GSEB Class 12 Maths Solutions Chapter 4 નિશ્ચાયક Exercise 4.6

Question 1. Solve the following system of linear equations: \(x + 2y = 2\), \(2x + 3y = 3\).
Answer: The given system of linear equations is:
\(x + 2y = 2\)
\(2x + 3y = 3\)
To solve this, we can form a matrix \(A\) from the coefficients of the variables, which is \(A = \begin{pmatrix} 1 & 2 \\ 2 & 3 \end{pmatrix}\).
Next, we find the determinant of matrix \(A\), denoted as \(|A|\).
\(|A| = \left| \begin{array}{cc} 1 & 2 \\ 2 & 3 \end{array} \right| = (1 \times 3) - (2 \times 2) = 3 - 4 = -1\).
Since \(|A| = -1\), which is not equal to zero, the system of equations has a unique solution. This also means that the system of equations is consistent.
In simple words: We have two equations with two unknown numbers. By making a special grid of the numbers in front of the letters and doing a quick calculation called a determinant, we found a result that wasn't zero. This tells us there's only one perfect pair of numbers that makes both equations true, and so the equations work together nicely.

Exam Tip: To check for consistency and a unique solution in a system of linear equations, always calculate the determinant of the coefficient matrix. If it's non-zero, the system is consistent and has a unique solution.

Question 2. Solve the following system of linear equations: \(2x - y = 5\), \(x + y = 4\).
Answer: The given system of linear equations is:
\(2x - y = 5\)
\(x + y = 4\)
We can represent the coefficients of these equations in a matrix \(A\), where \(A = \left[ \begin{array}{cc} 2 & -1 \\ 1 & 1 \end{array} \right]\).
Now, we calculate the determinant of matrix \(A\):
\(|A| = \left| \begin{array}{cc} 2 & -1 \\ 1 & 1 \end{array} \right| = (2 \times 1) - (-1 \times 1) = 2 - (-1) = 2 + 1 = 3\).
Since the determinant \(|A|\) is \(3\), which is not equal to zero, the system of equations has a unique solution. Therefore, the system of equations is consistent.
In simple words: Given two equations, we first organize the numbers in front of the variables into a matrix. Then, we calculate the determinant of this matrix. If the determinant is not zero, it means there's a single set of values for \(x\) and \(y\) that satisfies both equations, showing the system is consistent.

Exam Tip: Remember that for a system of two linear equations in two variables, a non-zero determinant of the coefficient matrix directly implies a unique solution and consistency.

Question 3. Solve the following system of linear equations: \(x + 3y = 5\), \(2x + 6y = 8\).
Answer: The given system of linear equations is:
\(x + 3y = 5\)
\(2x + 6y = 8\)
Let's form the coefficient matrix \(A = \left[ \begin{array}{cc} 1 & 3 \\ 2 & 6 \end{array} \right]\).
Now, we calculate the determinant of \(A\):
\(|A| = \left| \begin{array}{cc} 1 & 3 \\ 2 & 6 \end{array} \right| = (1 \times 6) - (3 \times 2) = 6 - 6 = 0\).
Since \(|A| = 0\), the system either has no solution or infinitely many solutions. To determine which, we need to calculate \((\text{adj } A)B\).
First, find the adjoint of \(A\): \( \text{adj } A = \left[ \begin{array}{cc} 6 & -2 \\ -3 & 1 \end{array} \right]^{\text{T}} = \left[ \begin{array}{cc} 6 & -3 \\ -2 & 1 \end{array} \right]\).
Let the constant matrix be \(B = \left[ \begin{array}{c} 5 \\ 8 \end{array} \right]\).
Now, calculate \((\text{adj } A)B\):
\((\text{adj } A)B = \left[ \begin{array}{cc} 6 & -3 \\ -2 & 1 \end{array} \right] \left[ \begin{array}{c} 5 \\ 8 \end{array} \right] = \left[ \begin{array}{c} (6 \times 5) + (-3 \times 8) \\ (-2 \times 5) + (1 \times 8) \end{array} \right] = \left[ \begin{array}{c} 30 - 24 \\ -10 + 8 \end{array} \right] = \left[ \begin{array}{c} 6 \\ -2 \end{array} \right]\).
Since \((\text{adj } A)B \neq \left[ \begin{array}{c} 0 \\ 0 \end{array} \right]\) (it's \(\left[ \begin{array}{c} 6 \\ -2 \end{array} \right]\)), the given system of equations is inconsistent, meaning it has no solution.
In simple words: We looked at the numbers in the equations and found that the determinant was zero, which means there isn't a single clear answer. Then, we did another matrix calculation and found that it didn't equal zero. This tells us the equations don't work together at all; there's no solution that can make both of them true.

Exam Tip: When the determinant of the coefficient matrix is zero, you must further check \((\text{adj } A)B\). If \((\text{adj } A)B\) is a non-zero matrix, the system is inconsistent (no solution). If it's a zero matrix, the system is consistent but has infinitely many solutions.

Question 4. Solve the following system of linear equations: \(x + y + z = 1\), \(2x + 3y + 2z = 2\), \(ax + ay + 2az = 4\).
Answer: The given system of linear equations is:
\(x + y + z = 1\)
\(2x + 3y + 2z = 2\)
\(ax + ay + 2az = 4 \Rightarrow x + y + 2z = \frac{4}{a}\)
Let's define the matrices \(A\), \(X\), and \(B\):
\(A = \left[ \begin{array}{ccc} 1 & 1 & 1 \\ 2 & 3 & 2 \\ 1 & 1 & 2 \end{array} \right]\), \(X = \left[ \begin{array}{c} x \\ y \\ z \end{array} \right]\) and \(B = \left[ \begin{array}{c} 1 \\ 2 \\ 4/a \end{array} \right]\).
Now, let's find the determinant of \(A\):
\(|A| = 1( (3 \times 2) - (2 \times 1) ) - 1( (2 \times 2) - (2 \times 1) ) + 1( (2 \times 1) - (3 \times 1) )\)
\(|A| = 1(6 - 2) - 1(4 - 2) + 1(2 - 3)\)
\(|A| = 4 - 2 - 1 = 1\).
Since \(|A| = 1\), which is not equal to zero, the system of equations is consistent and has a unique solution.
In simple words: We have three equations with three unknown numbers. We formed a matrix from the numbers next to the letters and calculated its determinant. Because this determinant was not zero, we know that there is one specific solution that makes all three equations true, meaning the system works perfectly.

Exam Tip: For systems with three variables, expand the determinant along any row or column. Ensure careful calculation of cofactors. A non-zero determinant indicates a consistent system with a unique solution, regardless of the value of 'a' as long as it's not zero (to avoid division by zero in 4/a).

