A brief note on Determinants . Read it before a test. Try each question at the end before you open its answer.
Evaluating ∣ a b c d ∣ \displaystyle \begin{vmatrix} a & b \\ c & d \end{vmatrix} a c b d = ad - bc$.
Minor and cofactor: M i j M_{ij} M ij is the determinant left when row i i i and column j j j are deleted. C i j = ( − 1 ) i + j M i j C_{ij} = (-1)^{i+j}M_{ij} C ij = ( − 1 ) i + j M ij . Expansion: ∣ A ∣ |A| ∣ A ∣ is the sum of (entry times cofactor) along any row or column. Choose the one with the most zeros. Mixed sum: entries of one row times the cofactors of a different row add up to 0 0 0 . Triangular matrix: the determinant is the product of the diagonal. Properties Transpose: ∣ A T ∣ = ∣ A ∣ |A^T| = |A| ∣ A T ∣ = ∣ A ∣ . Swap two rows: the sign changes. Multiply a row by k k k : the determinant is multiplied by k k k . Add a multiple of a row to another: no change. Equal rows, proportional rows or a zero row: ∣ A ∣ = 0 |A| = 0 ∣ A ∣ = 0 . Scalar multiple and product: ∣ k A ∣ = k n ∣ A ∣ |kA| = k^n|A| ∣ k A ∣ = k n ∣ A ∣ and ∣ A B ∣ = ∣ A ∣ ∣ B ∣ |AB| = |A||B| ∣ A B ∣ = ∣ A ∣∣ B ∣ . Area and collinear points Area: 1 2 ∣ det ∣ \tfrac{1}{2}\left|\det\right| 2 1 ∣ det ∣ , with rows ( x i , y i , 1 ) (x_i, y_i, 1) ( x i , y i , 1 ) . Collinear: the same determinant is 0 0 0 . Line through two points: the determinant with rows ( x , y , 1 ) (x, y, 1) ( x , y , 1 ) , ( x 1 , y 1 , 1 ) (x_1, y_1, 1) ( x 1 , y 1 , 1 ) and ( x 2 , y 2 , 1 ) (x_2, y_2, 1) ( x 2 , y 2 , 1 ) is 0 0 0 . Area scale factor: a matrix T T T multiplies every area by ∣ det T ∣ |\det T| ∣ det T ∣ . A worked example Find $
∣ 1 4 2 3 1 5 2 2 1 ∣ \displaystyle \begin{vmatrix} 1 & 4 & 2 \\ 3 & 1 & 5 \\ 2 & 2 & 1 \end{vmatrix} 1 3 2 4 1 2 2 5 1 b y m a k i n g z e r o s i n c o l u m n by making zeros in column b y mak in g z er os in co l u mn 1$.
Clear column 1. R 2 → R 2 − 3 R 1 R_2 \to R_2 - 3R_1 R 2 → R 2 − 3 R 1 gives ( 0 , − 11 , − 1 ) (0, -11, -1) ( 0 , − 11 , − 1 ) . R 3 → R 3 − 2 R 1 R_3 \to R_3 - 2R_1 R 3 → R 3 − 2 R 1 gives ( 0 , − 6 , − 3 ) (0, -6, -3) ( 0 , − 6 , − 3 ) . Neither operation changes the determinant. Expand along column 1. The value is $1 \cdot ∣ − 11 − 1 − 6 − 3 ∣ \displaystyle \begin{vmatrix} -11 & -1 \\ -6 & -3 \end{vmatrix} − 11 − 6 − 1 − 3 $.
Finish. 33 − 6 = 27 33 - 6 = 27 33 − 6 = 27 . Practice with new numbers
A program to try The program checks the rules of this note on one matrix: the transpose, a swap, adding a multiple of a row, and the scalar multiple 2 A 2A 2 A .
Run it. Which operations leave the determinant at 27?
1. Change 3 * B[1] to 5 * B[2]. Does the value change?
Mistakes to avoid Adding the products in a d − b c ad - bc a d − b c . The second product is subtracted. Writing ∣ k A ∣ = k ∣ A ∣ |kA| = k|A| ∣ k A ∣ = k ∣ A ∣ . It is k n ∣ A ∣ k^n|A| k n ∣ A ∣ . Forgetting the half. The area of a triangle is half of the determinant. Dropping the minus sign. The cofactor sign is ( − 1 ) i + j (-1)^{i+j} ( − 1 ) i + j . Believing that adding a multiple of a row changes the value. It does not. Check yourself Find $ ∣ 3 − 2 4 5 ∣ \displaystyle \begin{vmatrix} 3 & -2 \\ 4 & 5 \end{vmatrix} 3 4 − 2 5 $.
Stuck? Show a hint Show the answerWhat is the determinant of $ [ 2 5 1 0 3 7 0 0 4 ] \displaystyle \begin{bmatrix} 2 & 5 & 1 \\ 0 & 3 & 7 \\ 0 & 0 & 4 \end{bmatrix} 2 0 0 5 3 0 1 7 4 $?
Stuck? Show a hint Show the answerA A A is 3 × 3 3 \times 3 3 × 3 with ∣ A ∣ = 2 |A| = 2 ∣ A ∣ = 2 . Find ∣ − A ∣ |-A| ∣ − A ∣ . Stuck? Show a hint Show the answerFind the area of the triangle with vertices ( 0 , 0 ) (0, 0) ( 0 , 0 ) , ( 6 , 0 ) (6, 0) ( 6 , 0 ) and ( 2 , 5 ) (2, 5) ( 2 , 5 ) . Stuck? Show a hint Show the answerAre ( 0 , 1 ) (0, 1) ( 0 , 1 ) , ( 2 , 5 ) (2, 5) ( 2 , 5 ) and ( 3 , 7 ) (3, 7) ( 3 , 7 ) collinear? Stuck? Show a hint Show the answerSources NCERT, Mathematics Textbook for Class XII, Part I, Chapter 4 (Determinants). OpenStax, Precalculus 2e, Section 9.8 (Solving Systems with Cramer's Rule).