Class 12 NCERT Maths – Chapter 4 (Determinants) – Exercise 4.5 Solutions

In this post, we apply what we’ve learned about determinants and matrices to examine the consistency of a system of linear equations.
Remember the rules from Ex 4.5:

• If |A| \neq 0, the system has a unique solution and is consistent.
• If |A| = 0 and (adj A)B \neq O, the system has no solution and is inconsistent.
• If |A| = 0 and (adj A)B = O, the system may or may not be consistent.

Let’s dive into the first two questions of Exercise 4.5!

Q 1: Examine the consistency of the system of equations:

x + 2y = 2
2x + 3y = 3

Solution:
First, let’s write the given system of equations in the matrix form AX = B.

Let the coefficient matrix be A, the variable matrix be X, and the constant matrix be B:

A = \begin{bmatrix}1&2\\2&3\end{bmatrix}, X = \begin{bmatrix}x\\y\end{bmatrix}, \text{and } B = \begin{bmatrix}2\\3\end{bmatrix}

To check for consistency, we need to find the determinant of matrix A (denoted as |A|).
|A| = \begin{vmatrix}1&2\\2&3\end{vmatrix}

|A| = (1)(3) - (2)(2)

|A| = 3 - 4 = -1

Since |A| = -1, we can see that |A| \neq 0.

This means that matrix A is a non-singular matrix.

Final Answer: Because A is a non-singular matrix (|A| \neq 0), its inverse exists.
Therefore, the given system of equations has a unique solution and is consistent.

Q 2: Examine the consistency of the system of equations:

2x - y = 5
x + y = 4

Solution:
We will start by expressing the system of equations in the matrix form AX = B.

Here, the matrices are defined as follows:
A = \begin{bmatrix}2&-1\\1&1\end{bmatrix}, X = \begin{bmatrix}x\\y\end{bmatrix}, \text{and } B = \begin{bmatrix}5\\4\end{bmatrix}

Next, calculate the determinant of the coefficient matrix A:
|A| = \begin{vmatrix}2&-1\\1&1\end{vmatrix}

|A| = (2)(1) - (-1)(1)

|A| = 2 - (-1) = 2 + 1 = 3

Since |A| = 3, we observe that |A| \neq 0.
Therefore, matrix A is a non-singular matrix.

Final Answer: Because A is a non-singular matrix (|A| \neq 0), its inverse exists.
Therefore, the given system of equations has a unique solution and is consistent.

Teacher’s Pro-Tip: Whenever you are asked to examine consistency for a system of equations, your very first reflex should always be to calculate |A|. That single number dictates your entire path forward!

Q 3: Examine the consistency of the system of equations:

x + 3y = 5
2x + 6y = 8

Solution:
First, let’s write the given system of equations in the matrix form AX = B.

Let the coefficient matrix be A, the variable matrix be X, and the constant matrix be B:
A = \begin{bmatrix}1&3\\2&6\end{bmatrix}, X = \begin{bmatrix}x\\y\end{bmatrix}, \text{and } B = \begin{bmatrix}5\\8\end{bmatrix}

To check for consistency, we need to find the determinant of matrix A (denoted as |A|).

|A| = \begin{vmatrix}1&3\\2&6\end{vmatrix}

|A| = (1)(6) - (3)(2)
|A| = 6 - 6 = 0

Since |A| = 0, we must find (adj A)B to determine if the system is consistent or inconsistent.

First, find adj A:
adj A = \begin{bmatrix}6&-3\\-2&1\end{bmatrix}

Next, multiply adj A by B:
(adj A)B = \begin{bmatrix}6&-3\\-2&1\end{bmatrix}\begin{bmatrix}5\\8\end{bmatrix} = \begin{bmatrix}30 - 24\\-10 + 8\end{bmatrix} = \begin{bmatrix}6\\-2\end{bmatrix}

Since (adj A)B \neq O (the zero matrix), the system has no solution.

Final Answer: Because |A| = 0 and (adj A)B \neq O, the given system of equations has no solution and is inconsistent.

Q 4: Examine the consistency of the system of equations:

x + y + z = 1
2x + 3y + 2z = 2
ax + ay + 2az = 4

Solution:
First, let’s write the given system of equations in the matrix form AX = B.

