answer:Let x be the number of units of product A and y be the number of units of product B. We can set up the following system of inequalities based on the given information: 3x + 4y ≤ 20 (Machine X constraint) 2x + 1y ≤ 10 (Machine Y constraint) x ≥ 0 (Non-negativity constraint for product A) y ≥ 0 (Non-negativity constraint for product B) We want to find the maximum number of units of each product that can be produced, so we need to find the feasible region of this system of inequalities. To do this, we can graph the inequalities: 1. 3x + 4y ≤ 20 To find the boundary line, set the inequality to an equation: 3x + 4y = 20 When x = 0, y = 5 When y = 0, x = 20/3 ≈ 6.67 Plot the points (0, 5) and (6.67, 0) and draw the line connecting them. Shade the region below the line, as we want values less than or equal to 20. 2. 2x + 1y ≤ 10 To find the boundary line, set the inequality to an equation: 2x + 1y = 10 When x = 0, y = 10 When y = 0, x = 5 Plot the points (0, 10) and (5, 0) and draw the line connecting them. Shade the region below the line, as we want values less than or equal to 10. 3. x ≥ 0 This is the vertical line at x = 0. Shade the region to the right of the line. 4. y ≥ 0 This is the horizontal line at y = 0. Shade the region above the line. The feasible region is the area where all shaded regions overlap. In this case, it is a quadrilateral with vertices at (0, 0), (0, 5), (5, 0), and (2, 3). Now we need to check the number of units of each product at each vertex: (0, 0): 0 units of A and 0 units of B (0, 5): 0 units of A and 5 units of B (5, 0): 5 units of A and 0 units of B (2, 3): 2 units of A and 3 units of B The maximum number of units that can be produced is at the vertex (2, 3), where 2 units of product A and 3 units of product B can be produced.

answer:We can write the given system of linear equations as a matrix equation: begin{bmatrix} 1 & 2 & -1 3 & -4 & 5 1 & -1 & 3 end{bmatrix} begin{bmatrix} x y z end{bmatrix} = begin{bmatrix} 6 7 4 end{bmatrix} Let's denote the matrix on the left as A, the column vector of variables as X, and the column vector on the right as B. Then, we have: AX = B To solve for X, we need to find the inverse of matrix A, denoted as A^{-1}. Then, we can multiply both sides of the equation by A^{-1}: A^{-1}AX = A^{-1}B Since A^{-1}A = I (the identity matrix), we get: IX = A^{-1}B And since IX = X, we have: X = A^{-1}B Now, we need to find the inverse of matrix A: A = begin{bmatrix} 1 & 2 & -1 3 & -4 & 5 1 & -1 & 3 end{bmatrix} Using the matrix inversion method, we first find the determinant of A: det(A) = 1(-12 - 5) - 2(15 - 5) - (-1)(3 + 4) = -17 - 20 + 7 = -30 Now, we find the matrix of cofactors of A: C = begin{bmatrix} -12 & -5 & -7 15 & -5 & -10 3 & 4 & 7 end{bmatrix} Next, we find the adjugate of A by transposing the matrix of cofactors: text{adj}(A) = begin{bmatrix} -12 & 15 & 3 -5 & -5 & 4 -7 & -10 & 7 end{bmatrix} Finally, we find the inverse of A by dividing the adjugate by the determinant: A^{-1} = frac{1}{det(A)} cdot text{adj}(A) = frac{1}{-30} begin{bmatrix} -12 & 15 & 3 -5 & -5 & 4 -7 & -10 & 7 end{bmatrix} = begin{bmatrix} 0.4 & -0.5 & -0.1 1/6 & 1/6 & -1/6 7/30 & 1/3 & -7/30 end{bmatrix} Now, we can find the solution X by multiplying A^{-1} and B: X = A^{-1}B = begin{bmatrix} 0.4 & -0.5 & -0.1 1/6 & 1/6 & -1/6 7/30 & 1/3 & -7/30 end{bmatrix} begin{bmatrix} 6 7 4 end{bmatrix} = begin{bmatrix} 1 -1 2 end{bmatrix} So, the solution to the system of linear equations is: x = 1, quad y = -1, quad z = 2

