NCERT Solutions Exercise 4.5: Determinants

Q1: Examine the consistency of the system of equations.
 x + 2y = 2
 2x + 3y = 3
Ans: The given system of equations is:
x + 2y = 2
2x + 3y = 3
The given system of equations can be written in the form of AX = B, where

Compute the determinant of the coefficient matrix A:
det(A) = |1 2
  2 3| = 1×3 - 2×2 = 3 - 4 = -1 ≠ 0.
Since det(A) ≠ 0, A is non-singular and A^-1 exists.
Therefore the system has a unique solution and is consistent.
Q2: Examine the consistency of the system of equations.
2x - y = 5 x
x + y = 4
Ans: The given system of equations is:
2x - y = 5 x
+ y = 4
The given system of equations can be written in the form of AX = B, where

From the coefficient matrix shown above, the determinant (as computed in the image) is non-zero, so A is non-singular.
Therefore A^-1 exists and the system has a unique solution.
Hence the system is consistent.
Q3: Examine the consistency of the system of equations.
x + 3y = 5
2x + 6y = 8
Ans: The given system of equations is:
x + 3y = 5
2x + 6y = 8
The given system of equations can be written in the form of AX = B, where

Compute the determinant of the coefficient matrix A:
det(A) = |1 3
  2 6| = 1×6 - 3×2 = 6 - 6 = 0.
So A is a singular matrix. Check consistency by comparing proportionality of equations:
The left-hand sides satisfy 2×(x + 3y) = 2x + 6y, but the constants give 2×5 = 10, whereas the second equation has 8 on the right-hand side.
Since the left-hand sides are proportional but the right-hand sides are not, the system is inconsistent (no solution).
Q4: Examine the consistency of the system of equations.
x + y + z = 1 
2x + 3y + 2z = 2 
ax + ay + 2az = 4
Ans: The given system of equations is:
x + y + z = 1
2x + 3y + 2z = 2
ax + ay + 2az = 4
This system can be written in the form of AX = B, where

Compute the determinant of A. Factor a from the third row to get:
det(A) = a × det |1 1 1
  2 3 2
  1 1 2|.
Evaluate the 3×3 determinant: det |1 1 1; 2 3 2; 1 1 2| = 1.
Hence det(A) = a × 1 = a.
Therefore:
- If a ≠ 0, det(A) ≠ 0, so A is non-singular, A^-1 exists and the system has a unique solution (so it is consistent).
- If a = 0, the third equation becomes 0 = 4 which is impossible; hence the system is inconsistent for a = 0.
Q5: Examine the consistency of the system of equations.
3x - y - 2z = 2
2y - z = -1
3x - 5y = 3
Ans: The given system of equations is:
3x - y - 2z = 2
2y - z = -1
3x - 5y = 3
This system can be written in the form of AX = B, where

Compute det(A):
det(A) = |3 -1 -2
  0 2 -1
  3 -5 0| = 0 (calculation: 3×(2×0 - (-1)(-5)) - (-1)×(0×0 - (-1)×3) + (-2)×(0×(-5) - 2×3) = -15 + 3 + 12 = 0).
Thus A is singular. Check consistency by elimination:
From the second equation, z = 2y + 1.
Substitute into the first: 3x - y - 2(2y + 1) = 2 ⇒ 3x - y - 4y - 2 = 2 ⇒ 3x - 5y = 4.
The third equation gives 3x - 5y = 3.
These two results for 3x - 5y are contradictory (4 ≠ 3), so the system is inconsistent (no solution).
Q6: Examine the consistency of the system of equations.
5x - y + 4z = 5
2x + 3y + 5z = 2
5x - 2y + 6z = -1
Ans: The given system of equations is:
5x - y + 4z = 5
2x + 3y + 5z = 2
5x - 2y + 6z = -1
This system can be written in the form of AX = B, where

Compute the determinant of A:
det(A) = 5×|3 5
  -2 6| - (-1)×|2 5
  5 6| + 4×|2 3
  5 -2| = 140 - 13 - 76 = 51 ≠ 0.
Since det(A) ≠ 0, A is non-singular and A^-1 exists.
Therefore the system has a unique solution and is consistent.
Q7: Solve system of linear equations, using matrix method.

Ans: The given system of equations can be written in the form of AX = B, where

As shown in the image, A is non-singular and its inverse exists.
Compute the solution by X = A^-1B (details of the inverse and multiplication are shown in the images):

The unique solution is obtained from the multiplication shown above.
Q8: Solve system of linear equations, using matrix method.

Ans: The given system of equations can be written in the form of AX = B, where

As shown in the image, A is non-singular and its inverse exists.
Compute the solution by X = A^-1B (the steps for finding A^-1 and the product are provided in the image):

The unique solution follows from the calculation displayed above.
Q9: Solve system of linear equations, using matrix method.