Question 5. Solve the following system of linear equations: \(3x - y - 2z = 2\), \(2y - z = -1\), \(3x - 5y = 3\).
Answer: The given system of equations is:
\(3x - y - 2z = 2\)
\(0x + 2y - z = -1\)
\(3x - 5y + 0z = 3\)
Let's define the matrices \(A\), \(X\), and \(B\):
\(A = \left[ \begin{array}{ccc} 3 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0 \end{array} \right]\), \(X = \left[ \begin{array}{c} x \\ y \\ z \end{array} \right]\) and \(B = \left[ \begin{array}{c} 2 \\ -1 \\ 3 \end{array} \right]\).
Now, let's find the determinant of \(A\):
\(|A| = 3( (2 \times 0) - (-1 \times -5) ) - (-1)( (0 \times 0) - (-1 \times 3) ) + (-2)( (0 \times -5) - (2 \times 3) )\)
\(|A| = 3(0 - 5) + 1(0 - (-3)) - 2(0 - 6)\)
\(|A| = 3(-5) + 1(3) - 2(-6)\)
\(|A| = -15 + 3 + 12 = 0\).
Since \(|A| = 0\), we need to check \((\text{adj } A)B\) to see if the system is consistent or inconsistent.
First, let's find the cofactors:
\(C_{11} = +(0 - 5) = -5\)
\(C_{12} = -(0 - (-3)) = -3\)
\(C_{13} = +(0 - 6) = -6\)
\(C_{21} = -((-1 \times 0) - (-2 \times -5)) = -(0 - 10) = 10\)
\(C_{22} = +( (3 \times 0) - (-2 \times 3) ) = +(0 - (-6)) = 6\)
\(C_{23} = -( (3 \times -5) - (-1 \times 3) ) = -(-15 - (-3)) = -(-12) = 12\)
\(C_{31} = +( (-1 \times -1) - (-2 \times 2) ) = +(1 - (-4)) = 5\)
\(C_{32} = -( (3 \times -1) - (-2 \times 0) ) = -(-3 - 0) = 3\)
\(C_{33} = +( (3 \times 2) - (-1 \times 0) ) = +(6 - 0) = 6\)
The cofactor matrix is \(\left[ \begin{array}{ccc} -5 & -3 & -6 \\ 10 & 6 & 12 \\ 5 & 3 & 6 \end{array} \right]\).
The adjoint of \(A\) is the transpose of the cofactor matrix:
\( \text{adj } A = \left[ \begin{array}{ccc} -5 & 10 & 5 \\ -3 & 6 & 3 \\ -6 & 12 & 6 \end{array} \right]\).
Now, calculate \((\text{adj } A)B\):
\((\text{adj } A)B = \left[ \begin{array}{ccc} -5 & 10 & 5 \\ -3 & 6 & 3 \\ -6 & 12 & 6 \end{array} \right] \left[ \begin{array}{c} 2 \\ -1 \\ 3 \end{array} \right]\)
\( = \left[ \begin{array}{c} (-5 \times 2) + (10 \times -1) + (5 \times 3) \\ (-3 \times 2) + (6 \times -1) + (3 \times 3) \\ (-6 \times 2) + (12 \times -1) + (6 \times 3) \end{array} \right]\)
\( = \left[ \begin{array}{c} -10 - 10 + 15 \\ -6 - 6 + 9 \\ -12 - 12 + 18 \end{array} \right] = \left[ \begin{array}{c} -5 \\ -3 \\ -6 \end{array} \right]\).
Since \((\text{adj } A)B \neq \left[ \begin{array}{c} 0 \\ 0 \\ 0 \end{array} \right]\), the given system of equations is inconsistent.
In simple words: First, we arranged the numbers from the equations into a matrix and found that its determinant was zero. This means there's no unique answer. To check further, we found a special related matrix called the adjoint and multiplied it by the constant terms. Since the result was not a matrix of all zeros, it means these equations do not have any solution at all, making the system inconsistent.

Exam Tip: When dealing with a 3x3 matrix and \(|A| = 0\), a common error is to stop there. Always proceed to calculate \((\text{adj } A)B\). This step is crucial to distinguish between an inconsistent system (no solution) and a consistent system with infinitely many solutions.

Question 6. Solve the following system of linear equations: \(5x - y + 4z = 5\), \(2x + 3y + 5z = 2\), \(5x - 2y + 6z = -1\).
Answer: The given system of equations is:
\(5x - y + 4z = 5\)
\(2x + 3y + 5z = 2\)
\(5x - 2y + 6z = -1\)
Let's define the matrices \(A\), \(X\), and \(B\):
\(A = \left[ \begin{array}{ccc} 5 & -1 & 4 \\ 2 & 3 & 5 \\ 5 & -2 & 6 \end{array} \right]\), \(X = \left[ \begin{array}{c} x \\ y \\ z \end{array} \right]\) and \(B = \left[ \begin{array}{c} 5 \\ 2 \\ -1 \end{array} \right]\).
Now, let's find the determinant of \(A\):
\(|A| = 5( (3 \times 6) - (5 \times -2) ) - (-1)( (2 \times 6) - (5 \times 5) ) + 4( (2 \times -2) - (3 \times 5) )\)
\(|A| = 5(18 - (-10)) + 1(12 - 25) + 4(-4 - 15)\)
\(|A| = 5(18 + 10) + 1(-13) + 4(-19)\)
\(|A| = 5(28) - 13 - 76\)
\(|A| = 140 - 13 - 76 = 140 - 89 = 51\).
Since \(|A| = 51\), which is not equal to zero, the given system of equations is consistent and has a unique solution.
In simple words: For these three equations, we set up a matrix using the numbers in front of \(x\), \(y\), and \(z\). Then we found the determinant of this matrix. Since the determinant was \(51\) (not zero), it means there is only one specific set of values for \(x\), \(y\), and \(z\) that will solve all three equations at once, making the system consistent.

Exam Tip: A positive or negative non-zero determinant for a square matrix signifies that its inverse exists, which in turn guarantees a unique solution for the corresponding system of linear equations.

Question 7. Solve the following system of linear equations: \(5x + 2y = 4\), \(7x + 3y = 5\).
Answer: The given system of equations is:
\(5x + 2y = 4\)
\(7x + 3y = 5\)
The system can be written in matrix form \(AX = B\), where:
\(A = \left[ \begin{array}{cc} 5 & 2 \\ 7 & 3 \end{array} \right]\), \(X = \left[ \begin{array}{c} x \\ y \end{array} \right]\), and \(B = \left[ \begin{array}{c} 4 \\ 5 \end{array} \right]\).
First, calculate the determinant of \(A\):
\(|A| = (5 \times 3) - (2 \times 7) = 15 - 14 = 1\).
Since \(|A| = 1 \neq 0\), an inverse \(A^{-1}\) exists, and the system has a unique solution.
Next, find \(A^{-1}\):
\( \text{adj } A = \left[ \begin{array}{cc} 3 & -2 \\ -7 & 5 \end{array} \right]\).
\(A^{-1} = \frac{1}{|A|} (\text{adj } A) = \frac{1}{1} \left[ \begin{array}{cc} 3 & -2 \\ -7 & 5 \end{array} \right] = \left[ \begin{array}{cc} 3 & -2 \\ -7 & 5 \end{array} \right]\).
Now, solve for \(X\) using \(X = A^{-1}B\):
\(X = \left[ \begin{array}{cc} 3 & -2 \\ -7 & 5 \end{array} \right] \left[ \begin{array}{c} 4 \\ 5 \end{array} \right] = \left[ \begin{array}{c} (3 \times 4) + (-2 \times 5) \\ (-7 \times 4) + (5 \times 5) \end{array} \right]\)
\( = \left[ \begin{array}{c} 12 - 10 \\ -28 + 25 \end{array} \right] = \left[ \begin{array}{c} 2 \\ -3 \end{array} \right]\).
Thus, \(x = 2\) and \(y = -3\).
In simple words: We organized the numbers from the equations into matrices. Then, we calculated the determinant of the main coefficient matrix, which showed us a unique solution exists. We then found the inverse of this matrix and multiplied it by the constant terms to get the exact values for \(x\) and \(y\).