Let the coefficient matrix be A, the variable matrix be X, and the constant matrix be B:
A = \begin{bmatrix}1&1&1\\2&3&2\\a&a&2a\end{bmatrix}, X = \begin{bmatrix}x\\y\\z\end{bmatrix}, \text{and } B = \begin{bmatrix}1\\2\\4\end{bmatrix}

To check for consistency, we need to find the determinant of matrix A.
|A| = \begin{vmatrix}1&1&1\\2&3&2\\a&a&2a\end{vmatrix}

Expanding along the first row:
|A| = 1(6a - 2a) - 1(4a - 2a) + 1(2a - 3a)
|A| = 4a - 2a - a = a

Assuming a \neq 0, we observe that |A| \neq 0.
Therefore, matrix A is a non-singular matrix.

Final Answer: Because A is a non-singular matrix (for a \neq 0), its inverse exists.
Therefore, the given system of equations has a unique solution and is consistent.

Q 5: Examine the consistency of the system of equations:

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

Solution:
First, let’s write the given system of equations in the matrix form AX = B.

Let the coefficient matrix be A, the variable matrix be X, and the constant matrix be B:
A = \begin{bmatrix}3&-1&-2\\0&2&-1\\3&-5&0\end{bmatrix}, X = \begin{bmatrix}x\\y\\z\end{bmatrix}, \text{and } B = \begin{bmatrix}2\\-1\\3\end{bmatrix}

To check for consistency, calculate the determinant of matrix A:
|A| = \begin{vmatrix}3&-1&-2\\0&2&-1\\3&-5&0\end{vmatrix}

Expanding along the first column:
|A| = 3(0 - 5) - 0 + 3(1 - (-4))
|A| = -15 + 0 + 15 = 0

Since |A| = 0, we need to calculate (adj A)B.
Calculating the cofactors to find adj A:

• A_{11} = -5, A_{12} = -3, A_{13} = -6
• A_{21} = 10, A_{22} = 6, A_{23} = 12
• A_{31} = 5, A_{32} = 3, A_{33} = 6

adj A = \begin{bmatrix}-5&10&5\\-3&6&3\\-6&12&6\end{bmatrix}

Now, multiply (adj A) by B:

(adj A)B = \begin{bmatrix}-5&10&5\\-3&6&3\\-6&12&6\end{bmatrix}\begin{bmatrix}2\\-1\\3\end{bmatrix} = \begin{bmatrix}-10 - 10 + 15\\-6 - 6 + 9\\-12 - 12 + 18\end{bmatrix} = \begin{bmatrix}-5\\-3\\-6\end{bmatrix}

Since (adj A)B \neq O, the system has no solution.

Final Answer: Because |A| = 0 and (adj A)B \neq O, the given system of equations has no solution and is inconsistent.

Q 6: Examine the consistency of the system of equations:

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

Solution:
First, express the system of equations in the matrix form AX = B.
A = \begin{bmatrix}5&-1&4\\2&3&5\\5&-2&6\end{bmatrix}, X = \begin{bmatrix}x\\y\\z\end{bmatrix}, \text{and } B = \begin{bmatrix}5\\2\\-1\end{bmatrix}

Calculate the determinant of matrix A:
|A| = \begin{vmatrix}5&-1&4\\2&3&5\\5&-2&6\end{vmatrix}

|A| = 5(18 - (-10)) - (-1)(12 - 25) + 4(-4 - 15)
|A| = 5(28) + 1(-13) + 4(-19)
|A| = 140 - 13 - 76 = 51

Since |A| = 51, we observe that |A| \neq 0.
Therefore, matrix A is a non-singular matrix.

Final Answer: Because A is a non-singular matrix (|A| \neq 0), its inverse exists.
Therefore, the given system of equations has a unique solution and is consistent.

Q 7: Solve the system of linear equations using matrix method:

5x + 2y = 4
7x + 3y = 5

Solution:
First, let’s write the given system of equations in the matrix form AX = B.

Let the coefficient matrix be A, the variable matrix be X, and the constant matrix be B:
A = \begin{bmatrix}5&2\\7&3\end{bmatrix}, X = \begin{bmatrix}x\\y\end{bmatrix}, \text{and } B = \begin{bmatrix}4\\5\end{bmatrix}

To solve for X, we first check if the matrix A is invertible by calculating |A|.
|A| = (5)(3) - (2)(7) = 15 - 14 = 1

Since |A| = 1 \neq 0, the inverse exists, and the unique solution is X = A^{-1}B.

