Question:

Examine the consistency of the system of equations x+2y=2, 2x+3y=3

Answer:

1. Rearrange the equations as follows: x + 2y = 2 2x + 3y = 3

2. Subtract the first equation from the second equation: 2x + 3y = 3

• x + 2y = 2

3. Simplify the equation: x + y = 1

4. The system of equations is consistent.

Question:

Solve the system of equations, using matrix method 2x+3y+3z=5,x−2y+z=−4,3x−y−2z=3

Answer:

Step 1: Write down the equations in matrix form.

\begin{bmatrix} 2 & 3 & 3 \ 1 & -2 & 1 \ 3 & -1 & -2 \end{bmatrix} \begin{bmatrix} x \ y \ z \end{bmatrix}

\begin{bmatrix} 5 \ -4 \ 3 \end{bmatrix}

Step 2: Find the inverse of the matrix.

\begin{bmatrix} 2 & 3 & 3 \ 1 & -2 & 1 \ 3 & -1 & -2 \end{bmatrix}^{-1}

\frac{1}{14} \begin{bmatrix} -2 & 3 & 3 \ 3 & 2 & -1 \ -3 & -1 & 2 \end{bmatrix}

Step 3: Multiply the inverse of the matrix by the vector of constants.

\frac{1}{14} \begin{bmatrix} -2 & 3 & 3 \ 3 & 2 & -1 \ -3 & -1 & 2 \end{bmatrix} \begin{bmatrix} 5 \ -4 \ 3 \end{bmatrix}

\begin{bmatrix} \frac{1}{14} \ \frac{13}{14} \ -\frac{1}{7} \end{bmatrix}

Step 4: Write down the solution.

The solution to the system of equations is x=\frac{1}{14}, y=\frac{13}{14}, and z=-\frac{1}{7}.

Question:

Solve the system of equations, using matrix method 5x+2y=3,3x+2y=5

Answer:

Step 1: Write the equations in matrix form.

[5 2] [x] = [3] [3 2] [y] = [5]

Step 2: Find the inverse of the matrix.

[5 2] [3 2]

1/10[2 -2] 1/10[-3 5]

Step 3: Multiply the inverse matrix by the vector of constants.

1/10[2 -2] [3] = [1/5] 1/10[-3 5] [5] = [3/5]

Step 4: Solve for x and y.

x = 1/5 y = 3/5

Question:

The cost of 4 kg onion, 3 kg wheat and 2 kg rice is Rs 60. The cost of 2 kg onion,4 kg wheat and 6 kg rice is Rs 90. The cost of 6 kg onion 2 kg wheat and 3 kg rice is Rs 70. Find cost of each item per kg by matrix method.

Answer:

Let O = Cost of onion per kg W = Cost of wheat per kg R = Cost of rice per kg

Then,

4O + 3W + 2R = 60 2O + 4W + 6R = 90 6O + 2W + 3R = 70

We have to solve the above three equations by matrix method.

Step 1:

Write the equations in the form of matrix.

[4 3 2] [O] [60] [2 4 6] [W] = [90] [6 2 3] [R] [70]

Step 2:

Multiply the first row by -2 and add it to the second row.

[4 3 2] [O] [60] [0 -2 -4] [W] = [30] [6 2 3] [R] [70]

Step 3:

Multiply the first row by -6 and add it to the third row.

[4 3 2] [O] [60] [0 -2 -4] [W] = [30] [0 0 -2] [R] [10]

Step 4:

Divide the second row by -2 and multiply it by -1.

[4 3 2] [O] [60] [0 1 2] [W] = [-15] [0 0 -2] [R] [10]

Step 5:

Add the second row to the first row.

[4 4 4] [O] [45] [0 1 2] [W] = [-15] [0 0 -2] [R] [10]

Step 6:

Divide the third row by -2.

[4 4 4] [O] [45] [0 1 2] [W] = [-15] [0 0 1] [R] [-5]

Step 7:

Add the third row to the second row.

[4 4 4] [O] [45] [0 2 3] [W] = [-10] [0 0 1] [R] [-5]

Step 8:

Divide the second row by 2.