Exam Tip: When using the matrix inversion method \(X = A^{-1}B\), clearly show the steps for finding \(|A|\), \( \text{adj } A\), and \(A^{-1}\) before performing the final multiplication to solve for \(X\).

Question 8. Solve the following system of linear equations: \(2x - y = -2\), \(3x + 4y = 3\).
Answer: The given system of equations is:
\(2x - y = -2\)
\(3x + 4y = 3\)
This system can be written in the matrix form \(AX = B\), or \(X = A^{-1}B\), where:
\(A = \left[ \begin{array}{cc} 2 & -1 \\ 3 & 4 \end{array} \right]\), \(X = \left[ \begin{array}{c} x \\ y \end{array} \right]\), and \(B = \left[ \begin{array}{c} -2 \\ 3 \end{array} \right]\).
First, we calculate the determinant of \(A\):
\(|A| = (2 \times 4) - (-1 \times 3) = 8 - (-3) = 8 + 3 = 11\).
Since \(|A| = 11 \neq 0\), the inverse \(A^{-1}\) exists, which means the system has a unique solution.
Next, we find the adjoint of \(A\):
\( \text{adj } A = \left[ \begin{array}{cc} 4 & -(-1) \\ -3 & 2 \end{array} \right] = \left[ \begin{array}{cc} 4 & 1 \\ -3 & 2 \end{array} \right]\).
Now, we calculate the inverse \(A^{-1}\):
\(A^{-1} = \frac{1}{|A|} (\text{adj } A) = \frac{1}{11} \left[ \begin{array}{cc} 4 & 1 \\ -3 & 2 \end{array} \right]\).
Finally, we solve for \(X\) using \(X = A^{-1}B\):
\(X = \frac{1}{11} \left[ \begin{array}{cc} 4 & 1 \\ -3 & 2 \end{array} \right] \left[ \begin{array}{c} -2 \\ 3 \end{array} \right]\)
\( = \frac{1}{11} \left[ \begin{array}{c} (4 \times -2) + (1 \times 3) \\ (-3 \times -2) + (2 \times 3) \end{array} \right]\)
\( = \frac{1}{11} \left[ \begin{array}{c} -8 + 3 \\ 6 + 6 \end{array} \right] = \frac{1}{11} \left[ \begin{array}{c} -5 \\ 12 \end{array} \right] = \left[ \begin{array}{c} -5/11 \\ 12/11 \end{array} \right]\).
Therefore, \(x = -\frac{5}{11}\) and \(y = \frac{12}{11}\).
In simple words: We converted the two equations into matrix form. After calculating the determinant of the coefficient matrix, we found it was not zero, indicating a unique answer. We then found the inverse of this matrix and multiplied it by the constant terms to arrive at the specific numerical values for \(x\) and \(y\).

Exam Tip: Remember to express fractional answers clearly. Double-check your determinant and adjoint calculations, as a single error can propagate through the entire solution.

Question 9. Solve the following system of linear equations: \(4x - 3y = 3\), \(3x - 5y = 7\).
Answer: The given system of equations is:
\(4x - 3y = 3\)
\(3x - 5y = 7\)
This system of equations can be written in matrix form \(AX = B\), or \(X = A^{-1}B\), where:
\(A = \left[ \begin{array}{cc} 4 & -3 \\ 3 & -5 \end{array} \right]\), \(X = \left[ \begin{array}{c} x \\ y \end{array} \right]\), and \(B = \left[ \begin{array}{c} 3 \\ 7 \end{array} \right]\).
First, we calculate the determinant of \(A\):
\(|A| = (4 \times -5) - (-3 \times 3) = -20 - (-9) = -20 + 9 = -11\).
Since \(|A| = -11 \neq 0\), the inverse \(A^{-1}\) exists, which means the system has a unique solution.
Next, we find the adjoint of \(A\):
\( \text{adj } A = \left[ \begin{array}{cc} -5 & -(-3) \\ -3 & 4 \end{array} \right] = \left[ \begin{array}{cc} -5 & 3 \\ -3 & 4 \end{array} \right]\).
Now, we calculate the inverse \(A^{-1}\):
\(A^{-1} = \frac{1}{|A|} (\text{adj } A) = \frac{1}{-11} \left[ \begin{array}{cc} -5 & 3 \\ -3 & 4 \end{array} \right]\).
Finally, we solve for \(X\) using \(X = A^{-1}B\):
\(X = \frac{1}{-11} \left[ \begin{array}{cc} -5 & 3 \\ -3 & 4 \end{array} \right] \left[ \begin{array}{c} 3 \\ 7 \end{array} \right]\)
\( = \frac{1}{-11} \left[ \begin{array}{c} (-5 \times 3) + (3 \times 7) \\ (-3 \times 3) + (4 \times 7) \end{array} \right]\)
\( = \frac{1}{-11} \left[ \begin{array}{c} -15 + 21 \\ -9 + 28 \end{array} \right] = \frac{1}{-11} \left[ \begin{array}{c} 6 \\ 19 \end{array} \right] = \left[ \begin{array}{c} -6/11 \\ -19/11 \end{array} \right]\).
Therefore, \(x = -\frac{6}{11}\) and \(y = -\frac{19}{11}\).
In simple words: We wrote the equations as matrices. The determinant of the coefficient matrix was calculated and found to be non-zero, confirming a unique solution. We then computed the inverse of this matrix and multiplied it by the constant terms to determine the exact values of \(x\) and \(y\).

Exam Tip: Pay attention to negative signs during determinant calculations and matrix multiplications. A common mistake is mismanaging these signs, especially when calculating the adjoint matrix elements.