Find adj A:
adj A = \begin{bmatrix}3&-2\\-7&5\end{bmatrix}

A^{-1} = \frac{1}{|A|}(adj A) = \begin{bmatrix}3&-2\\-7&5\end{bmatrix}

Multiply A^{-1} by B:

X = \begin{bmatrix}3&-2\\-7&5\end{bmatrix}\begin{bmatrix}4\\5\end{bmatrix} = \begin{bmatrix}12 - 10\\-28 + 25\end{bmatrix} = \begin{bmatrix}2\\-3\end{bmatrix}

Final Answer: The unique solution for the given system of equations is x = 2 and y = -3.

Q 8: Solve the system of linear equations using matrix method:

2x - y = -2
3x + 4y = 3

Solution:
First, let’s write the given system of equations in the matrix form AX = B.
A = \begin{bmatrix}2&-1\\3&4\end{bmatrix}, X = \begin{bmatrix}x\\y\end{bmatrix}, \text{and } B = \begin{bmatrix}-2\\3\end{bmatrix}

Calculate the determinant of matrix A:
|A| = (2)(4) - (-1)(3) = 8 + 3 = 11

Since |A| = 11 \neq 0, the inverse exists, and the unique solution is X = A^{-1}B.

Find adj A:
adj A = \begin{bmatrix}4&1\\-3&2\end{bmatrix}

A^{-1} = \frac{1}{11}\begin{bmatrix}4&1\\-3&2\end{bmatrix}

Multiply A^{-1} by B:

X = \frac{1}{11}\begin{bmatrix}4&1\\-3&2\end{bmatrix}\begin{bmatrix}-2\\3\end{bmatrix} = \frac{1}{11}\begin{bmatrix}-8 + 3\\6 + 6\end{bmatrix} = \frac{1}{11}\begin{bmatrix}-5\\12\end{bmatrix} = \begin{bmatrix}-\frac{5}{11}\\\frac{12}{11}\end{bmatrix}

Final Answer: The unique solution for the given system of equations is x = -\frac{5}{11} and y = \frac{12}{11}.

Q 9: Solve the system of linear equations using matrix method:

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

Solution:
First, let’s write the given system of equations in the matrix form AX = B.
A = \begin{bmatrix}4&-3\\3&-5\end{bmatrix}, X = \begin{bmatrix}x\\y\end{bmatrix}, \text{and } B = \begin{bmatrix}3\\7\end{bmatrix}

Calculate the determinant of matrix A:
|A| = (4)(-5) - (-3)(3) = -20 + 9 = -11

Since |A| = -11 \neq 0, the inverse exists, and the unique solution is given by X = A^{-1}B.

Find adj A:
adj A = \begin{bmatrix}-5&3\\-3&4\end{bmatrix}

A^{-1} = -\frac{1}{11}\begin{bmatrix}-5&3\\-3&4\end{bmatrix}

Multiply A^{-1} by B:

X = -\frac{1}{11}\begin{bmatrix}-5&3\\-3&4\end{bmatrix}\begin{bmatrix}3\\7\end{bmatrix} = -\frac{1}{11}\begin{bmatrix}-15 + 21\\-9 + 28\end{bmatrix} = -\frac{1}{11}\begin{bmatrix}6\\19\end{bmatrix} = \begin{bmatrix}-\frac{6}{11}\\-\frac{19}{11}\end{bmatrix}

Final Answer: The unique solution for the given system of equations is x = -\frac{6}{11} and y = -\frac{19}{11}.

Q 10: Solve the system of linear equations using matrix method:

5x + 2y = 3
3x + 2y = 5

Solution:
First, let’s write the given system of equations in the matrix form AX = B.

Let the coefficient matrix be A, the variable matrix be X, and the constant matrix be B:
A = \begin{bmatrix}5&2\\3&2\end{bmatrix}, X = \begin{bmatrix}x\\y\end{bmatrix}, \text{and } B = \begin{bmatrix}3\\5\end{bmatrix}

To solve for X, we first check if the matrix A is invertible by calculating its determinant |A|.
|A| = (5)(2) - (2)(3) = 10 - 6 = 4

Since |A| = 4 \neq 0, the matrix A is non-singular, its inverse exists,
and the unique solution is given by X = A^{-1}B.

