Maharashtra Board Class 12 Maths Part 1 Chapter 2 Matrices 2.6 Solutions

Question 1. Solve the following equations by the method of inversion:
(i) x + 2y = 2, 2x + 3y = 3
Answer:
Solution:
The given equations can be written in the matrix form as: \[ \begin{pmatrix} 1 & 2 \\ 2 & 3 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 2 \\ 3 \end{pmatrix} \] This is of the form \(AX = B\), where \[ A= \begin{pmatrix} 1 & 2 \\ 2 & 3 \end{pmatrix}, X = \begin{pmatrix} x \\ y \end{pmatrix} \text{ and } B = \begin{pmatrix} 2 \\ 3 \end{pmatrix} \] Let us find \(A^{-1}\).
\(|A| = \begin{vmatrix} 1 & 2 \\ 2 & 3 \end{vmatrix} = 3-4 = -1 \neq 0\)
\(\therefore A^{-1}\) exists.
We write \(AA^{-1}=I\)
\(\begin{pmatrix} 1 & 2 \\ 2 & 3 \end{pmatrix} A^{-1} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\)
By \(R_2 - 2R_1\), we get \[ \begin{pmatrix} 1 & 2 \\ 0 & -1 \end{pmatrix} A^{-1} = \begin{pmatrix} 1 & 0 \\ -2 & 1 \end{pmatrix} \] By \((-1)R_2\), we get \[ \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} A^{-1} = \begin{pmatrix} 1 & 0 \\ 2 & -1 \end{pmatrix} \] By \(R_1 - 2R_2\), we get \[ \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} A^{-1} = \begin{pmatrix} -3 & 2 \\ 2 & -1 \end{pmatrix} \] \(\therefore A^{-1} = \begin{pmatrix} -3 & 2 \\ 2 & -1 \end{pmatrix}\)
Now, premultiply \(AX = B\) by \(A^{-1}\), we get
\(A^{-1}(AX)=A^{-1}B\)
\(\therefore (A^{-1}A)X=A^{-1}B\)
\(\therefore IX = A^{-1}B\)
\(\therefore X = A^{-1}B\)
\(\therefore \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} -3 & 2 \\ 2 & -1 \end{pmatrix} \begin{pmatrix} 2 \\ 3 \end{pmatrix}\)
\(\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} -6+6 \\ 4-3 \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \end{pmatrix}\)
By equality of matrices,
\(x = 0, y = 1\) is the required solution.
In simple words: This method solves simultaneous linear equations by first converting them into a matrix equation \( (AX=B) \), finding the inverse of the coefficient matrix \( (A^{-1}) \) using row operations, and then multiplying both sides by the inverse to find the variable matrix \( (X) \).

🎯 Exam Tip: Mastery of matrix inversion using elementary row operations is crucial for solving linear equations. Pay close attention to arithmetic and step-by-step transformations to avoid calculation errors.