Question 10. Solve the following system of linear equations: \(5x + 2y = 3\), \(3x + 2y = 5\).
Answer: The given system of equations is:
\(5x + 2y = 3\)
\(3x + 2y = 5\)
This system of equations can be written in matrix form \(AX = B\), or \(X = A^{-1}B\), where:
\(A = \left[ \begin{array}{cc} 5 & 2 \\ 3 & 2 \end{array} \right]\), \(X = \left[ \begin{array}{c} x \\ y \end{array} \right]\), and \(B = \left[ \begin{array}{c} 3 \\ 5 \end{array} \right]\).
First, we calculate the determinant of \(A\):
\(|A| = (5 \times 2) - (2 \times 3) = 10 - 6 = 4\).
Since \(|A| = 4 \neq 0\), the inverse \(A^{-1}\) exists, which means the system has a unique solution.
Next, we find the adjoint of \(A\):
\( \text{adj } A = \left[ \begin{array}{cc} 2 & -2 \\ -3 & 5 \end{array} \right]\).
Now, we calculate the inverse \(A^{-1}\):
\(A^{-1} = \frac{1}{|A|} (\text{adj } A) = \frac{1}{4} \left[ \begin{array}{cc} 2 & -2 \\ -3 & 5 \end{array} \right]\).
Finally, we solve for \(X\) using \(X = A^{-1}B\):
\(X = \frac{1}{4} \left[ \begin{array}{cc} 2 & -2 \\ -3 & 5 \end{array} \right] \left[ \begin{array}{c} 3 \\ 5 \end{array} \right]\)
\( = \frac{1}{4} \left[ \begin{array}{c} (2 \times 3) + (-2 \times 5) \\ (-3 \times 3) + (5 \times 5) \end{array} \right]\)
\( = \frac{1}{4} \left[ \begin{array}{c} 6 - 10 \\ -9 + 25 \end{array} \right] = \frac{1}{4} \left[ \begin{array}{c} -4 \\ 16 \end{array} \right] = \left[ \begin{array}{c} -1 \\ 4 \end{array} \right]\).
Therefore, \(x = -1\) and \(y = 4\).
In simple words: We converted the equations into matrices and found the determinant of the main coefficient matrix. Since it was not zero, we knew a unique answer existed. We then calculated the inverse of that matrix and multiplied it by the constant terms to solve for \(x\) and \(y\).

Exam Tip: Be careful with fractional multiplication when computing \(A^{-1}B\). Distribute the scalar factor \(\frac{1}{|A|}\) to each element of the resulting matrix product at the end.

Question 11. Solve the following system of linear equations: \(2x + y + z = 1\), \(x - 2y - z = \frac{3}{2}\), \(3y - 5z = 9\).
Answer: The given system of equations is:
\(2x + y + z = 1\)
\(x - 2y - z = \frac{3}{2}\)
\(0x + 3y - 5z = 9\)
This system can be written in matrix form \(AX = B\), or \(X = A^{-1}B\), where:
\(A = \left[ \begin{array}{ccc} 2 & 1 & 1 \\ 1 & -2 & -1 \\ 0 & 3 & -5 \end{array} \right]\), \(X = \left[ \begin{array}{c} x \\ y \\ z \end{array} \right]\), and \(B = \left[ \begin{array}{c} 1 \\ 3/2 \\ 9 \end{array} \right]\).
First, we calculate the determinant of \(A\):
\(|A| = 2( (-2 \times -5) - (-1 \times 3) ) - 1( (1 \times -5) - (-1 \times 0) ) + 1( (1 \times 3) - (-2 \times 0) )\)
\(|A| = 2(10 - (-3)) - 1(-5 - 0) + 1(3 - 0)\)
\(|A| = 2(13) - 1(-5) + 1(3)\)
\(|A| = 26 + 5 + 3 = 34\).
Since \(|A| = 34 \neq 0\), the inverse \(A^{-1}\) exists, and the system has a unique solution.
Next, we find the cofactors:
\(C_{11} = +((-2)(-5) - (-1)(3)) = +(10 + 3) = 13\)
\(C_{12} = -((1)(-5) - (-1)(0)) = -(-5 - 0) = 5\)
\(C_{13} = +((1)(3) - (-2)(0)) = +(3 - 0) = 3\)
\(C_{21} = -((1)(-5) - (1)(3)) = -(-5 - 3) = -(-8) = 8\)
\(C_{22} = +((2)(-5) - (1)(0)) = +(-10 - 0) = -10\)
\(C_{23} = -((2)(3) - (1)(0)) = -(6 - 0) = -6\)
\(C_{31} = +((1)(-1) - (1)(-2)) = +(-1 + 2) = 1\)
\(C_{32} = -((2)(-1) - (1)(1)) = -(-2 - 1) = -(-3) = 3\)
\(C_{33} = +((2)(-2) - (1)(1)) = +(-4 - 1) = -5\)
The cofactor matrix is \(\left[ \begin{array}{ccc} 13 & 5 & 3 \\ 8 & -10 & -6 \\ 1 & 3 & -5 \end{array} \right]\).
The adjoint of \(A\) is the transpose of the cofactor matrix:
\( \text{adj } A = \left[ \begin{array}{ccc} 13 & 8 & 1 \\ 5 & -10 & 3 \\ 3 & -6 & -5 \end{array} \right]\).
Now, we calculate the inverse \(A^{-1}\):
\(A^{-1} = \frac{1}{|A|} (\text{adj } A) = \frac{1}{34} \left[ \begin{array}{ccc} 13 & 8 & 1 \\ 5 & -10 & 3 \\ 3 & -6 & -5 \end{array} \right]\).
Finally, we solve for \(X\) using \(X = A^{-1}B\):
\(X = \frac{1}{34} \left[ \begin{array}{ccc} 13 & 8 & 1 \\ 5 & -10 & 3 \\ 3 & -6 & -5 \end{array} \right] \left[ \begin{array}{c} 1 \\ 3/2 \\ 9 \end{array} \right]\)
\( = \frac{1}{34} \left[ \begin{array}{c} (13 \times 1) + (8 \times 3/2) + (1 \times 9) \\ (5 \times 1) + (-10 \times 3/2) + (3 \times 9) \\ (3 \times 1) + (-6 \times 3/2) + (-5 \times 9) \end{array} \right]\)
\( = \frac{1}{34} \left[ \begin{array}{c} 13 + 12 + 9 \\ 5 - 15 + 27 \\ 3 - 9 - 45 \end{array} \right] = \frac{1}{34} \left[ \begin{array}{c} 34 \\ 17 \\ -51 \end{array} \right] = \left[ \begin{array}{c} 1 \\ 1/2 \\ -3/2 \end{array} \right]\).
Therefore, \(x = 1\), \(y = \frac{1}{2}\), and \(z = -\frac{3}{2}\).
In simple words: We organized the three equations into a matrix form and then calculated the determinant of the coefficient matrix. Since the determinant was not zero, we knew there was a unique answer. We then found the cofactor matrix, its transpose (the adjoint), and finally the inverse matrix. Multiplying the inverse by the constant terms gave us the exact numerical values for \(x\), \(y\), and \(z\).

Exam Tip: When dealing with fractions in the constant matrix \(B\), be careful with multiplication. It's often helpful to keep the scalar \(\frac{1}{|A|}\) outside until the matrix multiplication is complete, then distribute it.

