CBSE Class 12 Mathematics HOTs Determinants Set 07

Question. The management committee of a residential colony decided to award some of its members (say \(x\)) for honesty, some (say \(y\)) for helping others and some others (say \(z\)) for supervising the workers to keep the colony neat and clean. The sum of all the awardees is 12. Three times the sum of awardees for cooperation and supervision added to two times the numbers of awardees for honesty is 33. If the sum of the number of awardees for honesty and supervision is twice the number of awardees for helping others, using matrix method, find the number of awardees of each category. Apart from these values, namely, honesty, cooperation and supervision, suggest one more value which the management of the colony must include for awards. (AI 2013)
Answer: According to the question, we have
\( x + y + z = 12 \)
\( 2x + 3(y + z) = 33 \Rightarrow 2x + 3y + 3z = 33 \)
\( x + z = 2y \Rightarrow x - 2y + z = 0 \)
The system of equations can be written as \( AX = B \)
where, \( A = \begin{bmatrix} 1 & 1 & 1 \\ 2 & 3 & 3 \\ 1 & -2 & 1 \end{bmatrix} \); \( X = \begin{bmatrix} x \\ y \\ z \end{bmatrix} \); \( B = \begin{bmatrix} 12 \\ 33 \\ 0 \end{bmatrix} \)
\( |A| = \begin{vmatrix} 1 & 1 & 1 \\ 2 & 3 & 3 \\ 1 & -2 & 1 \end{vmatrix} = 1(3 + 6) - 1(2 - 3) + 1(-4 - 3) = 9 + 1 - 7 = 3 \neq 0 \)
Since \( A^{-1} \) exists, so system of equations has a unique solution and it is given by \( X = A^{-1}B \)
Now, \( A_{11} = 9 \), \( A_{12} = 1 \), \( A_{13} = -7 \),
\( A_{21} = -3 \), \( A_{22} = 0 \), \( A_{23} = 3 \),
\( A_{31} = 0 \), \( A_{32} = -1 \), \( A_{33} = 1 \)
\( \text{adj } A = \begin{bmatrix} 9 & -3 & 0 \\ 1 & 0 & -1 \\ -7 & 3 & 1 \end{bmatrix} \)
\( A^{-1} = \frac{1}{|A|}(\text{adj } A) = \frac{1}{3} \begin{bmatrix} 9 & -3 & 0 \\ 1 & 0 & -1 \\ -7 & 3 & 1 \end{bmatrix} \)
Now, \( X = A^{-1} B \)
\( \Rightarrow \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \frac{1}{3} \begin{bmatrix} 9 & -3 & 0 \\ 1 & 0 & -1 \\ -7 & 3 & 1 \end{bmatrix} \begin{bmatrix} 12 \\ 33 \\ 0 \end{bmatrix} = \frac{1}{3} \begin{bmatrix} 9 \\ 12 \\ 15 \end{bmatrix} = \begin{bmatrix} 3 \\ 4 \\ 5 \end{bmatrix} \)
\( \Rightarrow x = 3, y = 4, z = 5 \).
The management of the colony can include the awards for those members of the colony who help for keeping the environment of the colony free from pollution (or discipline).