Question 1. Solve the following equations by the method of inversion:
(ii) 2x + y = 5, 3x + 5y = -3
Answer:
Solution:
Matrix form of the given system of equations is \[ \begin{pmatrix} 2 & 1 \\ 3 & 5 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 5 \\ -3 \end{pmatrix} \] This is of the form \(AX = B\),
where \(A = \begin{pmatrix} 2 & 1 \\ 3 & 5 \end{pmatrix}, X = \begin{pmatrix} x \\ y \end{pmatrix} \text{ and } B = \begin{pmatrix} 5 \\ -3 \end{pmatrix}\)
To determine \(X\), we have to find \(A^{-1}\).
\(|A| = \begin{vmatrix} 2 & 1 \\ 3 & 5 \end{vmatrix} = 10-3 = 7 \neq 0\)
\(\therefore A^{-1}\) exists.
Consider \(AA^{-1} = I\)
\(\begin{pmatrix} 2 & 1 \\ 3 & 5 \end{pmatrix} A^{-1} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\)
Applying \(R_1 \leftrightarrow R_2\), we get \[ \begin{pmatrix} 3 & 5 \\ 2 & 1 \end{pmatrix} A^{-1} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \] Applying \(R_1 \to R_1 - R_2\), we get \[ \begin{pmatrix} 1 & 4 \\ 2 & 1 \end{pmatrix} A^{-1} = \begin{pmatrix} -1 & 1 \\ 1 & 0 \end{pmatrix} \] Applying \(R_2 \to R_2 - 2R_1\), we get \[ \begin{pmatrix} 1 & 4 \\ 0 & -7 \end{pmatrix} A^{-1} = \begin{pmatrix} -1 & 1 \\ 3 & -2 \end{pmatrix} \] Applying \(R_2 \to \left(-\frac{1}{7}\right) R_2\), we get \[ \begin{pmatrix} 1 & 4 \\ 0 & 1 \end{pmatrix} A^{-1} = \begin{pmatrix} -1 & 1 \\ -\frac{3}{7} & \frac{2}{7} \end{pmatrix} \] Applying \(R_1 \to R_1 - 4R_2\), we get \[ \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} A^{-1} = \begin{pmatrix} -1 - 4\left(-\frac{3}{7}\right) & 1 - 4\left(\frac{2}{7}\right) \\ -\frac{3}{7} & \frac{2}{7} \end{pmatrix} = \begin{pmatrix} -1 + \frac{12}{7} & 1 - \frac{8}{7} \\ -\frac{3}{7} & \frac{2}{7} \end{pmatrix} = \begin{pmatrix} \frac{5}{7} & -\frac{1}{7} \\ -\frac{3}{7} & \frac{2}{7} \end{pmatrix} \] \(\therefore A^{-1} = \begin{pmatrix} \frac{5}{7} & -\frac{1}{7} \\ -\frac{3}{7} & \frac{2}{7} \end{pmatrix}\)
Pre-multiplying \(AX = B\) by \(A^{-1}\), we get
\(A^{-1}(AX) = A^{-1}B\)
\(\therefore (A^{-1}A)X = A^{-1}B\)
\(\therefore IX = A^{-1}B\)
\(\therefore X = A^{-1}B\)
\(\therefore \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} \frac{5}{7} & -\frac{1}{7} \\ -\frac{3}{7} & \frac{2}{7} \end{pmatrix} \begin{pmatrix} 5 \\ -3 \end{pmatrix}\)
\(\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} \frac{25}{7} + \frac{3}{7} \\ -\frac{15}{7} - \frac{6}{7} \end{pmatrix} = \begin{pmatrix} \frac{28}{7} \\ -\frac{21}{7} \end{pmatrix} = \begin{pmatrix} 4 \\ -3 \end{pmatrix}\)
\(\therefore\) By equality of matrices,
\(x = 4\) and \(y = -3\).
In simple words: Similar to the previous problem, this involves transforming the system of equations into a matrix form, calculating the inverse of the coefficient matrix through elementary row operations, and then multiplying to find the unknown variables.

🎯 Exam Tip: Fractional elements in matrices can be challenging. Practice careful calculation and simplification of fractions at each step to maintain accuracy in your inverse matrix determination.

