\rvert = \lvert A\rvert\lvert B\rvert ∣ A B ∣ = ∣ A ∣ ∣ B ∣ ∣ k A ∣ = k n ∣ A ∣ \lvert kA\rvert = k^n\lvert A\rvert ∣ k A ∣ = k n ∣ A ∣ n × n n \times n n × n
This continues from matrices: types and multiplication .
Evaluating determinants
3 × 3 expansion along the first row:
∣ a 1 b 1 c 1 a 2 b 2 c 2 a 3 b 3 c 3 ∣ = a 1 ( b 2 c 3 − b 3 c 2 ) − b 1 ( a 2 c 3 − a 3 c 2 ) + c 1 ( a 2 b 3 − a 3 b 2 ) \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} = a_1(b_2c_3 - b_3c_2) - b_1(a_2c_3 - a_3c_2) + c_1(a_2b_3 - a_3b_2) a 1 a 2 a 3 b 1 b 2 b 3 c 1 c 2 c 3 = a 1 ( b 2 c 3 − b 3 c 2 ) − b 1 ( a 2 c 3 − a 3 c 2 ) + c 1 ( a 2 b 3 − a 3 b 2 )
Expand along the row or column with the most zeros to save work.
Minors and cofactors
The minor M i j M_{ij} M ij
The cofactor is C i j = ( − 1 ) i + j M i j C_{ij} = (-1)^{i + j}M_{ij} C ij = ( − 1 ) i + j M ij
∣ A ∣ \lvert A\rvert ∣ A ∣ The sum of entries of a row times the cofactors of a different row is 0.
Properties of determinants
Property Effect ∣ A T ∣ = ∣ A ∣ \lvert A^T\rvert = \lvert A\rvert ∣ A T ∣ = ∣ A ∣ Rows and columns behave the same Swap two rows Sign changes Two identical rows Determinant is 0 Multiply a row by k Determinant multiplied by k Add a multiple of one row to another No change A row is a sum of two terms Split into two determinants Triangular matrix Product of diagonal entries ∣ A B ∣ \lvert AB\rvert ∣ A B ∣ ∣ A ∣ ∣ B ∣ \lvert A\rvert\lvert B\rvert ∣ A ∣ ∣ B ∣ ∣ k A ∣ \lvert kA\rvert ∣ k A ∣ n × n n \times n n × n k n ∣ A ∣ k^n\lvert A\rvert k n ∣ A ∣
Adjoint and inverse
adj A = [ C i j ] T , A − 1 = 1 ∣ A ∣ adj A \text{adj}A = [C_{ij}]^T, \quad A^{-1} = \frac{1}{\lvert A\rvert}\text{adj}A adj A = [ C ij ] T , A − 1 = ∣ A ∣ 1 adj A
For a 2 × 2 2 \times 2 2 × 2 [ a b c d ] \begin{bmatrix} a & b \\ c & d \end{bmatrix} [ a c b d ] [ d − b − c a ] \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} [ d − c − b a ]
Result (n × n n \times n n × n ∣ A ∣ ≠ 0 \lvert A\rvert \ne 0 ∣ A ∣ = 0 Value A ( adj A ) A(\text{adj}A) A ( adj A ) ∣ A ∣ I \lvert A\rvert I ∣ A ∣ I ∣ adj A ∣ \lvert\text{adj}A\rvert ∣ adj A ∣ ∣ A ∣ n − 1 \lvert A\rvert^{n - 1} ∣ A ∣ n − 1 adj ( adj A ) \text{adj}(\text{adj}A) adj ( adj A ) ∣ A ∣ n − 2 A \lvert A\rvert^{n - 2}A ∣ A ∣ n − 2 A ∣ A − 1 ∣ \lvert A^{-1}\rvert ∣ A − 1 ∣ 1 ∣ A ∣ \frac{1}{\lvert A\rvert} ∣ A ∣ 1 ( A B ) − 1 (AB)^{-1} ( A B ) − 1 B − 1 A − 1 B^{-1}A^{-1} B − 1 A − 1 ( A T ) − 1 (A^T)^{-1} ( A T ) − 1 ( A − 1 ) T (A^{-1})^T ( A − 1 ) T
A matrix with ∣ A ∣ = 0 \lvert A\rvert = 0 ∣ A ∣ = 0 singular and has no inverse.