Question. Two institutions decided to award their employees for the three values of resourcefulness, competence and determination in the form of prizes at the rate of ₹ \(x\), ₹ \(y\) and ₹ \(z\) respectively per person. The first institution decided to award respectively 4, 3 and 2 employees with a total prize money of ₹ 37,000 and the second institution decided to award respectively 5, 3 and 4 employees with a total prize money of ₹ 47,000. If all the three prizes per person together amount to ₹ 12,000 then using matrix method find the value of \(x\), \(y\) and \(z\). What values are described in the question? (Delhi 2013C)
Answer: According to the question, we have
\( x + y + z = 12000 \)
\( 4x + 3y + 2z = 37000 \)
\( 5x + 3y + 4z = 47000 \)
The system of equations can be written as \( AX = B \)
where, \( A = \begin{bmatrix} 1 & 1 & 1 \\ 4 & 3 & 2 \\ 5 & 3 & 4 \end{bmatrix} \); \( X = \begin{bmatrix} x \\ y \\ z \end{bmatrix} \); \( B = \begin{bmatrix} 12000 \\ 37000 \\ 47000 \end{bmatrix} \)
\( |A| = \begin{vmatrix} 1 & 1 & 1 \\ 4 & 3 & 2 \\ 5 & 3 & 4 \end{vmatrix} = 1(12 - 6) - 1(16 - 10) + 1(12 - 15) = 6 - 6 - 3 = -3 \neq 0 \)
Since \( A^{-1} \) exists, so system of equations has a unique solution and it is given by \( X = A^{-1}B \)
Now, \( A_{11} = 6 \), \( A_{12} = -6 \), \( A_{13} = -3 \),
\( A_{21} = -1 \), \( A_{22} = -1 \), \( A_{23} = 2 \),
\( A_{31} = -1 \), \( A_{32} = 2 \), \( A_{33} = -1 \)
\( \text{adj } A = \begin{bmatrix} 6 & -1 & -1 \\ -6 & -1 & 2 \\ -3 & 2 & -1 \end{bmatrix} \)
\( A^{-1} = \frac{1}{|A|}(\text{adj } A) = -\frac{1}{3} \begin{bmatrix} 6 & -1 & -1 \\ -6 & -1 & 2 \\ -3 & 2 & -1 \end{bmatrix} = \frac{1}{3} \begin{bmatrix} -6 & 1 & 1 \\ 6 & 1 & -2 \\ 3 & -2 & 1 \end{bmatrix} \)
Now, \( X = A^{-1} B \)
\( \Rightarrow \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \frac{1}{3} \begin{bmatrix} -6 & 1 & 1 \\ 6 & 1 & -2 \\ 3 & -2 & 1 \end{bmatrix} \begin{bmatrix} 12000 \\ 37000 \\ 47000 \end{bmatrix} = -\frac{1}{3} \begin{bmatrix} -12000 \\ -15000 \\ -9000 \end{bmatrix} = \begin{bmatrix} 4000 \\ 5000 \\ 3000 \end{bmatrix} \)
\( \Rightarrow x = 4000, y = 5000, z = 3000 \).
The values described in this question are resourcefulness, competence and determination.

Question. Two factories decided to award their employees for three values of (a) adaptable to new techniques, (b) careful and alert in difficult situations and (c) keeping calm in tense situations, at the rate of ₹ \(x\), ₹ \(y\) and ₹ \(z\) per person respectively. The first factory decided to honour respectively 2, 4 and 3 employees with a total prize money of ₹ 29,000. The second factory decided to honour respectively 5, 2 and 3 employees with the prize money of ₹ 30,500. If the three prizes per person together cost ₹ 9,500; then
(i) Represent the above situation by a matrix equation and form linear equations using matrix multiplication.
(ii) Solve these equations using matrices.