Question 1. Solve the following equations by the method of inversion:
(iii) 2x - y + z = 1, x + 2y + 3z = 8 and 3x + y - 4z = 1
Answer:
Solution:
The given equations can be written in the matrix form as: \[ \begin{pmatrix} 2 & -1 & 1 \\ 1 & 2 & 3 \\ 3 & 1 & -4 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 1 \\ 8 \\ 1 \end{pmatrix} \] This is of the form \(AX = B\), where \[ A= \begin{pmatrix} 2 & -1 & 1 \\ 1 & 2 & 3 \\ 3 & 1 & -4 \end{pmatrix}, X = \begin{pmatrix} x \\ y \\ z \end{pmatrix} \text{ and } B = \begin{pmatrix} 1 \\ 8 \\ 1 \end{pmatrix} \] Let us find \(A^{-1}\).
\(|A| = \begin{vmatrix} 2 & -1 & 1 \\ 1 & 2 & 3 \\ 3 & 1 & -4 \end{vmatrix}\)
\( = 2(-8-3) + 1(-4-9) + 1(1-6)\)
\( = 2(-11) + 1(-13) + 1(-5)\)
\( = -22 - 13 - 5 = -40 \neq 0\)
\(\therefore A^{-1}\) exists.
We write \(AA^{-1}=I\)
\(\begin{pmatrix} 2 & -1 & 1 \\ 1 & 2 & 3 \\ 3 & 1 & -4 \end{pmatrix} A^{-1} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}\)
By \(R_1 \leftrightarrow R_2\), we get \[ \begin{pmatrix} 1 & 2 & 3 \\ 2 & -1 & 1 \\ 3 & 1 & -4 \end{pmatrix} A^{-1} = \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix} \] By \(R_2 - 2R_1\) and \(R_3 - 3R_1\), we get \[ \begin{pmatrix} 1 & 2 & 3 \\ 0 & -5 & -5 \\ 0 & -5 & -13 \end{pmatrix} A^{-1} = \begin{pmatrix} 0 & 1 & 0 \\ 1 & -2 & 0 \\ 0 & -3 & 1 \end{pmatrix} \] By \(\left(-\frac{1}{5}\right) R_2\), we get \[ \begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 1 \\ 0 & -5 & -13 \end{pmatrix} A^{-1} = \begin{pmatrix} 0 & 1 & 0 \\ -\frac{1}{5} & \frac{2}{5} & 0 \\ 0 & -3 & 1 \end{pmatrix} \] By \(R_1 - 2R_2\) and \(R_3 + 5R_2\), we get \[ \begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & -8 \end{pmatrix} A^{-1} = \begin{pmatrix} \frac{2}{5} & \frac{1}{5} & 0 \\ -\frac{1}{5} & \frac{2}{5} & 0 \\ -1 & -1 & 1 \end{pmatrix} \] By \(\left(-\frac{1}{8}\right) R_3\), we get \[ \begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix} A^{-1} = \begin{pmatrix} \frac{2}{5} & \frac{1}{5} & 0 \\ -\frac{1}{5} & \frac{2}{5} & 0 \\ \frac{1}{8} & \frac{1}{8} & -\frac{1}{8} \end{pmatrix} \] By \(R_1 - R_3\) and \(R_2 - R_3\), we get \[ \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} A^{-1} = \begin{pmatrix} \frac{2}{5} - \frac{1}{8} & \frac{1}{5} - \frac{1}{8} & 0 - \left(-\frac{1}{8}\right) \\ -\frac{1}{5} - \frac{1}{8} & \frac{2}{5} - \frac{1}{8} & 0 - \left(-\frac{1}{8}\right) \\ \frac{1}{8} & \frac{1}{8} & -\frac{1}{8} \end{pmatrix} = \begin{pmatrix} \frac{16-5}{40} & \frac{8-5}{40} & \frac{1}{8} \\ \frac{-8-5}{40} & \frac{16-5}{40} & \frac{1}{8} \\ \frac{1}{8} & \frac{1}{8} & -\frac{1}{8} \end{pmatrix} = \begin{pmatrix} \frac{11}{40} & \frac{3}{40} & \frac{1}{8} \\ -\frac{13}{40} & \frac{11}{40} & \frac{1}{8} \\ \frac{1}{8} & \frac{1}{8} & -\frac{1}{8} \end{pmatrix} \] \(\therefore A^{-1} = \frac{1}{40} \begin{pmatrix} 11 & 3 & 5 \\ -13 & 11 & 5 \\ 5 & 5 & -5 \end{pmatrix}\)
Now, premultiply \(AX = B\) by \(A^{-1}\), we get
\(A^{-1}(AX) = A^{-1}B\)
\(\therefore (A^{-1}A)X = A^{-1}B\)
\(\therefore IX = A^{-1}B\)
\(\therefore X = A^{-1}B\)
\(\therefore \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \frac{1}{40} \begin{pmatrix} 11 & 3 & 5 \\ -13 & 11 & 5 \\ 5 & 5 & -5 \end{pmatrix} \begin{pmatrix} 1 \\ 8 \\ 1 \end{pmatrix}\)
\(= \frac{1}{40} \begin{pmatrix} 11+24+5 \\ -13+88+5 \\ 5+40-5 \end{pmatrix} = \frac{1}{40} \begin{pmatrix} 40 \\ 80 \\ 40 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}\)
\(\therefore \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}\)
By equality of matrices,
\(x = 1, y = 2, z = 1\) is the required solution.
In simple words: For a system of three linear equations, the matrix inversion method involves expressing the system as \( AX=B \), finding the inverse of the 3x3 coefficient matrix, and then multiplying to determine the values of \( x, y, \) and \( z \).