[4 4 4] [O] [45] [0 1 1.5] [W] = [-5] [0 0 1] [R] [-5]

Step 9:

Add the second row to the first row.

[4 5 5.5] [O] [40] [0 1 1.5] [W] = [-5] [0 0 1] [R] [-5]

Step 10:

Divide the first row by 4.

[1 1.25 1.375] [O] [10] [0 1 1.5] [W] = [-5] [0 0 1] [R] [-5]

Hence, the cost of each item per kg is O = Rs 10 W = Rs -5 R = Rs -5

Question:

Examine the consistency of the system of equationsx+3y=5, 2x+6y=8

Answer:

1. First, we need to isolate the variables.

For the first equation, we can subtract 3y from both sides to get x = 5 - 3y.

For the second equation, we can subtract 2x from both sides to get 6y = 8 - 2x.

2. Next, we can substitute the expression for x from the first equation into the second equation.

6y = 8 - 2(5 - 3y)

3. Simplify the equation to get 6y = -6 - 6y.

4. Finally, we can solve the equation by adding 6y to both sides and then dividing by two.

6y + 6y = -6 + 6y

2(6y) = -6

6y = -3

Therefore, the system of equations has a unique solution of x = -3 and y = 1/2.

Question:

Examine the consistency of the system of equations3x−y−2z=2,2y−z=−1,3x−5y=3

Answer:

1. 3x - y - 2z = 2
2. 2y - z = -1
3. 3x - 5y = 3

Subtract equation (2) from equation (1): 3x - y - 2z = 2 -2y + z = 1

5x - 5y = 3

Subtract 3x from both sides of equation (3): 5x - 5y = 3 -3x

2x - 5y = 3

Divide both sides of equation (4) by 2: 2x - 5y = 3 2 2

x - 5y = 3/2

Subtract 5y from both sides of equation (5): x - 5y = 3/2 -5y

x = 3/2 + 5y

Substitute x = 3/2 + 5y into equation (1): 3(3/2 + 5y) - y - 2z = 2 9/2 + 15y - y - 2z = 2 9/2 + 14y - 2z = 2

Substitute x = 3/2 + 5y and y = 14y - 2z into equation (2): 2(14y - 2z) - z = -1 28y - 4z - z = -1 28y - 5z = -1

The system of equations is consistent.

Question:

Examine the consistency of the system of equations 2x−y=5, x+y=4

Answer:

1. Rearrange the first equation to solve for y: 2x - 5 = y

2. Substitute the expression for y in the second equation: x + (2x - 5) = 4

3. Simplify the equation: 3x - 5 = 4

4. Add 5 to both sides of the equation: 3x = 9

5. Divide both sides of the equation by 3: x = 3

6. Substitute the value of x in the first equation: 2(3) - y = 5

7. Simplify the equation: 6 - y = 5

8. Add y to both sides of the equation: 6 = 5 + y

9. Subtract 5 from both sides of the equation: y = 1

10. The system of equations is consistent, since both equations have the same solution: x = 3 and y = 1.

Question:

Examine the consistency of the system of equations5x−y+4z=5,2x+3y+5z=2,5x−2y+6z=−1

Answer:

1. Begin by rearranging the equations into the standard form of Ax + By + Cz = D.

5x - y + 4z = 5 2x + 3y + 5z = 2 5x - 2y + 6z = -1

5x - y + 4z = 5 2x + 3y + 5z = 2 -3x - 3y + 2z = -6

2. Calculate the determinant of the coefficient matrix.

|5 -1 4| |2 3 5| |-3 -3 2|

Determinant = (552) + (1*-35) + (43*-3) - (45-3) - (132) - (5*-3*2)

Determinant = -25

3. The determinant is not zero, so the system is consistent and has a unique solution.

Question:

Solve the system of equations, using matrix method x−y+2z=7,3x+4y−5z=−5,2x−y+3z=12

Answer:

Step 1: Write down the system of equations in matrix form.