(iii) Which values are reflected in this question? (AI 2013C)
Answer: (i) According to the question, we have
\( x + y + z = 9500 \)
\( 2x + 4y + 3z = 29000 \)
\( 5x + 2y + 3z = 30500 \)
The system of equations can be written as \( AX = B \)
where, \( A = \begin{bmatrix} 1 & 1 & 1 \\ 2 & 4 & 3 \\ 5 & 2 & 3 \end{bmatrix} \); \( X = \begin{bmatrix} x \\ y \\ z \end{bmatrix} \); \( B = \begin{bmatrix} 9500 \\ 29000 \\ 30500 \end{bmatrix} \)

(ii) \( |A| = \begin{vmatrix} 1 & 1 & 1 \\ 2 & 4 & 3 \\ 5 & 2 & 3 \end{vmatrix} = 1(12 - 6) - 1(6 - 15) + 1(4 - 20) = 6 + 9 - 16 = -1 \neq 0 \)
Since \( A^{-1} \) exists, so system of equations has a unique solution and it is given by \( X = A^{-1}B \)
Now, \( A_{11} = 6 \), \( A_{12} = 9 \), \( A_{13} = -16 \),
\( A_{21} = -1 \), \( A_{22} = -2 \), \( A_{23} = 3 \),
\( A_{31} = -1 \), \( A_{32} = -1 \), \( A_{33} = 2 \)
\( \text{adj } A = \begin{bmatrix} 6 & -1 & -1 \\ 9 & -2 & -1 \\ -16 & 3 & 2 \end{bmatrix} \)
\( A^{-1} = \frac{1}{|A|} \text{adj } A = \frac{1}{-1} \begin{bmatrix} 6 & -1 & -1 \\ 9 & -2 & -1 \\ -16 & 3 & 2 \end{bmatrix} = \begin{bmatrix} -6 & 1 & 1 \\ -9 & 2 & 1 \\ 16 & -3 & -2 \end{bmatrix} \)
Now, \( X = A^{-1}B \)
\( \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} -6 & 1 & 1 \\ -9 & 2 & 1 \\ 16 & -3 & -2 \end{bmatrix} \begin{bmatrix} 9500 \\ 29000 \\ 30500 \end{bmatrix} = \begin{bmatrix} 2500 \\ 3000 \\ 4000 \end{bmatrix} \)
\( \Rightarrow x = 2500, y = 3000, z = 4000 \).

(iii) The factories honour the most, those employees who are keeping calm in tense situations.

Question. Using matrices, solve the following system of linear equations :
\( x - y + 2z = 7 \), \( 3x + 4y - 5z = -5 \)
\( 2x - y + 3z = 12 \) (Delhi 2012)
Answer: We have,
\( x - y + 2z = 7 \)
\( 3x + 4y - 5z = -5 \)
\( 2x - y + 3z = 12 \)
The given system of equations can be written as \( AX = B \)
where, \( A = \begin{bmatrix} 1 & -1 & 2 \\ 3 & 4 & -5 \\ 2 & -1 & 3 \end{bmatrix} \), \( X = \begin{bmatrix} x \\ y \\ z \end{bmatrix} \), \( B = \begin{bmatrix} 7 \\ -5 \\ 12 \end{bmatrix} \)
Here \( |A| = \begin{vmatrix} 1 & -1 & 2 \\ 3 & 4 & -5 \\ 2 & -1 & 3 \end{vmatrix} = 1(12 - 5) - 3(-3 + 2) + 2(5 - 8) = 7 + 3 - 6 = 4 \neq 0 \).
Since \( A^{-1} \) exists, so system of equations has a unique solution given by \( X = A^{-1}B \)
Now, \( A_{11} = 7 \), \( A_{12} = -19 \), \( A_{13} = -11 \), \( A_{21} = 1 \), \( A_{22} = -1 \), \( A_{23} = -1 \), \( A_{31} = -3 \), \( A_{32} = 11 \), \( A_{33} = 7 \)
\( \text{adj } A = \begin{bmatrix} 7 & 1 & -3 \\ -19 & -1 & 11 \\ -11 & -1 & 7 \end{bmatrix} \)
