Determinants and Cramer's Rule
Solving linear systems with matrices
Determinants and Cramer's Rule
Solving linear systems with matrices
Math - Grade 9-12
- 1
Find the determinant of the matrix [[4, 7], [2, 5]].
For a 2 by 2 matrix [[a, b], [c, d]], the determinant is ad - bc.
The determinant is 4(5) - 7(2) = 20 - 14 = 6. - 2
Find the determinant of the matrix [[-3, 6], [4, -2]].
The determinant is (-3)(-2) - 6(4) = 6 - 24 = -18. - 3
Find the determinant of the matrix [[1, 2, 3], [0, 4, 5], [0, 0, 6]].
For a triangular matrix, multiply the entries on the main diagonal.
The determinant is 1(4)(6) = 24 because the matrix is triangular, so the determinant is the product of the diagonal entries. - 4
Find the determinant of the matrix [[2, 1, 0], [3, -1, 4], [1, 2, 5]].
Use cofactor expansion along the row with the zero.
Expanding along the first row gives 2((-1)(5) - 4(2)) - 1(3(5) - 4(1)) + 0 = 2(-13) - 11 = -37. The determinant is -37. - 5
Use Cramer's Rule to solve the system: 2x + y = 7 and x - y = 2.
Replace the x-column with the constants to find Dx, and replace the y-column with the constants to find Dy.
The coefficient determinant is D = 2(-1) - 1(1) = -3. The x determinant is Dx = 7(-1) - 1(2) = -9, so x = Dx/D = 3. The y determinant is Dy = 2(2) - 7(1) = -3, so y = Dy/D = 1. The solution is x = 3 and y = 1. - 6
Use Cramer's Rule to solve the system: 3x - 2y = 4 and x + y = 5.
The coefficient determinant is D = 3(1) - (-2)(1) = 5. The x determinant is Dx = 4(1) - (-2)(5) = 14, so x = 14/5. The y determinant is Dy = 3(5) - 4(1) = 11, so y = 11/5. The solution is x = 14/5 and y = 11/5. - 7
A system has coefficient matrix [[5, 2], [10, 4]]. Explain why Cramer's Rule cannot give a unique solution for this system.
Check whether the two rows are multiples of each other.
The determinant is 5(4) - 2(10) = 20 - 20 = 0. Since the coefficient determinant is 0, Cramer's Rule cannot be used to find a unique solution. - 8
Use Cramer's Rule to solve the system: 4x + 3y = 18 and 2x - y = 0.
The coefficient determinant is D = 4(-1) - 3(2) = -10. The x determinant is Dx = 18(-1) - 3(0) = -18, so x = 9/5. The y determinant is Dy = 4(0) - 18(2) = -36, so y = 18/5. The solution is x = 9/5 and y = 18/5. - 9
Find the determinant of the matrix [[0, 3, -1], [2, 1, 4], [5, 0, 2]] using cofactor expansion along the first row.
Remember the signs across the first row are positive, negative, positive.
Expanding along the first row gives 0 - 3(2(2) - 4(5)) + (-1)(2(0) - 1(5)). This equals -3(4 - 20) + (-1)(-5) = 48 + 5 = 53. The determinant is 53. - 10
Use Cramer's Rule to solve the system: x + y + z = 6, 2x - y + z = 3, and x + 2y - z = 2.
Set up D, Dx, Dy, and Dz carefully before dividing.
The coefficient determinant is D = 6. Replacing the x-column gives Dx = 6, so x = 1. Replacing the y-column gives Dy = 12, so y = 2. Replacing the z-column gives Dz = 18, so z = 3. The solution is x = 1, y = 2, and z = 3. - 11
For the system ax + by = e and cx + dy = f, write the Cramer's Rule formulas for x and y, assuming the coefficient determinant is not 0.
The coefficient determinant is D = ad - bc. The formula for x is x = (ed - bf)/(ad - bc), and the formula for y is y = (af - ec)/(ad - bc). - 12
Use determinants to decide whether the system 6x - 9y = 12 and -2x + 3y = -4 has a unique solution. Then explain your conclusion.
Compare the coefficients and constants in the two equations.
The coefficient determinant is 6(3) - (-9)(-2) = 18 - 18 = 0. Since the determinant is 0, the system does not have a unique solution. The second equation is a multiple of the first, so the system has infinitely many solutions.