Given a matrix A = [2 4 6; 1 3 5; 7 9 11], find the determinant of A.
We can use the expansion method to find the determinant of A. Expansion along the first row: det(A) = 2 * det([3 5; 9 11]) - 4 * det([1 5; 7 11]) + 6 * det([1 3; 7 9]) Now, let’s calculate the determinants of the smaller matrices. det([3 5; 9 11]) = (3 * 11) - (5 * 9) = 33 - 45 = -12 det([1 5; 7 11]) = (1 * 11) - (5 * 7) = 11 - 35 = -24 det([1 3; 7 9]) = (1 * 9) - (3 * 7) = 9 - 21 = -12 Plugging these values back into the expansion formula, we have: det(A) = 2 * (-12) - 4 * (-24) + 6 * (-12) = -24 + 96 - 72 = 0 Therefore, the determinant of matrix A is 0.
Find the adjoint of the matrix B = [3 4 5; 1 2 3; 6 7 8].
To find the adjoint of a matrix, we need to find the cofactor of each element and then form the adjoint matrix by taking the transpose of the cofactor matrix. Let’s calculate the cofactors first: C11 = (-1)^(1+1) * det([2 3; 7 8]) = 1 * (16 - 21) = -5 C12 = (-1)^(1+2) * det([1 3; 6 8]) = -1 * (8 - 18) = 10 C13 = (-1)^(1+3) * det([1 2; 6 7]) = 1 * (7 - 12) = -5 C21 = (-1)^(2+1) * det([4 5; 7 8]) = -1 * (32 - 35) = 3 C22 = (-1)^(2+2) * det([3 5; 6 8]) = 1 * (24 - 30) = -6 C23 = (-1)^(2+3) * det([3 4; 6 7]) = -1 * (21 - 24) = 3 C31 = (-1)^(3+1) * det([4 5; 2 3]) = 1 * (12 - 10) = 2 C32 = (-1)^(3+2) * det([3 5; 1 3]) = -1 * (9 - 5) = 4 C33 = (-1)^(3+3) * det([3 4; 1 2]) = 1 * (6 - 4) = 2 Using these cofactors, we can form the adjoint matrix: adj(B) = [C11 C21 C31; C12 C22 C32; C13 C23 C33] = [-5 3 2; 10 -6 4; -5 3 2] Therefore, the adjoint of matrix B is [-5 3 2; 10 -6 4; -5 3 2].
For a matrix C = [2 1; 3 -2], find the value of k for which kC = C^(T).
To find the value of k, we need to equate kC to the transpose of C and solve for k. kC = [k * 2 k * 1; k * 3 k * (-2)] C^(T) = [2 3; 1 -2] Setting kC equal to C^(T), we have: [k * 2 k * 1; k * 3 k * (-2)] = [2 3; 1 -2] Comparing corresponding elements, we get: k * 2 = 2 (1) k * 1 = 3 (2) k * 3 = 1 (3) k * (-2) = -2 (4) From equation (2), we find that k = 3. Now, let’s substitute k = 3 into equations (1), (3), and (4) to check if they satisfy: Equation (1): k * 2 - 2 = 2
3 * 2 - 2 = 4 - 2 = 2 (satisfied) Equation (3): k * 3 - 1 = 1
3 * 3 - 1 = 9 - 1 = 8 (not satisfied) Equation (4): k * (-2) - (-2) = -2
3 * (-2) + 2 = -6 + 2 = -4 (not satisfied) Therefore, there is no value of k for which kC is equal to C^(T).
Solve the system of equations using determinants:
2x + 3y = 5
4x - 2y = 9
We can represent the system of equations using matrices: [A] * [X] = [B] where [A] = [2 3; 4 -2], [X] = [x; y], and [B] = [5; 9]. To solve for [X], we can use the formula: [X] = [A]^-1 * [B] where [A]^-1 is the inverse of matrix [A]. First, let’s find the determinant of [A]: det([A]) = (2 * -2) - (3 * 4) = -4 - 12 = -16 Since the determinant is not zero, matrix [A] is invertible. Next, let’s find the adjoint of [A]: adj([A]) = [C]^T where [C] is the cofactor matrix of [A] obtained by calculating the cofactor of each element. C11 = (-1)^(1+1) * det([-2]) = -1 * (-2) = 2 C12 = (-1)^(1+2) * det([4]) = 1 * 4 = 4 C21 = (-1)^(2+1) * det([3]) = -1 * 3 = -3 C22 = (-1)^(2+2) * det([2]) = 1 * 2 = 2 Thus, the adjoint matrix is: adj([A]) = [C]^T = [2 -3; 4 2] Finally, we can find the inverse of [A]: [A]^-1 = (1/det([A])) * adj([A]) = (1/-16) * [2 -3; 4 2] To find [X], we can now substitute the values: [X] = [A]^-1 * [B] = [(1/-16) * [2 -3; 4 2]] * [5; 9] Simplifying further, we get: [X] = [(1/-16) * [2 * 5 + (-3) * 9; 4 * 5 + 2 * 9]] = [(1/-16) * [-7; 26]] = [-7/-16; 26/-16] = [7/16; -13/8] Therefore, the solution to the system of equations is x = 7/16 and y = -13/8.