\( A^{-1} = \frac{1}{|A|}(\text{adj } A) = \frac{1}{4} \begin{bmatrix} 7 & 1 & -3 \\ -19 & -1 & 11 \\ -11 & -1 & 7 \end{bmatrix} \)
Now, \( X = A^{-1}B \)
\( \Rightarrow \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \frac{1}{4} \begin{bmatrix} 7 & 1 & -3 \\ -19 & -1 & 11 \\ -11 & -1 & 7 \end{bmatrix} \begin{bmatrix} 7 \\ -5 \\ 12 \end{bmatrix} = \frac{1}{4} \begin{bmatrix} 8 \\ 4 \\ 12 \end{bmatrix} = \begin{bmatrix} 2 \\ 1 \\ 3 \end{bmatrix} \)
\( \Rightarrow x = 2, y = 1, z = 3 \)

Question. Using matrices, solve the following system of equations :
\( 2x + 3y + 3z = 5 \), \( x - 2y + z = -4 \),
\( 3x - y - 2z = 3 \) (AI 2012, Delhi 2007)
Answer: We have,
\( 2x + 3y + 3z = 5 \)
\( x - 2y + z = -4 \)
\( 3x - y - 2z = 3 \)
The given system of equations can be written as \( AX = B \)
where, \( A = \begin{bmatrix} 2 & 3 & 3 \\ 1 & -2 & 1 \\ 3 & -1 & -2 \end{bmatrix} \); \( X = \begin{bmatrix} x \\ y \\ z \end{bmatrix} \), and \( B = \begin{bmatrix} 5 \\ -4 \\ 3 \end{bmatrix} \)
Now, \( |A| = \begin{vmatrix} 2 & 3 & 3 \\ 1 & -2 & 1 \\ 3 & -1 & -2 \end{vmatrix} = 2(4 + 1) - 3(-2 - 3) + 3(-1 + 6) = 10 + 15 + 15 = 40 \neq 0 \)
Since \( A^{-1} \) exists, so system of equations has a unique solution and it is given by \( X = A^{-1}B \)
Now, \( A_{11} = 5 \), \( A_{12} = 5 \), \( A_{13} = 5 \), \( A_{21} = 3 \), \( A_{22} = -13 \), \( A_{23} = 11 \), \( A_{31} = 9 \), \( A_{32} = 1 \), \( A_{33} = -7 \)
\( \text{adj } A = \begin{bmatrix} 5 & 3 & 9 \\ 5 & -13 & 1 \\ 5 & 11 & -7 \end{bmatrix} \)
\( A^{-1} = \frac{1}{|A|} \text{adj } A = \frac{1}{40} \begin{bmatrix} 5 & 3 & 9 \\ 5 & -13 & 1 \\ 5 & 11 & -7 \end{bmatrix} \)
Now, \( X = A^{-1}B = \frac{1}{40} \begin{bmatrix} 5 & 3 & 9 \\ 5 & -13 & 1 \\ 5 & 11 & -7 \end{bmatrix} \begin{bmatrix} 5 \\ -4 \\ 3 \end{bmatrix} = \frac{1}{40} \begin{bmatrix} 40 \\ 80 \\ -40 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ -1 \end{bmatrix} \)
\( \Rightarrow x = 1, y = 2, z = -1 \)

Question. Using matrices, solve the following system of equations :
\( 3x + 4y + 7z = 4 \); \( 2x - y + 3z = -3 \);
\( x + 2y - 3z = 8 \) (AI 2012)
Answer: We have,
\( 3x + 4y + 7z = 4 \)
\( 2x - y + 3z = -3 \)
\( x + 2y - 3z = 8 \)
The system of equations can be written as \( AX = B \)
where, \( A = \begin{bmatrix} 3 & 4 & 7 \\ 2 & -1 & 3 \\ 1 & 2 & -3 \end{bmatrix} \); \( X = \begin{bmatrix} x \\ y \\ z \end{bmatrix} \), and \( B = \begin{bmatrix} 4 \\ -3 \\ 8 \end{bmatrix} \)
Now, \( |A| = \begin{vmatrix} 3 & 4 & 7 \\ 2 & -1 & 3 \\ 1 & 2 & -3 \end{vmatrix} = 3(3 - 6) - 4(-6 - 3) + 7(4 + 1) = -9 + 36 + 35 = 62 \neq 0 \)
Since \( A^{-1} \) exists, so system of equations has a unique solution given by \( X = A^{-1}B \)
