SAT SUBJECT TEST MATH LEVEL 2
REVIEW OF MAJOR TOPICS
Numbers and Operations
Answers and Explanations
Addition, Subtraction, and Scalar Multiplication
1. (C) First find the sum of the matrices to the left of the equals sign: . Since the first row of the matrix to the right of the equals sign is (3 2), K must be 4. Since (J M ) is the bottom row, J = –2 and M = 4. Therefore, K + J + M = 6.
2. (E) In order for these matrices to be equal, .
Therefore, x = 2x2 and y = 6y – 10. Solving the first equation yields x = 0, and y = 2.
3. (B) To solve for X , first subtract from both sides of the equation.
1. (B) By definition, AB = = [−37 21]
2. (C) By definition, the first row, second column of the product is (x )(–x ) + (1)(1) = –x2 + 1.
3. (B) X must have as many rows as A has columns, which is 3. X must have as many columns as B does, which is 2.
4. (B) Matrix multiplication is row by column. Since the answer must be a 3 by 1 matrix, the only possible answer choice is B.
Determinants and Inverses of Square Matrices
1. (B) By definition, the determinant of is (p )(1) – (3)(–2) = p + 6.
2. (B) Enter the 3 by 3 matrix on the left side of the equation into your graphing calculator and evaluate its determinant (zero). The determinant on the right side of the equation is x2 – 20. Therefore x = ± ±4.47.
3. * (C) To find X , multiply both sides of the equation by on the right. Enter both matrices in your calculator, key the product on your graphing calculator, and key MATH/ENTER/ENTER to convert the decimal answer to a fraction.
Solving Systems of Equations
1. (B) First, write the system in standard form: . The matrix form of this equation is .
2. * (D) This is the matrix form AX = B of a system of equations. Multiply both sides of the equation by A–1 on the left to get the solution X = = A–1B. Enter the 2 by 2 matrix, A , and the 2 by 1 matrix, B , into your graphing calculator. Return to the home screen and enter A–1B = .