🎯 Exam Tip: Solving 3x3 matrix inversions requires careful and systematic application of elementary row operations. Double-check each step's arithmetic to avoid cascading errors that can invalidate the final solution.

Question 1. Solve the following equations by the method of inversion:
(iv) x + y + z = 1, x - y + z = 2 and x + y - z = 3
Answer:
Solution:
Matrix form of the given system of equations is \[ \begin{pmatrix} 1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} \] This is of the form \(AX = B\),
where \(A = \begin{pmatrix} 1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{pmatrix}, X = \begin{pmatrix} x \\ y \\ z \end{pmatrix} \text{ and } B = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}\)
Consider \(AA^{-1} = I\)
\(\begin{pmatrix} 1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{pmatrix} A^{-1} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}\)
Applying \(R_2 \to R_2 - R_1\) and \(R_3 \to R_3 - R_1\), we get \[ \begin{pmatrix} 1 & 1 & 1 \\ 0 & -2 & 0 \\ 0 & 0 & -2 \end{pmatrix} A^{-1} = \begin{pmatrix} 1 & 0 & 0 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{pmatrix} \] Applying \(R_2 \to \left(-\frac{1}{2}\right) R_2\), we get \[ \begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & -2 \end{pmatrix} A^{-1} = \begin{pmatrix} 1 & 0 & 0 \\ \frac{1}{2} & -\frac{1}{2} & 0 \\ -1 & 0 & 1 \end{pmatrix} \] Applying \(R_1 \to R_1 - R_2\), we get \[ \begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & -2 \end{pmatrix} A^{-1} = \begin{pmatrix} \frac{1}{2} & \frac{1}{2} & 0 \\ \frac{1}{2} & -\frac{1}{2} & 0 \\ -1 & 0 & 1 \end{pmatrix} \] Applying \(R_3 \to \left(-\frac{1}{2}\right) R_3\), we get \[ \begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} A^{-1} = \begin{pmatrix} \frac{1}{2} & \frac{1}{2} & 0 \\ \frac{1}{2} & -\frac{1}{2} & 0 \\ \frac{1}{2} & 0 & -\frac{1}{2} \end{pmatrix} \] Applying \(R_1 \to R_1 - R_3\), we get \[ \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} A^{-1} = \begin{pmatrix} \frac{1}{2} - \frac{1}{2} & \frac{1}{2} - 0 & 0 - \left(-\frac{1}{2}\right) \\ \frac{1}{2} & -\frac{1}{2} & 0 \\ \frac{1}{2} & 0 & -\frac{1}{2} \end{pmatrix} = \begin{pmatrix} 0 & \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & -\frac{1}{2} & 0 \\ \frac{1}{2} & 0 & -\frac{1}{2} \end{pmatrix} \] \(\therefore A^{-1} = \begin{pmatrix} 0 & \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & -\frac{1}{2} & 0 \\ \frac{1}{2} & 0 & -\frac{1}{2} \end{pmatrix}\)
Pre-multiplying \(AX = B\) by \(A^{-1}\), we get
\(A^{-1}(AX) = A^{-1}B\)
\(\therefore (A^{-1}A)X = A^{-1}B\)
\(\therefore IX = A^{-1}B\)
\(\therefore X = A^{-1}B\)
\(\therefore \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 & \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & -\frac{1}{2} & 0 \\ \frac{1}{2} & 0 & -\frac{1}{2} \end{pmatrix} \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}\)
\(\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0+1+\frac{3}{2} \\ \frac{1}{2}-1+0 \\ \frac{1}{2}+0-\frac{3}{2} \end{pmatrix} = \begin{pmatrix} \frac{5}{2} \\ -\frac{1}{2} \\ -1 \end{pmatrix}\)
\(\therefore\) By equality of matrices,
\(x = \frac{5}{2}, y = -\frac{1}{2}\) and \(z = -1\).
In simple words: This solution uses the matrix inversion method for a 3x3 system, where the inverse matrix is calculated via row operations and then multiplied by the constant matrix to find the solution vector \( (x, y, z) \).