Now, \( A_{11} = -3 \), \( A_{12} = 9 \), \( A_{13} = 5 \), \( A_{21} = 26 \), \( A_{22} = -16 \), \( A_{23} = -2 \), \( A_{31} = 19 \), \( A_{32} = 5 \), \( A_{33} = -11 \)
\( \text{adj } A = \begin{bmatrix} -3 & 26 & 19 \\ 9 & -16 & 5 \\ 5 & -2 & -11 \end{bmatrix} \)
\( A^{-1} = \frac{1}{|A|} \cdot \text{adj } A = \frac{1}{62} \begin{bmatrix} -3 & 26 & 19 \\ 9 & -16 & 5 \\ 5 & -2 & -11 \end{bmatrix} \)
Now, \( X = A^{-1}B = \frac{1}{62} \begin{bmatrix} -3 & 26 & 19 \\ 9 & -16 & 5 \\ 5 & -2 & -11 \end{bmatrix} \begin{bmatrix} 4 \\ -3 \\ 8 \end{bmatrix} = \frac{1}{62} \begin{bmatrix} 62 \\ 124 \\ -62 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ -1 \end{bmatrix} \)
\( \Rightarrow x = 1, y = 2, z = -1 \)

Question. Using matrices, solve the following system of equations :
\( x + y - z = 3 \); \( 2x + 3y + z = 10 \);
\( 3x - y - 7z = 1 \) (AI 2012)
Answer: We have,
\( x + y - z = 3 \)
\( 2x + 3y + z = 10 \)
\( 3x - y - 7z = 1 \)
The system of equations can be written as \( AX = B \)
where, \( A = \begin{bmatrix} 1 & 1 & -1 \\ 2 & 3 & 1 \\ 3 & -1 & -7 \end{bmatrix} \); \( X = \begin{bmatrix} x \\ y \\ z \end{bmatrix} \), and \( B = \begin{bmatrix} 3 \\ 10 \\ 1 \end{bmatrix} \)
Now, \( |A| = \begin{vmatrix} 1 & 1 & -1 \\ 2 & 3 & 1 \\ 3 & -1 & -7 \end{vmatrix} = 1(-21 + 1) - 1(-14 - 3) - 1(-2 - 9) = -20 + 17 + 11 = 8 \neq 0 \)
Since \( A^{-1} \) exists, so system of equation has a unique solution given by \( X = A^{-1}B \)
Now, \( A_{11} = -20 \), \( A_{12} = 17 \), \( A_{13} = -11 \), \( A_{21} = 8 \), \( A_{22} = -4 \), \( A_{23} = 4 \), \( A_{31} = 4 \), \( A_{32} = -3 \), \( A_{33} = 1 \)
\( \text{adj } A = \begin{bmatrix} -20 & 8 & 4 \\ 17 & -4 & -3 \\ -11 & 4 & 1 \end{bmatrix} \)
\( A^{-1} = \frac{1}{|A|} \cdot \text{adj } A = \frac{1}{8} \begin{bmatrix} -20 & 8 & 4 \\ 17 & -4 & -3 \\ -11 & 4 & 1 \end{bmatrix} \)
Now, \( X = A^{-1} B = \frac{1}{8} \begin{bmatrix} -20 & 8 & 4 \\ 17 & -4 & -3 \\ -11 & 4 & 1 \end{bmatrix} \begin{bmatrix} 3 \\ 10 \\ 1 \end{bmatrix} = \frac{1}{8} \begin{bmatrix} 24 \\ 8 \\ 8 \end{bmatrix} = \begin{bmatrix} 3 \\ 1 \\ 1 \end{bmatrix} \)
\( \Rightarrow x = 3, y = 1, z = 1 \)

Question. If \( A = \begin{bmatrix} 1 & 2 & 1 \\ -1 & 1 & 1 \\ 1 & -3 & 1 \end{bmatrix} \), find \( A^{-1} \) and hence solve the system of equations :
\( x + 2y + z = 4 \), \( -x + y + z = 0 \),
\( x - 3y + z = 4 \) (Delhi 2012C)
Answer: Here, \( A = \begin{bmatrix} 1 & 2 & 1 \\ -1 & 1 & 1 \\ 1 & -3 & 1 \end{bmatrix} \)
\( \Rightarrow |A| = \begin{vmatrix} 1 & 2 & 1 \\ -1 & 1 & 1 \\ 1 & -3 & 1 \end{vmatrix} = 1(1 + 3) - 2(-1 - 1) + 1(3 - 1) = 10 \neq 0 \)
Since \( A^{-1} \) exists.