Question 12. Solve the following system of linear equations: \(x - y + z = 4\), \(2x + y - 3z = 0\), \(x + y + z = 2\).
Answer: The given system of equations is:
\(x - y + z = 4\)
\(2x + y - 3z = 0\)
\(x + y + z = 2\)
This system can be written in matrix form \(AX = B\), or \(X = A^{-1}B\), where:
\(A = \left[ \begin{array}{ccc} 1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1 \end{array} \right]\), \(X = \left[ \begin{array}{c} x \\ y \\ z \end{array} \right]\), and \(B = \left[ \begin{array}{c} 4 \\ 0 \\ 2 \end{array} \right]\).
First, we calculate the determinant of \(A\):
\(|A| = 1( (1 \times 1) - (-3 \times 1) ) - (-1)( (2 \times 1) - (-3 \times 1) ) + 1( (2 \times 1) - (1 \times 1) )\)
\(|A| = 1(1 - (-3)) + 1(2 - (-3)) + 1(2 - 1)\)
\(|A| = 1(1 + 3) + 1(2 + 3) + 1(1)\)
\(|A| = 4 + 5 + 1 = 10\).
Since \(|A| = 10 \neq 0\), the inverse \(A^{-1}\) exists, and the system has a unique solution.
Next, we find the cofactors:
\(C_{11} = +(1 - (-3)) = 4\)
\(C_{12} = -(2 - (-3)) = -5\)
\(C_{13} = +(2 - 1) = 1\)
\(C_{21} = -((-1)(1) - (1)(1)) = -(-1 - 1) = 2\)
\(C_{22} = +((1)(1) - (1)(1)) = +(1 - 1) = 0\)
\(C_{23} = -((1)(1) - (-1)(1)) = -(1 - (-1)) = -2\)
\(C_{31} = +((-1)(-3) - (1)(1)) = +(3 - 1) = 2\)
\(C_{32} = -((1)(-3) - (1)(2)) = -(-3 - 2) = 5\)
\(C_{33} = +((1)(1) - (-1)(2)) = +(1 - (-2)) = 3\)
The cofactor matrix is \(\left[ \begin{array}{ccc} 4 & -5 & 1 \\ 2 & 0 & -2 \\ 2 & 5 & 3 \end{array} \right]\).
The adjoint of \(A\) is the transpose of the cofactor matrix:
\( \text{adj } A = \left[ \begin{array}{ccc} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{array} \right]\).
Now, we calculate the inverse \(A^{-1}\):
\(A^{-1} = \frac{1}{|A|} (\text{adj } A) = \frac{1}{10} \left[ \begin{array}{ccc} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{array} \right]\).
Finally, we solve for \(X\) using \(X = A^{-1}B\):
\(X = \frac{1}{10} \left[ \begin{array}{ccc} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{array} \right] \left[ \begin{array}{c} 4 \\ 0 \\ 2 \end{array} \right]\)
\( = \frac{1}{10} \left[ \begin{array}{c} (4 \times 4) + (2 \times 0) + (2 \times 2) \\ (-5 \times 4) + (0 \times 0) + (5 \times 2) \\ (1 \times 4) + (-2 \times 0) + (3 \times 2) \end{array} \right]\)
\( = \frac{1}{10} \left[ \begin{array}{c} 16 + 0 + 4 \\ -20 + 0 + 10 \\ 4 + 0 + 6 \end{array} \right] = \frac{1}{10} \left[ \begin{array}{c} 20 \\ -10 \\ 10 \end{array} \right] = \left[ \begin{array}{c} 2 \\ -1 \\ 1 \end{array} \right]\).
Therefore, \(x = 2\), \(y = -1\), and \(z = 1\).
In simple words: We set up the given equations as a matrix problem. We found the determinant of the coefficient matrix, which was not zero, meaning there's a unique answer. After finding the cofactor matrix, its transpose (the adjoint), and then the inverse matrix, we multiplied the inverse by the constant terms to determine the precise values for \(x\), \(y\), and \(z\).

Exam Tip: Ensure that the signs are correct when calculating the cofactors, particularly the alternating \((-1)^{i+j}\) factor. A simple sign error can lead to incorrect adjoint and inverse matrices.

Question 13. Solve the following system of linear equations: \(2x + y + 3z = 5\), \(x - 2y + z = -4\), \(3x - y - 2z = 3\).
Answer: The given system of equations is:
\(2x + y + 3z = 5\)
\(x - 2y + z = -4\)
\(3x - y - 2z = 3\)
This system can be written in matrix form \(AX = B\), or \(X = A^{-1}B\), where:
\(A = \left[ \begin{array}{ccc} 2 & 1 & 3 \\ 1 & -2 & 1 \\ 3 & -1 & -2 \end{array} \right]\), \(X = \left[ \begin{array}{c} x \\ y \\ z \end{array} \right]\), and \(B = \left[ \begin{array}{c} 5 \\ -4 \\ 3 \end{array} \right]\).
First, we calculate the determinant of \(A\):
\(|A| = 2( (-2 \times -2) - (1 \times -1) ) - 1( (1 \times -2) - (1 \times 3) ) + 3( (1 \times -1) - (-2 \times 3) )\)
\(|A| = 2(4 - (-1)) - 1(-2 - 3) + 3(-1 - (-6))\)
\(|A| = 2(4 + 1) - 1(-5) + 3(-1 + 6)\)
\(|A| = 2(5) + 5 + 3(5)\)
\(|A| = 10 + 5 + 15 = 40\).
Since \(|A| = 40 \neq 0\), the inverse \(A^{-1}\) exists, and the system has a unique solution.
Next, we find the cofactors:
\(C_{11} = +(4+1) = 5\)
\(C_{12} = -(-2-3) = 5\)
\(C_{13} = +(-1+6) = 5\)
\(C_{21} = -(-2-(-3)) = -(-2+3) = -1\)
\(C_{22} = +(-4-9) = -13\)
\(C_{23} = -(-2-3) = 11\)
\(C_{31} = +(1-(-6)) = +(1+6) = 7\)
\(C_{32} = -(2-3) = -(-1) = 1\)
\(C_{33} = +(-4-1) = -7\)
The cofactor matrix is \(\left[ \begin{array}{ccc} 5 & 5 & 5 \\ -1 & -13 & 11 \\ 7 & 1 & -7 \end{array} \right]\).
The adjoint of \(A\) is the transpose of the cofactor matrix:
\( \text{adj } A = \left[ \begin{array}{ccc} 5 & -1 & 7 \\ 5 & -13 & 1 \\ 5 & 11 & -7 \end{array} \right]\).
Now, we calculate the inverse \(A^{-1}\):
\(A^{-1} = \frac{1}{|A|} (\text{adj } A) = \frac{1}{40} \left[ \begin{array}{ccc} 5 & -1 & 7 \\ 5 & -13 & 1 \\ 5 & 11 & -7 \end{array} \right]\).
Finally, we solve for \(X\) using \(X = A^{-1}B\):
\(X = \frac{1}{40} \left[ \begin{array}{ccc} 5 & -1 & 7 \\ 5 & -13 & 1 \\ 5 & 11 & -7 \end{array} \right] \left[ \begin{array}{c} 5 \\ -4 \\ 3 \end{array} \right]\)
\( = \frac{1}{40} \left[ \begin{array}{c} (5 \times 5) + (-1 \times -4) + (7 \times 3) \\ (5 \times 5) + (-13 \times -4) + (1 \times 3) \\ (5 \times 5) + (11 \times -4) + (-7 \times 3) \end{array} \right]\)
\( = \frac{1}{40} \left[ \begin{array}{c} 25 + 4 + 21 \\ 25 + 52 + 3 \\ 25 - 44 - 21 \end{array} \right] = \frac{1}{40} \left[ \begin{array}{c} 50 \\ 80 \\ -40 \end{array} \right] = \left[ \begin{array}{c} 50/40 \\ 80/40 \\ -40/40 \end{array} \right] = \left[ \begin{array}{c} 5/4 \\ 2 \\ -1 \end{array} \right]\).
Therefore, \(x = \frac{5}{4}\), \(y = 2\), and \(z = -1\).
In simple words: We took the three given equations and set them up as a matrix problem. We then found the determinant of the coefficient matrix. Since the determinant was not zero, we knew a single correct answer existed. After finding the cofactor matrix, its transpose (the adjoint), and then the inverse matrix, we multiplied the inverse by the constant terms to determine the precise values for \(x\), \(y\), and \(z\).