Find the adjoint matrix, adj A:
adj A = \begin{bmatrix}2&-2\\-3&5\end{bmatrix}

A^{-1} = \frac{1}{|A|}(adj A) = \frac{1}{4}\begin{bmatrix}2&-2\\-3&5\end{bmatrix}

Now, multiply A^{-1} by B:
X = \frac{1}{4}\begin{bmatrix}2&-2\\-3&5\end{bmatrix}\begin{bmatrix}3\\5\end{bmatrix} = \frac{1}{4}\begin{bmatrix}6 - 10\\-9 + 25\end{bmatrix} = \frac{1}{4}\begin{bmatrix}-4\\16\end{bmatrix} = \begin{bmatrix}-1\\4\end{bmatrix}

Final Answer: The unique solution for the given system of equations is x = -1 and y = 4.

Q 11: Solve the system of linear equations using matrix method:

x + y + z = 1
x - 2y - z = \frac{3}{2}

3y - 5z = 9

Solution:
Express the system of equations in the matrix form AX = B.

A = \begin{bmatrix}1&1&1\\1&-2&-1\\0&3&-5\end{bmatrix}, X = \begin{bmatrix}x\\y\\z\end{bmatrix}, \text{and } B = \begin{bmatrix}1\\\frac{3}{2}\\9\end{bmatrix}

Calculate the determinant of matrix A:
|A| = 1(10 - (-3)) - 1(-5 - 0) + 1(3 - 0)
|A| = 1(13) + 5 + 3 = 21
Since |A| = 21 \neq 0, the inverse exists.

Now, calculate the cofactors to find adj A:

• A_{11} = 13, A_{12} = 5, A_{13} = 3
• A_{21} = -(-5 - 3) = 8, A_{22} = -5, A_{23} = -(3 - 0) = -3
• A_{31} = -1 - (-2) = 1, A_{32} = -(-1 - 1) = 2, A_{33} = -2 - 1 = -3

adj A = \begin{bmatrix}13&8&1\\5&-5&2\\3&-3&-3\end{bmatrix}

A^{-1} = \frac{1}{21}\begin{bmatrix}13&8&1\\5&-5&2\\3&-3&-3\end{bmatrix}

Multiply A^{-1} by B:

X = \frac{1}{21}\begin{bmatrix}13&8&1\\5&-5&2\\3&-3&-3\end{bmatrix}\begin{bmatrix}1\\\frac{3}{2}\\9\end{bmatrix} = \frac{1}{21}\begin{bmatrix}13 + 12 + 9\\5 - \frac{15}{2} + 18\\3 - \frac{9}{2} - 27\end{bmatrix} = \frac{1}{21}\begin{bmatrix}34\\\frac{31}{2}\\-\frac{57}{2}\end{bmatrix} = \begin{bmatrix}\frac{34}{21}\\\frac{31}{42}\\-\frac{19}{14}\end{bmatrix}

Final Answer: The unique solution for the given system of equations is x = \frac{34}{21}, y = \frac{31}{42}, and z = -\frac{19}{14}.

Q 12: Solve the system of linear equations using matrix method:

x - y + z = 4
2x + y - 3z = 0
x + y + z = 2

Solution:
Express the system of equations in the matrix form AX = B.

A = \begin{bmatrix}1&-1&1\\2&1&-3\\1&1&1\end{bmatrix}, X = \begin{bmatrix}x\\y\\z\end{bmatrix}, \text{and } B = \begin{bmatrix}4\\0\\2\end{bmatrix}

Calculate the determinant of matrix A:
|A| = 1(1 - (-3)) - (-1)(2 - (-3)) + 1(2 - 1)
|A| = 1(4) + 1(5) + 1(1) = 10

Since |A| = 10 \neq 0, the inverse exists.
Calculate the cofactors to find adj A:

• A_{11} = 4, A_{12} = -5, A_{13} = 1
• A_{21} = 2, A_{22} = 0, A_{23} = -2
• A_{31} = 2, A_{32} = 5, A_{33} = 3

adj A = \begin{bmatrix}4&2&2\\-5&0&5\\1&-2&3\end{bmatrix}

A^{-1} = \frac{1}{10}\begin{bmatrix}4&2&2\\-5&0&5\\1&-2&3\end{bmatrix}

Multiply A^{-1} by B:
X = \frac{1}{10}\begin{bmatrix}4&2&2\\-5&0&5\\1&-2&3\end{bmatrix}\begin{bmatrix}4\\0\\2\end{bmatrix} = \frac{1}{10}\begin{bmatrix}16 + 0 + 4\\-20 + 0 + 10\\4 + 0 + 6\end{bmatrix} = \frac{1}{10}\begin{bmatrix}20\\-10\\10\end{bmatrix} = \begin{bmatrix}2\\-1\\1\end{bmatrix}

Final Answer: The unique solution for the given system of equations is x = 2, y = -1, and z = 1.

