Determinants Exercise 06
Question:
Examine the consistency of the system of equations x+2y=2, 2x+3y=3
Answer:

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

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

Simplify the equation: x + y = 1

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:
 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.
 Next, we can substitute the expression for x from the first equation into the second equation.
6y = 8  2(5  3y)

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

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:
 3x  y  2z = 2
 2y  z = 1
 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:

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

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

Simplify the equation: 3x  5 = 4

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

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

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

Simplify the equation: 6  y = 5

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

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

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:
 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
 Calculate the determinant of the coefficient matrix.
5 1 4 2 3 5 3 3 2
Determinant = (552) + (1*35) + (43*3)  (453)  (132)  (5*3*2)
Determinant = 25
 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:
 Begin by writing the equations in matrix form:
$\begin{bmatrix} 1 & 1 & 1 \\ 2 & 1 & 3 \\ 1 & 1 & 1 \end{bmatrix}$ $\begin{bmatrix} x \\ y \\ z \end{bmatrix}$ $\begin{bmatrix} 4 \\ 0 \\ 2 \end{bmatrix}$
 Multiply the first row by 2 and subtract it from the second row:
$\begin{bmatrix} 1 & 1 & 1 \\ 0 & 3 & 5 \\ 1 & 1 & 1 \end{bmatrix}$ $\begin{bmatrix} x \\ y \\ z \end{bmatrix}$ $\begin{bmatrix} 4 \\ 4 \\ 2 \end{bmatrix}$
 Multiply the first row by 1 and add it to the third row:
$\begin{bmatrix} 1 & 1 & 1 \\ 0 & 3 & 5 \\ 0 & 2 & 0 \end{bmatrix}$ $\begin{bmatrix} x \\ y \\ z \end{bmatrix}$ $\begin{bmatrix} 4 \\ 4 \\ 2 \end{bmatrix}$
 Divide the second row by 3:
$\begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & \frac{5}{3} \\ 0 & 2 & 0 \end{bmatrix}$ $\begin{bmatrix} x \\ y \\ z \end{bmatrix}$ $\begin{bmatrix} 4 \\ \frac{4}{3} \\ 2 \end{bmatrix}$
 Add the second row to the first row:
$\begin{bmatrix} 1 & 0 & \frac{2}{3} \\ 0 & 1 & \frac{5}{3} \\ 0 & 2 & 0 \end{bmatrix}$ $\begin{bmatrix} x \\ y \\ z \end{bmatrix}$ $\begin{bmatrix} 4 \\ \frac{4}{3} \\ 2 \end{bmatrix}$
 Subtract the third row from the second row:
$\begin{bmatrix} 1 & 0 & \frac{2}{3} \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{bmatrix}$ $\begin{bmatrix} x \\ y \\ z \end{bmatrix}$ $\begin{bmatrix} 4 \\ \frac{4}{3} \\ 2 \end{bmatrix}$
 Solve for x:
x = 4
 Substitute x = 4 into the second equation:
y = $\frac{4}{3}$
 Substitute x = 4 and y = $\frac{4}{3}$ into the third equation:
z = $\frac{2}{3}$
 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 rowechelon 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]]
A1 = [[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 A1 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