Exam Tip: Be meticulous with arithmetic, especially when multiplying matrices. A small calculation error can lead to an entirely wrong final solution. Also, simplify fractions at the end.

Question 14. Solve the following system of linear equations: \(x - y + 2z = 7\), \(3x + 4y - 5z = -5\), \(2x - y + 3z = 12\).
Answer: The given system of equations is:
\(x - y + 2z = 7\)
\(3x + 4y - 5z = -5\)
\(2x - y + 3z = 12\)
This system can be written in matrix form \(AX = B\), or \(X = A^{-1}B\), where:
\(A = \left[ \begin{array}{ccc} 1 & -1 & 2 \\ 3 & 4 & -5 \\ 2 & -1 & 3 \end{array} \right]\), \(X = \left[ \begin{array}{c} x \\ y \\ z \end{array} \right]\), and \(B = \left[ \begin{array}{c} 7 \\ -5 \\ 12 \end{array} \right]\).
First, we calculate the determinant of \(A\):
\(|A| = 1( (4 \times 3) - (-5 \times -1) ) - (-1)( (3 \times 3) - (-5 \times 2) ) + 2( (3 \times -1) - (4 \times 2) )\)
\(|A| = 1(12 - 5) + 1(9 - (-10)) + 2(-3 - 8)\)
\(|A| = 1(7) + 1(9 + 10) + 2(-11)\)
\(|A| = 7 + 19 - 22 = 4\).
Since \(|A| = 4 \neq 0\), the inverse \(A^{-1}\) exists, and the system has a unique solution.
Next, we find the cofactors:
\(C_{11} = +(12 - 5) = 7\)
\(C_{12} = -(9 - (-10)) = -(9+10) = -19\)
\(C_{13} = +(-3 - 8) = -11\)
\(C_{21} = -((-1)(3) - (2)(-1)) = -(-3 - (-2)) = -(-3+2) = 1\)
\(C_{22} = +((1)(3) - (2)(2)) = +(3 - 4) = -1\)
\(C_{23} = -((1)(-1) - (2)(-1)) = -(-1 - (-2)) = -(-1+2) = -1\)
\(C_{31} = +((-1)(-5) - (2)(4)) = +(5 - 8) = -3\)
\(C_{32} = -((1)(-5) - (2)(3)) = -(-5 - 6) = -(-11) = 11\)
\(C_{33} = +((1)(4) - (-1)(3)) = +(4 - (-3)) = 7\)
The cofactor matrix is \(\left[ \begin{array}{ccc} 7 & -19 & -11 \\ 1 & -1 & -1 \\ -3 & 11 & 7 \end{array} \right]\).
The adjoint of \(A\) is the transpose of the cofactor matrix:
\( \text{adj } A = \left[ \begin{array}{ccc} 7 & 1 & -3 \\ -19 & -1 & 11 \\ -11 & -1 & 7 \end{array} \right]\).
Now, we calculate the inverse \(A^{-1}\):
\(A^{-1} = \frac{1}{|A|} (\text{adj } A) = \frac{1}{4} \left[ \begin{array}{ccc} 7 & 1 & -3 \\ -19 & -1 & 11 \\ -11 & -1 & 7 \end{array} \right]\).
Finally, we solve for \(X\) using \(X = A^{-1}B\):
\(X = \frac{1}{4} \left[ \begin{array}{ccc} 7 & 1 & -3 \\ -19 & -1 & 11 \\ -11 & -1 & 7 \end{array} \right] \left[ \begin{array}{c} 7 \\ -5 \\ 12 \end{array} \right]\)
\( = \frac{1}{4} \left[ \begin{array}{c} (7 \times 7) + (1 \times -5) + (-3 \times 12) \\ (-19 \times 7) + (-1 \times -5) + (11 \times 12) \\ (-11 \times 7) + (-1 \times -5) + (7 \times 12) \end{array} \right]\)
\( = \frac{1}{4} \left[ \begin{array}{c} 49 - 5 - 36 \\ -133 + 5 + 132 \\ -77 + 5 + 84 \end{array} \right] = \frac{1}{4} \left[ \begin{array}{c} 8 \\ 4 \\ 12 \end{array} \right] = \left[ \begin{array}{c} 2 \\ 1 \\ 3 \end{array} \right]\).
Therefore, \(x = 2\), \(y = 1\), and \(z = 3\).
In simple words: We organized the three equations into matrix form. After computing the determinant of the coefficient matrix and finding it to be non-zero, we confirmed that a unique solution exists. We then found the cofactor matrix, its transpose (the adjoint), and then the inverse matrix. Multiplying the inverse matrix by the constant terms matrix gave us the specific numerical values for \(x\), \(y\), and \(z\).

Exam Tip: For problems involving larger matrices, systematically calculate each cofactor and cross-check its sign before forming the adjoint matrix. This helps prevent errors in subsequent calculations.