Q 13: Solve the system of linear equations using matrix method:

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

Solution:
Express the system of equations in the matrix form AX = B.
A = \begin{bmatrix}2&3&3\\1&-2&1\\3&-1&-2\end{bmatrix}, X = \begin{bmatrix}x\\y\\z\end{bmatrix}, \text{and } B = \begin{bmatrix}5\\-4\\3\end{bmatrix}

Calculate the determinant of matrix A:
|A| = 2(4 - (-1)) - 3(-2 - 3) + 3(-1 - (-6))
|A| = 2(5) - 3(-5) + 3(5) = 10 + 15 + 15 = 40

Since |A| = 40 \neq 0, the inverse exists.
Calculate the cofactors to find adj A:

• A_{11} = 5, A_{12} = 5, A_{13} = 5
• A_{21} = 3, A_{22} = -13, A_{23} = 11
• A_{31} = 9, A_{32} = 1, A_{33} = -7

adj A = \begin{bmatrix}5&3&9\\5&-13&1\\5&11&-7\end{bmatrix}

A^{-1} = \frac{1}{40}\begin{bmatrix}5&3&9\\5&-13&1\\5&11&-7\end{bmatrix}

Multiply A^{-1} by B:
X = \frac{1}{40}\begin{bmatrix}5&3&9\\5&-13&1\\5&11&-7\end{bmatrix}\begin{bmatrix}5\\-4\\3\end{bmatrix} = \frac{1}{40}\begin{bmatrix}25 - 12 + 27\\25 + 52 + 3\\25 - 44 - 21\end{bmatrix} = \frac{1}{40}\begin{bmatrix}40\\80\\-40\end{bmatrix} = \begin{bmatrix}1\\2\\-1\end{bmatrix}

Final Answer: The unique solution for the given system of equations is x = 1, y = 2, and z = -1.

Q 14: Solve the system of linear equations using matrix method:

x - y + 2z = 7
3x + 4y - 5z = -5
2x - y + 3z = 12

Solution:
Express the system of equations in the matrix form AX = B.
A = \begin{bmatrix}1&-1&2\\3&4&-5\\2&-1&3\end{bmatrix}, X = \begin{bmatrix}x\\y\\z\end{bmatrix}, \text{and } B = \begin{bmatrix}7\\-5\\12\end{bmatrix}

Calculate the determinant of matrix A:
|A| = 1(12 - 5) - (-1)(9 - (-10)) + 2(-3 - 8)
|A| = 1(7) + 1(19) + 2(-11) = 7 + 19 - 22 = 4

Since |A| = 4 \neq 0, the inverse exists.

Calculate the cofactors to find adj A:

• A_{11} = 7, A_{12} = -19, A_{13} = -11
• A_{21} = 1, A_{22} = -1, A_{23} = -1
• A_{31} = -3, A_{32} = 11, A_{33} = 7

adj A = \begin{bmatrix}7&1&-3\\-19&-1&11\\-11&-1&7\end{bmatrix}

A^{-1} = \frac{1}{4}\begin{bmatrix}7&1&-3\\-19&-1&11\\-11&-1&7\end{bmatrix}

Multiply A^{-1} by B:

X = \frac{1}{4}\begin{bmatrix}7&1&-3\\-19&-1&11\\-11&-1&7\end{bmatrix}\begin{bmatrix}7\\-5\\12\end{bmatrix} = \frac{1}{4}\begin{bmatrix}49 - 5 - 36\\-133 + 5 + 132\\-77 + 5 + 84\end{bmatrix} = \frac{1}{4}\begin{bmatrix}8\\4\\12\end{bmatrix} = \begin{bmatrix}2\\1\\3\end{bmatrix}

Final Answer: The unique solution for the given system of equations is x = 2, y = 1, and z = 3.