🎯 Exam Tip: Be careful with fractional calculations during matrix inversion, especially when performing row operations. Verify intermediate steps to ensure the final result is accurate.

Question 2. Express the following equations in matrix form and solve them by method of reduction:
(i) x + 3y = 2, 3x + 5y = 4.
Answer:
Solution:
The given equations can be written in the matrix form as: \[ \begin{pmatrix} 1 & 3 \\ 3 & 5 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 2 \\ 4 \end{pmatrix} \] By \(R_2 - 3R_1\), we get \[ \begin{pmatrix} 1 & 3 \\ 0 & -4 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 2 \\ -2 \end{pmatrix} \] \(\therefore \begin{pmatrix} x+3y \\ 0-4y \end{pmatrix} = \begin{pmatrix} 2 \\ -2 \end{pmatrix}\)
By equality of matrices,
\(x + 3y = 2\) .....(1)
\(-4y = -2\) .....(2)
From (2), \(y = \frac{-2}{-4} = \frac{1}{2}\)
Substituting \(y = \frac{1}{2}\) in (1), we get
\(x + 3\left(\frac{1}{2}\right) = 2\)
\(x + \frac{3}{2} = 2\)
\(\therefore x = 2 - \frac{3}{2} = \frac{4-3}{2} = \frac{1}{2}\)
Hence, \(x = \frac{1}{2}, y = \frac{1}{2}\) is the required solution.
In simple words: The reduction method transforms the matrix equation into an upper triangular form through row operations. This allows for straightforward back-substitution to find the values of the variables.

🎯 Exam Tip: The reduction method is faster for solving systems of equations than inversion when only the solution is needed. Ensure row operations are applied correctly to both the coefficient and constant matrices simultaneously.

Question 2. Express the following equations in matrix form and solve them by method of reduction:
(ii) 3x - y = 1, 4x + y = 6.
Answer:
Solution:
The given equations can be written in the matrix form as: \[ \begin{pmatrix} 3 & -1 \\ 4 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 1 \\ 6 \end{pmatrix} \] Multiplying the first equation by 4 and the second equation by 3, the equations become: \[ \begin{pmatrix} 12 & -4 \\ 12 & 3 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 4 \\ 18 \end{pmatrix} \] By \(R_2 - R_1\), we get \[ \begin{pmatrix} 12 & -4 \\ 0 & 7 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 4 \\ 14 \end{pmatrix} \] \(\therefore \begin{pmatrix} 12x-4y \\ 0+7y \end{pmatrix} = \begin{pmatrix} 4 \\ 14 \end{pmatrix}\)
By equality of matrices,
\(12x - 4y = 4\) .....(1)
\(7y = 14\) .....(2)
From (2), \(y = \frac{14}{7} = 2\)
Substituting \(y = 2\) in (1), we get
\(12x - 4(2) = 4\)
\(12x - 8 = 4\)
\(12x = 4 + 8\)
\(12x = 12\)
\(\therefore x = 1\)
Hence, \(x = 1, y = 2\) is the required solution.
In simple words: This problem demonstrates the reduction method by first scaling the original equations and then applying a row operation to eliminate a variable, simplifying the system into an easily solvable form via back-substitution.

🎯 Exam Tip: Be mindful that scaling equations is a valid initial step, but ensure that subsequent row operations are applied consistently to the entire augmented matrix (coefficient and constant parts). Careful arithmetic is key.