Now, \( A_{11} = 4 \), \( A_{12} = 2 \), \( A_{13} = 2 \), \( A_{21} = -5 \), \( A_{22} = 0 \), \( A_{23} = 5 \), \( A_{31} = 1 \), \( A_{32} = -2 \), \( A_{33} = 3 \)
\( \text{adj } A = \begin{bmatrix} 4 & -5 & 1 \\ 2 & 0 & -2 \\ 2 & 5 & 3 \end{bmatrix} \)
\( A^{-1} = \frac{1}{|A|} \cdot \text{adj } A = \frac{1}{10} \begin{bmatrix} 4 & -5 & 1 \\ 2 & 0 & -2 \\ 2 & 5 & 3 \end{bmatrix} \)
The given system of equations is:
\( x + 2y + z = 4 \)
\( -x + y + z = 0 \)
\( x - 3y + z = 4 \)
The system of equations can be written as \( AX = B \)
where, \( A = \begin{bmatrix} 1 & 2 & 1 \\ -1 & 1 & 1 \\ 1 & -3 & 1 \end{bmatrix} \); \( X = \begin{bmatrix} x \\ y \\ z \end{bmatrix} \); \( B = \begin{bmatrix} 4 \\ 0 \\ 4 \end{bmatrix} \);
Since \( A^{-1} \) exists, therefore, system of equations has a unique solution given by:
\( X = A^{-1} B = \frac{1}{10} \begin{bmatrix} 4 & -5 & 1 \\ 2 & 0 & -2 \\ 2 & 5 & 3 \end{bmatrix} \begin{bmatrix} 4 \\ 0 \\ 4 \end{bmatrix} = \frac{1}{10} \begin{bmatrix} 20 \\ 0 \\ 20 \end{bmatrix} = \begin{bmatrix} 2 \\ 0 \\ 2 \end{bmatrix} \)
\( \Rightarrow x = 2, y = 0, z = 2 \)

Question. Determine the product \( \begin{bmatrix} -4 & 4 & 4 \\ -7 & 1 & 3 \\ 5 & -3 & -1 \end{bmatrix} \begin{bmatrix} 1 & -1 & 1 \\ 1 & -2 & -2 \\ 2 & 1 & 3 \end{bmatrix} \) and use it to solve the system of equations :
\( x - y + z = 4 \); \( x - 2y - 2z = 9 \);
\( 2x + y + 3z = 1 \) (Delhi 2012C, 2010C)
Answer: We have,
\( \begin{bmatrix} -4 & 4 & 4 \\ -7 & 1 & 3 \\ 5 & -3 & -1 \end{bmatrix} \begin{bmatrix} 1 & -1 & 1 \\ 1 & -2 & -2 \\ 2 & 1 & 3 \end{bmatrix} = \begin{bmatrix} 8 & 0 & 0 \\ 0 & 8 & 0 \\ 0 & 0 & 8 \end{bmatrix} = 8 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = 8I \)
\( \Rightarrow \frac{1}{8} \begin{bmatrix} -4 & 4 & 4 \\ -7 & 1 & 3 \\ 5 & -3 & -1 \end{bmatrix} \begin{bmatrix} 1 & -1 & 1 \\ 1 & -2 & -2 \\ 2 & 1 & 3 \end{bmatrix} = I \)
\( \Rightarrow \begin{bmatrix} 1 & -1 & 1 \\ 1 & -2 & -2 \\ 2 & 1 & 3 \end{bmatrix}^{-1} = \frac{1}{8} \begin{bmatrix} -4 & 4 & 4 \\ -7 & 1 & 3 \\ 5 & -3 & -1 \end{bmatrix} \)
Now the given system of equations is:
\( x - y + z = 4 \)
\( x - 2y - 2z = 9 \)
\( 2x + y + 3z = 1 \)
The system of equations can be written as \( AX = B \)
where, \( A = \begin{bmatrix} 1 & -1 & 1 \\ 1 & -2 & -2 \\ 2 & 1 & 3 \end{bmatrix} \); \( X = \begin{bmatrix} x \\ y \\ z \end{bmatrix} \); \( B = \begin{bmatrix} 4 \\ 9 \\ 1 \end{bmatrix} \);
Since \( A^{-1} \) exists, so system of equations has a unique solution given by:
\( X = A^{-1} B \)
\( \Rightarrow \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \frac{1}{8} \begin{bmatrix} -4 & 4 & 4 \\ -7 & 1 & 3 \\ 5 & -3 & -1 \end{bmatrix} \begin{bmatrix} 4 \\ 9 \\ 1 \end{bmatrix} = \frac{1}{8} \begin{bmatrix} 24 \\ -16 \\ -8 \end{bmatrix} = \begin{bmatrix} 3 \\ -2 \\ -1 \end{bmatrix} \)
\( \Rightarrow x = 3, y = -2, z = -1 \)

Question. Find \( A^{-1} \), where \( A = \begin{bmatrix} 1 & 2 & -3 \\ 2 & 3 & 2 \\ 3 & -3 & -4 \end{bmatrix} \).
Hence solve the system of equations
\( x + 2y - 3z = -4 \); \( 2x + 3y + 2z = 2 \);
\( 3x - 3y - 4z = 11 \). (Delhi 2012C)
Answer: Here, \( A = \begin{bmatrix} 1 & 2 & -3 \\ 2 & 3 & 2 \\ 3 & -3 & -4 \end{bmatrix} \)
\( \Rightarrow |A| = \begin{vmatrix} 1 & 2 & -3 \\ 2 & 3 & 2 \\ 3 & -3 & -4 \end{vmatrix} = 1(-12 + 6) - 2(-8 - 6) - 3(-6 - 9) = -6 + 28 + 45 = 67 \neq 0 \)
Since \( A^{-1} \) exists.
Now, \( A_{11} = -6 \), \( A_{12} = 14 \), \( A_{13} = -15 \), \( A_{21} = 17 \), \( A_{22} = 5 \), \( A_{23} = 9 \), \( A_{31} = 13 \), \( A_{32} = -8 \), \( A_{33} = -1 \)
\( \text{adj } A = \begin{bmatrix} -6 & 17 & 13 \\ 14 & 5 & -8 \\ -15 & 9 & -1 \end{bmatrix} \)
\( A^{-1} = \frac{1}{|A|} \cdot \text{adj } A = \frac{1}{67} \begin{bmatrix} -6 & 17 & 13 \\ 14 & 5 & -8 \\ -15 & 9 & -1 \end{bmatrix} \)
The given system of equations is:
\( x + 2y - 3z = -4 \)
\( 2x + 3y + 2z = 2 \)
\( 3x - 3y - 4z = 11 \)
The system of equations can be written as \( AX = B \)
where, \( A = \begin{bmatrix} 1 & 2 & -3 \\ 2 & 3 & 2 \\ 3 & -3 & -4 \end{bmatrix} \); \( X = \begin{bmatrix} x \\ y \\ z \end{bmatrix} \); \( B = \begin{bmatrix} -4 \\ 2 \\ 11 \end{bmatrix} \);
Since \( A^{-1} \) exists, so system of equations has a unique solution given by \( X = A^{-1} B \)
\( \Rightarrow \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \frac{1}{67} \begin{bmatrix} -6 & 17 & 13 \\ 14 & 5 & -8 \\ -15 & 9 & -1 \end{bmatrix} \begin{bmatrix} -4 \\ 2 \\ 11 \end{bmatrix} = \frac{1}{67} \begin{bmatrix} 201 \\ -134 \\ 67 \end{bmatrix} = \begin{bmatrix} 3 \\ -2 \\ 1 \end{bmatrix} \)
\( \Rightarrow x = 3, y = -2, z = 1 \)

Question. If \( A = \begin{bmatrix} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{bmatrix} \) and \( B = \begin{bmatrix} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{bmatrix} \) are two square matrices, find \( AB \) and hence solve the system of equations
\( x - y = 3 \), \( 2x + 3y + 4z = 17 \); and \( y + 2z = 7 \). (AI 2012C, 2010C)