[1 -1 2 7] [3 4 -5 -5] [2 -1 3 12]

Step 2: Perform row operations to reduce the matrix to reduced row echelon form.

[1 -1 2 7] → [1 -1 2 7] [3 4 -5 -5] → [0 1 -7 -2] [2 -1 3 12] → [0 0 1 5]

Step 3: Read the solution from the reduced row echelon form.

The solution is x = 5, y = -2, and z = 5.

Question:

Solve the system of equations, using matrix method x−y+z=4,2x+y−3z=0,x+y+z=2

Answer:

1. Begin by writing the equations in matrix form:

\begin{bmatrix} 1 & -1 & 1 \ 2 & 1 & -3 \ 1 & 1 & 1 \end{bmatrix} \begin{bmatrix} x \ y \ z \end{bmatrix}

\begin{bmatrix} 4 \ 0 \ 2 \end{bmatrix}

2. Multiply the first row by 2 and subtract it from the second row:

\begin{bmatrix} 1 & -1 & 1 \ 0 & 3 & -5 \ 1 & 1 & 1 \end{bmatrix} \begin{bmatrix} x \ y \ z \end{bmatrix}

\begin{bmatrix} 4 \ -4 \ 2 \end{bmatrix}

3. Multiply the first row by -1 and add it to the third row:

\begin{bmatrix} 1 & -1 & 1 \ 0 & 3 & -5 \ 0 & 2 & 0 \end{bmatrix} \begin{bmatrix} x \ y \ z \end{bmatrix}

\begin{bmatrix} 4 \ -4 \ 2 \end{bmatrix}

4. Divide the second row by 3:

\begin{bmatrix} 1 & -1 & 1 \ 0 & 1 & -\frac{5}{3} \ 0 & 2 & 0 \end{bmatrix} \begin{bmatrix} x \ y \ z \end{bmatrix}

\begin{bmatrix} 4 \ -\frac{4}{3} \ 2 \end{bmatrix}

5. Add the second row to the first row:

\begin{bmatrix} 1 & 0 & \frac{2}{3} \ 0 & 1 & -\frac{5}{3} \ 0 & 2 & 0 \end{bmatrix} \begin{bmatrix} x \ y \ z \end{bmatrix}

\begin{bmatrix} 4 \ -\frac{4}{3} \ 2 \end{bmatrix}

6. Subtract the third row from the second row:

\begin{bmatrix} 1 & 0 & \frac{2}{3} \ 0 & 1 & 0 \ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} x \ y \ z \end{bmatrix}

\begin{bmatrix} 4 \ -\frac{4}{3} \ 2 \end{bmatrix}

7. Solve for x:

x = 4

8. Substitute x = 4 into the second equation:

y = -\frac{4}{3}

9. Substitute x = 4 and y = -\frac{4}{3} into the third equation:

z = \frac{2}{3}

10. The solution is x = 4, y = -\frac{4}{3}, z = \frac{2}{3}.

Question:

Solve system of linear equations, using matrix method 5x+2y=4,7x+3y=5

Answer:

Step 1: Rewrite the equations in matrix form:

\begin{bmatrix} 5 & 2 \ 7 & 3 \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix}

\begin{bmatrix} 4 \ 5 \end{bmatrix}

Step 2: Multiply the inverse of the coefficient matrix by the constant matrix:

\begin{bmatrix} 5 & 2 \ 7 & 3 \end{bmatrix}^{-1} \begin{bmatrix} 5 & 2 \ 7 & 3 \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix}

\begin{bmatrix} 5 & 2 \ 7 & 3 \end{bmatrix}^{-1} \begin{bmatrix} 4 \ 5 \end{bmatrix}

Step 3: Calculate the inverse of the coefficient matrix:

\begin{bmatrix} 5 & 2 \ 7 & 3 \end{bmatrix}^{-1}

\frac{1}{35} \begin{bmatrix} 3 & -2 \ -7 & 5 \end{bmatrix}

Step 4: Multiply the inverse of the coefficient matrix by the constant matrix:

\frac{1}{35} \begin{bmatrix} 3 & -2 \ -7 & 5 \end{bmatrix} \begin{bmatrix} 4 \ 5 \end{bmatrix}

\begin{bmatrix} \frac{13}{35} \ \frac{2}{35} \end{bmatrix}

Step 5: Substitute the values of x and y into the original equations to check the solution:

5(\frac{13}{35})+2(\frac{2}{35})=4 7(\frac{13}{35})+3(\frac{2}{35})=5

The solution is x=13/35 and y=2/35.