Question 2. Express the following equations in matrix form and solve them by method of reduction:
(iii) x + 2y + z = 8, 2x + 3y - z = 11 and 3x - y - 2z = 5.
Answer:
Solution:
The given equations can be written in the matrix form as: \[ \begin{pmatrix} 1 & 2 & 1 \\ 2 & 3 & -1 \\ 3 & -1 & -2 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 8 \\ 11 \\ 5 \end{pmatrix} \] By \(R_2 - 2R_1\) and \(R_3 - 3R_1\), we get \[ \begin{pmatrix} 1 & 2 & 1 \\ 0 & -1 & -3 \\ 0 & -7 & -5 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 8 \\ -5 \\ -19 \end{pmatrix} \] By \(R_3 - 7R_2\), we get \[ \begin{pmatrix} 1 & 2 & 1 \\ 0 & -1 & -3 \\ 0 & 0 & 16 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 8 \\ -5 \\ 16 \end{pmatrix} \] \(\therefore \begin{pmatrix} x+2y+z \\ 0-y-3z \\ 0+0+16z \end{pmatrix} = \begin{pmatrix} 8 \\ -5 \\ 16 \end{pmatrix}\)
By equality of matrices,
\(x + 2y + z = 8\) .....(1)
\(-y - 3z = -5\) .....(2)
\(16z = 16\) .....(3)
From (3), \(z = \frac{16}{16} = 1\)
Substituting \(z = 1\) in (2), we get
\(-y - 3(1) = -5\)
\(-y - 3 = -5\)
\(-y = -5 + 3\)
\(-y = -2\)
\(\therefore y = 2\)
Substituting \(y = 2, z = 1\) in (1), we get
\(x + 2(2) + 1 = 8\)
\(x + 4 + 1 = 8\)
\(x + 5 = 8\)
\(\therefore x = 3\)
Hence, \(x = 3, y = 2, z = 1\) is the required solution.
In simple words: This problem demonstrates solving a 3x3 system of linear equations using the reduction method by systematically applying row operations to transform the augmented matrix into an upper triangular form, enabling straightforward back-substitution to find the variables.

🎯 Exam Tip: When reducing 3x3 matrices, focus on creating zeros in the lower triangle first. Careful organization of steps and accurate arithmetic are paramount to avoid errors in the final solution.

Question 2. Express the following equations in matrix form and solve them by method of reduction:
(iv) x + y + z = 1, 2x + 3y + 2z = 2 and x + y + 2z = 4.
Answer:
Solution:
The given equations can be written in the matrix form as: \[ \begin{pmatrix} 1 & 1 & 1 \\ 2 & 3 & 2 \\ 1 & 1 & 2 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ 4 \end{pmatrix} \] By \(R_2 - 2R_1\) and \(R_3 - R_1\), we get \[ \begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \\ 3 \end{pmatrix} \] \(\therefore \begin{pmatrix} x+y+z \\ 0+y+0 \\ 0+0+z \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \\ 3 \end{pmatrix}\)
By equality of matrices,
\(x + y + z = 1\) .....(1)
\(y = 0\)
\(z = 3\)
Substituting \(y = 0, z = 3\) in (1), we get
\(x + 0 + 3 = 1\)
\(x + 3 = 1\)
\(\therefore x = 1 - 3 = -2\)
Hence, \(x = -2, y = 0, z = 3\) is the required solution.
In simple words: This problem uses the matrix reduction method to solve a system of linear equations by transforming the coefficient matrix into an upper triangular form through row operations, which directly yields the values of the variables via back-substitution.

🎯 Exam Tip: When the coefficient matrix transforms into a simple identity-like matrix quickly, it simplifies the back-substitution process significantly. Always perform row operations on the entire augmented matrix to maintain consistency.