answer:To solve the given system of linear equations using block matrices, we first represent the system as a matrix equation: begin{bmatrix} 2 & 1 & 1 3 & -2 & -1 1 & 3 & 2 end{bmatrix} begin{bmatrix} x y z end{bmatrix} = begin{bmatrix} 9 4 7 end{bmatrix} Now, we can write this as AX = B, where A is the matrix of coefficients, X is the column vector of variables, and B is the column vector of constants. To find X, we need to compute the inverse of A and multiply it with B. So, we have X = A^{-1}B. First, let's find the inverse of matrix A: A = begin{bmatrix} 2 & 1 & 1 3 & -2 & -1 1 & 3 & 2 end{bmatrix} To find the inverse of A, we will use the formula A^{-1} = frac{1}{text{det}(A)} text{adj}(A), where text{det}(A) is the determinant of A and text{adj}(A) is the adjugate of A. First, let's find the determinant of A: text{det}(A) = 2(-2 cdot 2 - (-1) cdot 3) - 1(3 cdot 2 - (-1) cdot 1) + 1(3 cdot 3 - 1 cdot (-2)) = 2(4 + 3) - 1(6 + 1) + 1(9 + 2) = 14 - 7 + 11 = 18 Now, let's find the adjugate of A: text{adj}(A) = begin{bmatrix} (-2 cdot 2 - (-1) cdot 3) & -(3 cdot 2 - (-1) cdot 1) & (3 cdot 3 - 1 cdot (-2)) -(1 cdot 1 - 1 cdot 2) & (2 cdot 2 - 1 cdot 1) & -(2 cdot 3 - 1 cdot 1) (1 cdot (-1) - 1 cdot (-2)) & -(2 cdot (-1) - 1 cdot 1) & (2 cdot 1 - 1 cdot 1) end{bmatrix} = begin{bmatrix} 4 & -5 & 11 -1 & 3 & -5 1 & 1 & 1 end{bmatrix} Now, we can find the inverse of A: A^{-1} = frac{1}{text{det}(A)} text{adj}(A) = frac{1}{18} begin{bmatrix} 4 & -5 & 11 -1 & 3 & -5 1 & 1 & 1 end{bmatrix} Finally, we can find the solution X by multiplying A^{-1} with B: X = A^{-1}B = frac{1}{18} begin{bmatrix} 4 & -5 & 11 -1 & 3 & -5 1 & 1 & 1 end{bmatrix} begin{bmatrix} 9 4 7 end{bmatrix} = frac{1}{18} begin{bmatrix} 36 -18 18 end{bmatrix} = begin{bmatrix} 2 -1 1 end{bmatrix} So, the solution to the system of linear equations is x = 2, y = -1, and z = 1.

answer:To solve this system of linear equations, we can use the method of substitution, elimination, or matrices. In this case, we will use the method of matrices. First, we will write the given system of equations in matrix form: ``` [ 2 3 -4 | 1 ] [ 5 2 1 | 10 ] [ 3 -4 2 | 5 ] ``` Now, we will perform Gaussian elimination to obtain the row-echelon form: 1. Swap Row 1 and Row 2: ``` [ 5 2 1 | 10 ] [ 2 3 -4 | 1 ] [ 3 -4 2 | 5 ] ``` 2. Replace Row 2 with (-2/5) * Row 1 + Row 2: ``` [ 5 2 1 | 10 ] [ 0 2.6 -4.4 | -3 ] [ 3 -4 2 | 5 ] ``` 3. Replace Row 3 with (-3/5) * Row 1 + Row 3: ``` [ 5 2 1 | 10 ] [ 0 2.6 -4.4 | -3 ] [ 0 -6.6 1.4 | -11 ] ``` 4. Divide Row 2 by 2.6: ``` [ 5 2 1 | 10 ] [ 0 1 -1.692| -1.154] [ 0 -6.6 1.4 | -11 ] ``` 5. Replace Row 3 with (6.6) * Row 2 + Row 3: ``` [ 5 2 1 | 10 ] [ 0 1 -1.692| -1.154] [ 0 0 -9.6 | -18.6] ``` 6. Divide Row 3 by -9.6: ``` [ 5 2 1 | 10 ] [ 0 1 -1.692| -1.154] [ 0 0 1 | 1.938] ``` Now that we have the row-echelon form, we can perform back-substitution to find the values of x, y, and z: 1. z = 1.938 2. y = -1.154 + 1.692 * z = -1.154 + 1.692 * 1.938 ≈ 2.308 3. x = (10 - 2 * y - z) / 5 = (10 - 2 * 2.308 - 1.938) / 5 ≈ 1.077 So, the solution to the system of equations is approximately: [x, y, z] = [1.077, 2.308, 1.938]