Find the value of k for which the matrix D = [k 2; 3 6] is singular.
A matrix is singular if its determinant is equal to zero. Let’s calculate the determinant of matrix D: det([D]) = (k * 6) - (2 * 3) = 6k - 6 For the matrix D to be singular, the determinant must be equal to zero. Therefore, we have:
6k - 6 = 0 Solving for k, we get:
6k = 6 k = 6/6 k = 1 Therefore, the matrix D = [1 2; 3 6] is singular when k = 1.
If the determinant of a 3x3 matrix E is 7, find the determinant of E^-1.
The determinant of the inverse of a matrix is equal to the reciprocal of the determinant of the original matrix. Given det(E) = 7, let’s find det(E^-1). We have: det(E^-1) = 1 / det(E) = 1 / 7 Therefore, the determinant of E^-1 is 1/7.
Solve the equation |4x - 3| = 7.
To solve this equation, we will consider two cases:
4x = 10 Dividing by 4, we have: x = 2.5 Case 2: 4x - 3 = -7 Adding 3 to both sides, we get:
4x = -4 Dividing by 4, we have: x = -1 Therefore, the solutions to the equation |4x - 3| = 7 are x = 2.5 and x = -1.
Find the value of x that satisfies the equation: |4x - 6| = |3 - x|
To solve this equation, we will consider two cases:
5x - 6 = 3 Adding 6 to both sides, we have:
5x = 9 Dividing by 5, we have: x = 9/5 Case 2: 4x - 6 = -(3 - x) Expanding the expression inside the absolute value, we have:
4x - 6 = -3 + x Subtracting x from both sides, we get:
3x - 6 = -3 Adding 6 to both sides, we have:
3x = 3 Dividing by 3, we have: x = 1 Therefore, the value of x that satisfies the equation |4x - 6| = |3 - x| is x = 9/5 and x = 1.
Solve the equation |3x + 2| = |2x - 3| - 2.
To solve this equation, we will consider two cases:
3x + 2 = 2x - 3 - 2 Simplifying further, we get:
3x + 2 = 2x - 5 Subtracting 2x from both sides, we have: x + 2 = -5 Subtracting 2 from both sides, we get: x = -7 Case 2: 3x + 2 = -(|2x - 3| - 2) Expanding the expression on the right side, we have:
3x + 2 = -(2x - 3 - 2) Simplifying further, we get:
3x + 2 = -2x + 1 + 2
3x + 2 = -2x + 3 Adding 2x to both sides, we have:
5x + 2 = 3 Subtracting 2 from both sides, we get:
5x = 1 Dividing by 5, we have: x = 1/5 Therefore, the solutions to the equation |3x + 2| = |2x - 3| - 2 are x = -7 and x = 1/5.
Solve the equation |x + 1| + |x - 2| = 5.
To solve this equation, we will consider three cases:
2x - 1 = 5 Adding 1 to both sides, we have:
2x = 6 Dividing by 2, we have: x = 3 Case 2: x + 1 ≥ 0 and x - 2 < 0 In this case, the equation becomes: (x + 1) - (x - 2) = 5 Simplifying further, we get:
2 = 5 This is not a valid solution. Case 3: x + 1 < 0 and x - 2 < 0 In this case, the equation becomes: -(x + 1) - (x - 2) = 5 Simplifying further, we get: -2x + 1 = 5 Subtracting 1 from both sides, we have: -2x = 4 Dividing by -2, we have: x = -2 Therefore, the solutions to the equation |x + 1| + |x - 2| = 5 are x = 3 and x = -2.