Question 3. The total cost of 3 T.V. and 2 V.C.R. is Rs. 35000. The shopkeeper wants a profit of Rs. 1000 per T.V. and Rs. 500 per V.C.R. He sells 2 T.V. and 1 V.C.R. and he gets total revenue of Rs. 21500. Find the cost and selling price of T.V. and V.C.R.
Answer:
Solution:
Let the cost of each T.V. be \(x\) and each V.C.R. be \(y\).
Then the total cost of 3 T.V. and 2 V.C.R. is \((3x + 2y)\) which is given to be Rs. 35000.
\(\therefore 3x + 2y = 35000\)
The shopkeeper wants a profit of Rs. 1000 per T.V. and Rs. 500 per V.C.R.
The selling price of each T.V. is Rs. \((x + 1000)\) and of each V.C.R. is Rs. \((y + 500)\).
\(\therefore\) selling price of 2 T.V. and 1 V.C.R is
Rs. \([2(x + 1000) + (y + 500)]\) which is given to be Rs. 21500.
\(\therefore 2(x + 1000) + (y + 500) = 21500\)
\(\therefore 2x + 2000 + y + 500 = 21500\)
\(\therefore 2x + y = 19000\)
Hence, the system of linear equations is
\(3x + 2y = 35000\)
\(2x + y = 19000\)
The equations can be written in matrix form as: \[ \begin{pmatrix} 3 & 2 \\ 2 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 35000 \\ 19000 \end{pmatrix} \] By \(R_1 - 2R_2\), we get \[ \begin{pmatrix} -1 & 0 \\ 2 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} -3000 \\ 19000 \end{pmatrix} \] \(\therefore \begin{pmatrix} -x+0 \\ 2x+y \end{pmatrix} = \begin{pmatrix} -3000 \\ 19000 \end{pmatrix}\)
By equality of matrices,
\(-x = -3000\) .....(1)
\(2x + y = 19000\) .....(2)
From (1), \(x = 3000\)
Substituting \(x = 3000\) in (2), we get
\(2(3000) + y = 19000\)
\(6000 + y = 19000\)
\(\therefore y = 19000 - 6000 = 13000\)
Hence, the cost price of one T.V. is Rs. 3000 and of one V.C.R. is Rs. 13000 and the selling price of one T.V. is Rs. \((3000 + 1000) = \) Rs. 4000 and of one V.C.R. is Rs. \((13000 + 500) = \) Rs. 13500.
In simple words: This problem involves setting up a system of linear equations based on given cost, profit, and revenue information. The system is then solved using the matrix reduction method to find the cost and selling prices of the items.

🎯 Exam Tip: Carefully read word problems to correctly translate information into mathematical equations. Define variables clearly, and ensure all conditions (cost, profit, revenue) are incorporated accurately into the matrix setup.

Question 4. The sum of the cost of one Economics book, one Cooperation book, and one Account book is Rs. 420. The total cost of an Economic book, 2 Cooperation books, and an Account book is Rs. 480. Also, the total cost of an Economic book, 3 Cooperation books, and 2 Account books is Rs. 600. Find the cost of each book.
Answer:
Solution:
Let the cost of 1 Economic book, 1 Cooperation book and 1 Account book be Rs. \(x\), Rs. \(y\) and Rs. \(z\) respectively.
Then, from the given information
\(x + y + z = 420\)
\(x + 2y + z = 480\)
\(x + 3y + 2z = 600\)
These equations can be written in matrix form as: \[ \begin{pmatrix} 1 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 3 & 2 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 420 \\ 480 \\ 600 \end{pmatrix} \] By \(R_2 - R_1\) and \(R_3 - R_1\), we get \[ \begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & 0 \\ 0 & 2 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 420 \\ 60 \\ 180 \end{pmatrix} \] \(\therefore \begin{pmatrix} x+y+z \\ 0+y+0 \\ 0+2y+z \end{pmatrix} = \begin{pmatrix} 420 \\ 60 \\ 180 \end{pmatrix}\)
By equality of matrices,
\(x + y + z = 420\) .....(1)
\(y = 60\) .....(2)
\(2y + z = 180\) .....(3)
Substituting \(y = 60\) in (3), we get
\(2(60) + z = 180\)
\(120 + z = 180\)
\(\therefore z = 180 - 120 = 60\)
Substituting \(y = 60, z = 60\) in (1), we get
\(x + 60 + 60 = 420\)
\(x + 120 = 420\)
\(\therefore x = 420 - 120 = 300\)
Hence, the cost of each Economic book is Rs. 300, each Cooperation book is Rs. 60 and each Account book is Rs. 60.
In simple words: This problem translates a real-world scenario of book costs into a system of linear equations, which is then solved using the matrix reduction method to find the individual cost of each type of book.

🎯 Exam Tip: For word problems, accurately setting up the initial system of equations from the narrative is the most critical first step. Errors here will lead to incorrect solutions even with perfect matrix operations. Verify your equations before proceeding.

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