Question 15. If \( A = \begin{pmatrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{pmatrix} \), find \( A^{-1} \). Using \( A^{-1} \), solve the following system of linear equations: \( 2x - 3y + 5z = 11 \), \( 3x + 2y - 4z = -5 \), \( x + y - 2z = -3 \).
Answer: The provided system of linear equations can be represented in matrix form as \( AX = B \), where:
\( A = \begin{pmatrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{pmatrix} \), \( X = \begin{pmatrix} x \\ y \\ z \end{pmatrix} \) and \( B = \begin{pmatrix} 11 \\ -5 \\ -3 \end{pmatrix} \)
First, we determine the determinant of matrix A:
\( |A| = \begin{vmatrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{vmatrix} \)
\( = 2((2)(-2) - (-4)(1)) - (-3)((3)(-2) - (-4)(1)) + 5((3)(1) - (2)(1)) \)
\( = 2(-4 + 4) + 3(-6 + 4) + 5(3 - 2) \)
\( = 2(0) + 3(-2) + 5(1) \)
\( = 0 - 6 + 5 = -1 \)
Since \( |A| = -1 \neq 0 \), the inverse matrix \( A^{-1} \) exists, and the system has a distinct solution.
Next, we compute the cofactor matrix of A:
\( C_{11} = \begin{vmatrix} 2 & -4 \\ 1 & -2 \end{vmatrix} = -4 - (-4) = 0 \)
\( C_{12} = -\begin{vmatrix} 3 & -4 \\ 1 & -2 \end{vmatrix} = -(-6 - (-4)) = -(-2) = 2 \)
\( C_{13} = \begin{vmatrix} 3 & 2 \\ 1 & 1 \end{vmatrix} = 3 - 2 = 1 \)
\( C_{21} = -\begin{vmatrix} -3 & 5 \\ 1 & -2 \end{vmatrix} = -(6 - 5) = -1 \)
\( C_{22} = \begin{vmatrix} 2 & 5 \\ 1 & -2 \end{vmatrix} = -4 - 5 = -9 \)
\( C_{23} = -\begin{vmatrix} 2 & -3 \\ 1 & 1 \end{vmatrix} = -(2 - (-3)) = -(5) = -5 \)
\( C_{31} = \begin{vmatrix} -3 & 5 \\ 2 & -4 \end{vmatrix} = 12 - 10 = 2 \)
\( C_{32} = -\begin{vmatrix} 2 & 5 \\ 3 & -4 \end{vmatrix} = -(-8 - 15) = -(-23) = 23 \)
\( C_{33} = \begin{vmatrix} 2 & -3 \\ 3 & 2 \end{vmatrix} = 4 - (-9) = 13 \)
The cofactor matrix is: \( C = \begin{pmatrix} 0 & 2 & 1 \\ -1 & -9 & -5 \\ 2 & 23 & 13 \end{pmatrix} \)
Now, we find the adjoint of A, which is the transpose of the cofactor matrix:
\( \text{adj A} = C^T = \begin{pmatrix} 0 & -1 & 2 \\ 2 & -9 & 23 \\ 1 & -5 & 13 \end{pmatrix} \)
Then, calculate the inverse of A:
\( A^{-1} = \frac{1}{|A|} \text{adj A} = \frac{1}{-1} \begin{pmatrix} 0 & -1 & 2 \\ 2 & -9 & 23 \\ 1 & -5 & 13 \end{pmatrix} = \begin{pmatrix} 0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13 \end{pmatrix} \)
Finally, we solve for X using \( X = A^{-1}B \):
\( X = \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13 \end{pmatrix} \begin{pmatrix} 11 \\ -5 \\ -3 \end{pmatrix} \)
\( = \begin{pmatrix} (0)(11) + (1)(-5) + (-2)(-3) \\ (-2)(11) + (9)(-5) + (-23)(-3) \\ (-1)(11) + (5)(-5) + (-13)(-3) \end{pmatrix} \)
\( = \begin{pmatrix} 0 - 5 + 6 \\ -22 - 45 + 69 \\ -11 - 25 + 39 \end{pmatrix} \)
\( = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} \)
Thus, the solution is \( x = 1, y = 2, z = 3 \).
In simple words: First, we change the equations into a matrix form. Then, we find a special number called the determinant and check if we can solve the system. After that, we calculate many small numbers called cofactors, arrange them to get the adjoint matrix, and finally compute the inverse matrix. Multiplying the inverse matrix by the constant terms gives us the values for x, y, and z.

Exam Tip: Pay close attention to negative signs during cofactor calculations, as a single error can affect the entire solution. Also, double-check your matrix multiplications for accuracy.

Question 16. The cost of 4 kg onions, 3 kg wheat and 2 kg rice is Rs 60. The cost of 2 kg onions, 4 kg wheat and 6 kg rice Rs 90. The cost of 6 kg onions, 2 kg wheat and 3 kg rice is Rs 70. Find the cost of each item per kg by matrix method.
Answer: Let \( x, y, z \) be the cost per kg of onions, wheat, and rice, respectively. The given information can be written as a system of linear equations:
\( 4x + 3y + 2z = 60 \)
\( 2x + 4y + 6z = 90 \)
\( 6x + 2y + 3z = 70 \)
This system can be represented in matrix form \( AX = B \), where:
\( A = \begin{pmatrix} 4 & 3 & 2 \\ 2 & 4 & 6 \\ 6 & 2 & 3 \end{pmatrix} \), \( X = \begin{pmatrix} x \\ y \\ z \end{pmatrix} \) and \( B = \begin{pmatrix} 60 \\ 90 \\ 70 \end{pmatrix} \)
First, we calculate the determinant of A:
\( |A| = \begin{vmatrix} 4 & 3 & 2 \\ 2 & 4 & 6 \\ 6 & 2 & 3 \end{vmatrix} \)
\( = 4((4)(3) - (6)(2)) - 3((2)(3) - (6)(6)) + 2((2)(2) - (4)(6)) \)
\( = 4(12 - 12) - 3(6 - 36) + 2(4 - 24) \)
\( = 4(0) - 3(-30) + 2(-20) \)
\( = 0 + 90 - 40 = 50 \)
Since \( |A| = 50 \neq 0 \), the inverse matrix \( A^{-1} \) exists, and the system has a distinct solution.
Next, we determine the cofactor matrix of A:
\( C_{11} = \begin{vmatrix} 4 & 6 \\ 2 & 3 \end{vmatrix} = 12 - 12 = 0 \)
\( C_{12} = -\begin{vmatrix} 2 & 6 \\ 6 & 3 \end{vmatrix} = -(6 - 36) = 30 \)
\( C_{13} = \begin{vmatrix} 2 & 4 \\ 6 & 2 \end{vmatrix} = 4 - 24 = -20 \)
\( C_{21} = -\begin{vmatrix} 3 & 2 \\ 2 & 3 \end{vmatrix} = -(9 - 4) = -5 \)
\( C_{22} = \begin{vmatrix} 4 & 2 \\ 6 & 3 \end{vmatrix} = 12 - 12 = 0 \)
\( C_{23} = -\begin{vmatrix} 4 & 3 \\ 6 & 2 \end{vmatrix} = -(8 - 18) = 10 \)
\( C_{31} = \begin{vmatrix} 3 & 2 \\ 4 & 6 \end{vmatrix} = 18 - 8 = 10 \)
\( C_{32} = -\begin{vmatrix} 4 & 2 \\ 2 & 6 \end{vmatrix} = -(24 - 4) = -20 \)
\( C_{33} = \begin{vmatrix} 4 & 3 \\ 2 & 4 \end{vmatrix} = 16 - 6 = 10 \)
The cofactor matrix is: \( C = \begin{pmatrix} 0 & 30 & -20 \\ -5 & 0 & 10 \\ 10 & -20 & 10 \end{pmatrix} \)
Now, we form the adjoint of A by transposing the cofactor matrix:
\( \text{adj A} = C^T = \begin{pmatrix} 0 & -5 & 10 \\ 30 & 0 & -20 \\ -20 & 10 & 10 \end{pmatrix} \)
Then, calculate the inverse of A:
\( A^{-1} = \frac{1}{|A|} \text{adj A} = \frac{1}{50} \begin{pmatrix} 0 & -5 & 10 \\ 30 & 0 & -20 \\ -20 & 10 & 10 \end{pmatrix} \)
Finally, we solve for X using \( X = A^{-1}B \):
\( X = \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \frac{1}{50} \begin{pmatrix} 0 & -5 & 10 \\ 30 & 0 & -20 \\ -20 & 10 & 10 \end{pmatrix} \begin{pmatrix} 60 \\ 90 \\ 70 \end{pmatrix} \)
\( = \frac{1}{50} \begin{pmatrix} (0)(60) + (-5)(90) + (10)(70) \\ (30)(60) + (0)(90) + (-20)(70) \\ (-20)(60) + (10)(90) + (10)(70) \end{pmatrix} \)
\( = \frac{1}{50} \begin{pmatrix} 0 - 450 + 700 \\ 1800 + 0 - 1400 \\ -1200 + 900 + 700 \end{pmatrix} \)
\( = \frac{1}{50} \begin{pmatrix} 250 \\ 400 \\ 400 \end{pmatrix} = \begin{pmatrix} 5 \\ 8 \\ 8 \end{pmatrix} \)
Thus, the cost of onions is Rs 5/kg, wheat is Rs 8/kg, and rice is Rs 8/kg.
In simple words: We convert the problem into three equations and then put them into a matrix. We find the determinant and the inverse of this matrix. Multiplying the inverse matrix by the constant terms gives us the individual costs for onions, wheat, and rice.