Worked examples
Example 1: a 3 × 3 determinant
Evaluate ∣ 2 0 1 3 1 0 1 2 4 ∣ \begin{vmatrix} 2 & 0 & 1 \\ 3 & 1 & 0 \\ 1 & 2 & 4 \end{vmatrix} 2 3 1 0 1 2 1 0 4
Expanding along the first row: 2 ( 4 − 0 ) − 0 + 1 ( 6 − 1 ) = 8 + 5 = 13 2(4 - 0) - 0 + 1(6 - 1) = 8 + 5 = 13 2 ( 4 − 0 ) − 0 + 1 ( 6 − 1 ) = 8 + 5 = 13
Example 2: using properties
Evaluate ∣ 1 a b + c 1 b c + a 1 c a + b ∣ \begin{vmatrix} 1 & a & b + c \\ 1 & b & c + a \\ 1 & c & a + b \end{vmatrix} 1 1 1 a b c b + c c + a a + b
Add column 2 to column 3: every entry of column 3 becomes a + b + c a + b + c a + b + c
Example 3: inverse of a 2 × 2 matrix
Find the inverse of A = [ 3 1 5 2 ] A = \begin{bmatrix} 3 & 1 \\ 5 & 2 \end{bmatrix} A = [ 3 5 1 2 ]
∣ A ∣ = 6 − 5 = 1 \lvert A\rvert = 6 - 5 = 1 ∣ A ∣ = 6 − 5 = 1
A − 1 = [ 2 − 1 − 5 3 ] A^{-1} = \begin{bmatrix} 2 & -1 \\ -5 & 3 \end{bmatrix} A − 1 = [ 2 − 5 − 1 3 ]
Example 4: adjoint determinant
If A is 3 × 3 3 \times 3 3 × 3 ∣ A ∣ = 4 \lvert A\rvert = 4 ∣ A ∣ = 4 ∣ adj A ∣ \lvert\text{adj}A\rvert ∣ adj A ∣ ∣ 2 A ∣ \lvert 2A\rvert ∣ 2 A ∣
∣ adj A ∣ = 4 2 = 16 \lvert\text{adj}A\rvert = 4^2 = 16 ∣ adj A ∣ = 4 2 = 16 ∣ 2 A ∣ = 2 3 ⋅ 4 = 32 \lvert 2A\rvert = 2^3 \cdot 4 = 32 ∣ 2 A ∣ = 2 3 ⋅ 4 = 32
Summary table
Idea Key result Cofactor ( − 1 ) i + j M i j (-1)^{i + j}M_{ij} ( − 1 ) i + j M ij Inverse 1 ∣ A ∣ adj A \frac{1}{\lvert A\rvert}\text{adj}A ∣ A ∣ 1 adj A Exists when ∣ A ∣ ≠ 0 \lvert A\rvert \ne 0 ∣ A ∣ = 0 ∣ k A ∣ \lvert kA\rvert ∣ k A ∣ k n ∣ A ∣ k^n\lvert A\rvert k n ∣ A ∣ ∣ adj A ∣ \lvert\text{adj}A\rvert ∣ adj A ∣ ∣ A ∣ n − 1 \lvert A\rvert^{n - 1} ∣ A ∣ n − 1
Exam tips
Use row and column operations to create zeros before expanding.
If two rows become proportional after an operation, the answer is 0.
Remember the power n n n ∣ k A ∣ = k n ∣ A ∣ \lvert kA\rvert = k^n\lvert A\rvert ∣ k A ∣ = k n ∣ A ∣
Next, read Cramer's rule and consistency of linear equations , or see the NIMCET Algebra chapter .
Common mistakes
Forgetting the alternating signs of cofactors.
Writing ∣ k A ∣ = k ∣ A ∣ \lvert kA\rvert = k\lvert A\rvert ∣ k A ∣ = k ∣ A ∣ 3 × 3 3 \times 3 3 × 3
Forgetting to transpose the cofactor matrix when forming the adjoint.
Writing ( A B ) − 1 = A − 1 B − 1 (AB)^{-1} = A^{-1}B^{-1} ( A B ) − 1 = A − 1 B − 1
Frequently Asked Questions
What is the cofactor of an element?
Cᵢⱼ = (−1)ⁱ⁺ʲ Mᵢⱼ, where Mᵢⱼ is the minor obtained by deleting row i and column j.
A⁻¹ = adj A / |A|, provided |A| ≠ 0.
What is |adj A| for an n × n matrix?
|adj A| = |A|ⁿ⁻¹.
What happens to a determinant when two rows are swapped?
Its sign changes.
What is |kA| for a 3 × 3 matrix?
|kA| = k³|A|.