Ans: The given system of equations can be written in the form of AX = B, where

As shown in the working images, A is non-singular and A^-1 exists.
Perform X = A^-1B as demonstrated in the images to obtain the unique solution:




Q10: Solve system of linear equations, using matrix method.
5x + 2y = 3
3x + 2y = 5
Ans: The given system of equations can be written in the form of AX = B, where

Coefficient matrix A = |5 2
  3 2|. Compute det(A): det(A) = 5×2 - 2×3 = 10 - 6 = 4 ≠ 0.
Hence A is non-singular and A^-1 exists.
Compute A^-1 = (1/det(A)) × |2 -2
  -3 5| = (1/4) × |2 -2
  -3 5|.
Now X = A^-1B with B = |3
  5| gives:
x = (1/4)[2×3 + (-2)×5] = (1/4)(6 - 10) = (1/4)(-4) = -1.
y = (1/4)[(-3)×3 + 5×5] = (1/4)(-9 + 25) = (1/4)(16) = 4.
Therefore the solution is x = -1, y = 4.
Q11: Solve system of linear equations, using matrix method.

Ans: The given system of equations can be written in the form of AX = B, where

As shown in the image, A is non-singular and its inverse exists.
The solution is obtained by X = A^-1B; the detailed inverse and multiplication are displayed in the image:

Q12: Solve system of linear equations, using matrix method.
x - y + z = 4 
2x + y - 3z = 0 
x + y + z = 2
Ans: The given system of equations can be written in the form of AX = B, where

As the coefficient matrix is non-singular (see image), its inverse exists.
Solving (either by A^-1B or by substitution):
From x - y + z = 4 ⇒ x = 4 + y - z.
Substitute into x + y + z = 2 ⇒ (4 + y - z) + y + z = 2 ⇒ 4 + 2y = 2 ⇒ y = -1.
Then x = 4 + (-1) - z = 3 - z.
Substitute into 2x + y - 3z = 0 ⇒ 2(3 - z) + (-1) - 3z = 0 ⇒ 6 - 2z - 1 - 3z = 0 ⇒ 5 - 5z = 0 ⇒ z = 1.
Thus x = 3 - 1 = 2.
Therefore the solution is x = 2, y = -1, z = 1.

Q13: Solve system of linear equations, using matrix method.
2x + 3y + 3z = 5
x - 2y + z = -4
3x - y - 2z = 3
Ans: The given system of equations can be written in the form of AX = B, where

As shown, A is non-singular and A^-1 exists. Solve by substitution or using A^-1B.
Using substitution from x - 2y + z = -4 ⇒ x = -4 + 2y - z.
Substitute into the first and third equations to eliminate x:
First: 2(-4 + 2y - z) + 3y + 3z = 5 ⇒ -8 + 4y - 2z + 3y + 3z = 5 ⇒ 7y + z = 13. (1)
Third: 3(-4 + 2y - z) - y - 2z = 3 ⇒ -12 + 6y - 3z - y - 2z = 3 ⇒ 5y - 5z = 15 ⇒ y - z = 3. (2)
From (2), z = y - 3. Substitute into (1): 7y + (y - 3) = 13 ⇒ 8y - 3 = 13 ⇒ 8y = 16 ⇒ y = 2.
Then z = 2 - 3 = -1 and x = -4 + 2(2) - (-1) = -4 + 4 + 1 = 1.
Therefore the solution is x = 1, y = 2, z = -1.

Q14: Solve system of linear equations, using matrix method.
x - y + 2z = 7
3x + 4y - 5z = -5
2x - y + 3z = 12
Ans: The given system of equations can be written in the form of AX = B, where

As shown, A is non-singular and A^-1 exists. Solve by substitution or A^-1B.
From x - y + 2z = 7 ⇒ x = 7 + y - 2z.
Substitute into the other two equations:
3(7 + y - 2z) + 4y - 5z = -5 ⇒ 21 + 3y - 6z + 4y - 5z = -5 ⇒ 7y - 11z = -26. (A)
2(7 + y - 2z) - y + 3z = 12 ⇒ 14 + 2y - 4z - y + 3z = 12 ⇒ y - z = -2 ⇒ y = z - 2. (B)
Substitute (B) into (A): 7(z - 2) - 11z = -26 ⇒ 7z - 14 - 11z = -26 ⇒ -4z = -12 ⇒ z = 3.
Then y = 3 - 2 = 1 and x = 7 + 1 - 2×3 = 8 - 6 = 2.
Therefore the solution is x = 2, y = 1, z = 3.

Q15: If   
find A^-1. Using A^-1 solve the system of equations
Ans: 

Now, the given system of equations can be written in the form of AX = B, where


Compute A^-1 as demonstrated in the image and then obtain the solution by X = A^-1B; the steps and the resulting values are shown in the provided workings.
Q16: 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.
Ans: Let the cost per kg of onion, wheat and rice be Rs x, Rs y and Rs z respectively.
Then the system is:
4x + 3y + 2z = 60
2x + 4y + 6z = 90
6x + 2y + 3z = 70
Write in matrix form AX = B where A = |4 3 2
  2 4 6
  6 2 3|, X = |x
  y
  z|, B = |60
  90
  70|. (Seeand.)
Solving (by elimination or by computing A^-1 and multiplying) we obtain:
From elimination: Multiply second equation by 2 and subtract first×1 gives y + 2z = 24. (i)
Eliminate x using suitable multiples of equations 1 and 3 gives y = 8. Substituting in (i) yields 8 + 2z = 24 ⇒ z = 8.
Substitute y and z in the first equation: 4x + 3(8) + 2(8) = 60 ⇒ 4x + 24 + 16 = 60 ⇒ 4x = 20 ⇒ x = 5.
Hence the cost per kg is: onion = Rs 5, wheat = Rs 8, rice = Rs 8.