Question:

Solve the system of equations, using matrix method2x+y+z=1,x−2y−z=3​/2,3y−5z=9

Answer:

Step 1: Write the system of equations in matrix form.

\begin{bmatrix} 2 & 1 & 1 & 1 \ 1 & -2 & -1 & 3/2 \ 0 & 3 & -5 & 9 \end{bmatrix}

Step 2: Use row operations to reduce the matrix to reduced row-echelon form.

\begin{bmatrix} 2 & 1 & 1 & 1 \ 0 & -3 & -2 & 5/2 \ 0 & 0 & -1 & 3 \end{bmatrix}

Step 3: Read off the solutions from the reduced matrix.

x = 3, y = -2, z = 3

Question:

Solve the system of equations, using Matrix method 2x−y=−2,3x−4y=3

Answer:

Step 1: Write the system of equations in matrix form:

[2 -1] [x] = [-2] [3 -4] [y] = [3]

Step 2: Multiply the first equation by 3 and the second equation by 2:

[6 -3] [x] = [-6] [6 -8] [y] = [6]

Step 3: Subtract the first equation from the second equation:

[0 -5] [x] = [0] [0 -2] [y] = [0]

Step 4: Solve for x:

x = 0

Step 5: Substitute the value of x in the first equation and solve for y:

2(0) - y = -2 y = 2

Therefore, the solution of the system of equations is x = 0 and y = 2.

Question:

If A=$\left[\begin{array}{ccc}2& -3& 5\\ 3& 2& -4\\ 1& 1& -2\end{array}\right]$ find $\begin{array}{}{A}^{-1}\end{array}$. Using $\begin{array}{}{A}^{-1}\end{array}$ solve the system of equations 2x−3y+5z=11 3x+2y−4z=5 x+y−2z=3

Answer:

A = [[2, -3, 5], [3, 2, -4], [1, 1, -2]]

A-1 = [[-2/3, 5/3, -1/3], [3/7, -2/7, 4/7], [2/3, -1/3, 1/3]]

The system of equations can be written as:

[2 -3 5][x] = [11] [3 2 -4][y] = [5] [1 1 -2][z] = [3]

Multiplying A-1 on the left side of the equation yields:

[-2/3 5/3 -1/3][2 -3 5][x] = [-2/3 5/3 -1/3][11] [3/7 -2/7 4/7][3 2 -4][y] = [3/7 -2/7 4/7][5] [2/3 -1/3 1/3][1 1 -2][z] = [2/3 -1/3 1/3][3]

Solving for x, y, and z yields:

x = 4 y = -1 z = 2

Question:

Examine the consistency of the system of equationsx+y+z=1, 2x+3y+2z=2, ax+ay+2az=4

Answer:

Step 1: Rewrite the equations in matrix form:

[x y z] [1] [2 3 2] * [2] = [1] [a a 2a] [4]

Step 2: Determine the determinant of the matrix.

Determinant = (132a) - (2a2) - (1a2) = 6a - 4a - 2a = 0a

Step 3: Since the determinant is 0, the system is not consistent.

Question:

Solve the system of equations, using matrix method 4x−3y=3,3x−5y=7

Answer:

Step 1: Write the equations in matrix form.

[4 -3] [x] = [3] [3 -5] [y] = [7]

Step 2: Find the inverse of the coefficient matrix.

[4 -3] [3 -5]

1/22 5/22 -3/22 4/22

Step 3: Multiply the inverse of the coefficient matrix with the constant matrix.

[1/22 5/22] [3] = [x] [-3/22 4/22] [7] = [y]

Step 4: Solve for x and y.

x = 5/11 y = 2/11