Exam Tip: In word problems, clearly defining variables and accurately setting up the initial equations is crucial before proceeding with matrix calculations.

Question 17. Solve the following system of linear equations using the matrix method: \( x - 2y = 10 \), \( 2x - y - z = 8 \), \( -2y + z = 7 \).
Answer: The given system of linear equations can be expressed in matrix form as \( AX = B \), where:
\( A = \begin{pmatrix} 1 & -2 & 0 \\ 2 & -1 & -1 \\ 0 & -2 & 1 \end{pmatrix} \), \( X = \begin{pmatrix} x \\ y \\ z \end{pmatrix} \) and \( B = \begin{pmatrix} 10 \\ 8 \\ 7 \end{pmatrix} \)
First, we calculate the determinant of A:
\( |A| = \begin{vmatrix} 1 & -2 & 0 \\ 2 & -1 & -1 \\ 0 & -2 & 1 \end{vmatrix} \)
\( = 1((-1)(1) - (-1)(-2)) - (-2)((2)(1) - (-1)(0)) + 0(...) \)
\( = 1(-1 - 2) + 2(2 - 0) + 0 \)
\( = 1(-3) + 2(2) \)
\( = -3 + 4 = 1 \)
Since \( |A| = 1 \neq 0 \), the inverse matrix \( A^{-1} \) exists, and the system has a distinct solution.
Next, we compute the cofactor matrix of A:
\( C_{11} = \begin{vmatrix} -1 & -1 \\ -2 & 1 \end{vmatrix} = (-1)(1) - (-1)(-2) = -1 - 2 = -3 \)
\( C_{12} = -\begin{vmatrix} 2 & -1 \\ 0 & 1 \end{vmatrix} = -((2)(1) - (-1)(0)) = -(2 - 0) = -2 \)
\( C_{13} = \begin{vmatrix} 2 & -1 \\ 0 & -2 \end{vmatrix} = (2)(-2) - (-1)(0) = -4 - 0 = -4 \)
\( C_{21} = -\begin{vmatrix} -2 & 0 \\ -2 & 1 \end{vmatrix} = -((-2)(1) - (0)(-2)) = -(-2 - 0) = 2 \)
\( C_{22} = \begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix} = (1)(1) - (0)(0) = 1 - 0 = 1 \)
\( C_{23} = -\begin{vmatrix} 1 & -2 \\ 0 & -2 \end{vmatrix} = -((1)(-2) - (-2)(0)) = -(-2 - 0) = 2 \)
\( C_{31} = \begin{vmatrix} -2 & 0 \\ -1 & -1 \end{vmatrix} = (-2)(-1) - (0)(-1) = 2 - 0 = 2 \)
\( C_{32} = -\begin{vmatrix} 1 & 0 \\ 2 & -1 \end{vmatrix} = -((1)(-1) - (0)(2)) = -(-1 - 0) = 1 \)
\( C_{33} = \begin{vmatrix} 1 & -2 \\ 2 & -1 \end{vmatrix} = (1)(-1) - (-2)(2) = -1 - (-4) = -1 + 4 = 3 \)
The cofactor matrix is: \( C = \begin{pmatrix} -3 & -2 & -4 \\ 2 & 1 & 2 \\ 2 & 1 & 3 \end{pmatrix} \)
Now, we find the adjoint of A, which is the transpose of the cofactor matrix:
\( \text{adj A} = C^T = \begin{pmatrix} -3 & 2 & 2 \\ -2 & 1 & 1 \\ -4 & 2 & 3 \end{pmatrix} \)
Then, calculate the inverse of A:
\( A^{-1} = \frac{1}{|A|} \text{adj A} = \frac{1}{1} \begin{pmatrix} -3 & 2 & 2 \\ -2 & 1 & 1 \\ -4 & 2 & 3 \end{pmatrix} = \begin{pmatrix} -3 & 2 & 2 \\ -2 & 1 & 1 \\ -4 & 2 & 3 \end{pmatrix} \)
Finally, we solve for X using \( X = A^{-1}B \):
\( X = \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} -3 & 2 & 2 \\ -2 & 1 & 1 \\ -4 & 2 & 3 \end{pmatrix} \begin{pmatrix} 10 \\ 8 \\ 7 \end{pmatrix} \)
\( = \begin{pmatrix} (-3)(10) + (2)(8) + (2)(7) \\ (-2)(10) + (1)(8) + (1)(7) \\ (-4)(10) + (2)(8) + (3)(7) \end{pmatrix} \)
\( = \begin{pmatrix} -30 + 16 + 14 \\ -20 + 8 + 7 \\ -40 + 16 + 21 \end{pmatrix} \)
\( = \begin{pmatrix} 0 \\ -5 \\ -3 \end{pmatrix} \)
Thus, the solution to the system of equations is \( x = 0, y = -5, z = -3 \).
In simple words: First, we write down the equations as a matrix problem. Then, we find its determinant, which tells us if a unique solution exists. Next, we calculate the cofactor matrix and its transpose, called the adjoint matrix. Using these, we get the inverse matrix. Finally, multiplying the inverse matrix by the constant matrix helps us find the values of x, y, and z.

Exam Tip: When an equation is missing a variable, remember to include a zero coefficient for that variable in the matrix to maintain correct dimensions.