Q 15: If A = \begin{bmatrix}2&-3&5\\3&2&-4\\1&1&-2\end{bmatrix}, find A^{-1}. Using A^{-1} solve the system of equations:

2x - 3y + 5z = 11
3x + 2y - 4z = -5
x + y - 2z = -3

Solution:
First, we find the inverse of matrix A.

Calculate its determinant:
|A| = \begin{vmatrix}2&-3&5\\3&2&-4\\1&1&-2\end{vmatrix} = 2(-4 - (-4)) - (-3)(-6 - (-4)) + 5(3 - 2)

|A| = 2(0) + 3(-2) + 5(1) = 0 - 6 + 5 = -1

Since |A| = -1 \neq 0, the inverse A^{-1} exists.

Find the cofactors of A:

• A_{11} = 0, A_{12} = 2, A_{13} = 1
• A_{21} = -1, A_{22} = -9, A_{23} = -5
• A_{31} = 2, A_{32} = 23, A_{33} = 13

adj A = \begin{bmatrix}0&-1&2\\2&-9&23\\1&-5&13\end{bmatrix}

A^{-1} = \frac{1}{-1}\begin{bmatrix}0&-1&2\\2&-9&23\\1&-5&13\end{bmatrix} = \begin{bmatrix}0&1&-2\\-2&9&-23\\-1&5&-13\end{bmatrix}

Now, notice that the given system of equations can be written as AX = B, where A is the same matrix we just found the inverse for, and B = \begin{bmatrix}11\\-5\\-3\end{bmatrix}.

To solve it, we simply multiply our A^{-1} by B:
X = \begin{bmatrix}0&1&-2\\-2&9&-23\\-1&5&-13\end{bmatrix}\begin{bmatrix}11\\-5\\-3\end{bmatrix} = \begin{bmatrix}0 - 5 + 6\\-22 - 45 + 69\\-11 - 25 + 39\end{bmatrix} = \begin{bmatrix}1\\2\\3\end{bmatrix}

Final Answer: A^{-1} = \begin{bmatrix}0&1&-2\\-2&9&-23\\-1&5&-13\end{bmatrix}

The unique solution for the system of equations is x = 1, y = 2, and z = 3.

Q 16: The cost of 4 kg onion, 3 kg wheat and 2 kg rice is ₹60. The cost of 2 kg onion, 4 kg wheat and 6 kg rice is ₹90. The cost of 6 kg onion 2 kg wheat and 3 kg rice is ₹70. Find cost of each item per kg by matrix method.

Solution:
Let the cost per kg of onion, wheat, and rice be x, y, and z respectively.

We can form the following linear equations from the given word problem:
4x + 3y + 2z = 60
2x + 4y + 6z = 90
6x + 2y + 3z = 70

Express this system in the matrix form AX = B:

A = \begin{bmatrix}4&3&2\\2&4&6\\6&2&3\end{bmatrix}, X = \begin{bmatrix}x\\y\\z\end{bmatrix}, \text{and } B = \begin{bmatrix}60\\90\\70\end{bmatrix}

Calculate the determinant of matrix A:
|A| = 4(12 - 12) - 3(6 - 36) + 2(4 - 24)
|A| = 4(0) - 3(-30) + 2(-20) = 0 + 90 - 40 = 50

Since |A| = 50 \neq 0, the inverse exists.
Calculate the cofactors to find adj A:

• A_{11} = 0, A_{12} = 30, A_{13} = -20
• A_{21} = -5, A_{22} = 0, A_{23} = 10
• A_{31} = 10, A_{32} = -20, A_{33} = 10

adj A = \begin{bmatrix}0&-5&10\\30&0&-20\\-20&10&10\end{bmatrix}

A^{-1} = \frac{1}{50}\begin{bmatrix}0&-5&10\\30&0&-20\\-20&10&10\end{bmatrix}

Multiply A^{-1} by B:

X = \frac{1}{50}\begin{bmatrix}0&-5&10\\30&0&-20\\-20&10&10\end{bmatrix}\begin{bmatrix}60\\90\\70\end{bmatrix} = \frac{1}{50}\begin{bmatrix}0 - 450 + 700\\1800 + 0 - 1400\\-1200 + 900 + 700\end{bmatrix} = \frac{1}{50}\begin{bmatrix}250\\400\\400\end{bmatrix} = \begin{bmatrix}5\\8\\8\end{bmatrix}

Final Answer: The cost per kg is ₹5 for onions (x), ₹8 for wheat (y), and ₹8 for rice (z).