Answer: Here, \( A = \begin{bmatrix} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{bmatrix} \); \( B = \begin{bmatrix} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{bmatrix} \)
\( \Rightarrow AB = \begin{bmatrix} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{bmatrix} \begin{bmatrix} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{bmatrix} = \begin{bmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{bmatrix} = 6I \)
\( \Rightarrow A \left(\frac{1}{6} B\right) = I \Rightarrow A \) is invertible and \( A^{-1} = \frac{1}{6} B \)
Now the given system of equations is:
\( x - y = 3 \)
\( 2x + 3y + 4z = 17 \)
\( y + 2z = 7 \)
The system of equations can be written as \( AX = P \)
where, \( A = \begin{bmatrix} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{bmatrix} \); \( X = \begin{bmatrix} x \\ y \\ z \end{bmatrix} \); \( P = \begin{bmatrix} 3 \\ 17 \\ 7 \end{bmatrix} \);
Since \( A^{-1} \) exists, so system of equations has a unique solution given by \( X = A^{-1} P \)
\( \Rightarrow \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \frac{1}{6} BP = \frac{1}{6} \begin{bmatrix} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{bmatrix} \begin{bmatrix} 3 \\ 17 \\ 7 \end{bmatrix} = \frac{1}{6} \begin{bmatrix} 12 \\ -6 \\ 24 \end{bmatrix} = \begin{bmatrix} 2 \\ -1 \\ 4 \end{bmatrix} \)
\( \Rightarrow x = 2, y = -1, z = 4 \)

Question. If \( A = \begin{bmatrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{bmatrix} \), find \( A^{-1} \) and hence solve the system of equations
\( 2x - 3y + 5z = 11 \); \( 3x + 2y - 4z = -5 \);
\( x + y - 2z = -3 \) (AI 2012C)
Answer: Here, \( A = \begin{bmatrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{bmatrix} \)
\( \Rightarrow |A| = \begin{vmatrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{vmatrix} = 2(-4 + 4) + 3(-6 + 4) + 5(3 - 2) = -6 + 5 = -1 \neq 0 \)
Since \( A^{-1} \) exists.
Now, \( A_{11} = 0 \), \( A_{12} = 2 \), \( A_{13} = 1 \), \( A_{21} = -1 \), \( A_{22} = -9 \), \( A_{23} = -5 \), \( A_{31} = 2 \), \( A_{32} = 23 \), \( A_{33} = 13 \)
\( \text{adj } A = \begin{bmatrix} 0 & -1 & 2 \\ 2 & -9 & 23 \\ 1 & -5 & 13 \end{bmatrix} \)
\( A^{-1} = \frac{1}{|A|} \cdot \text{adj } A = -1 \begin{bmatrix} 0 & -1 & 2 \\ 2 & -9 & 23 \\ 1 & -5 & 13 \end{bmatrix} = \begin{bmatrix} 0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13 \end{bmatrix} \)
The given system of equations is:
\( 2x - 3y + 5z = 11 \)
\( 3x + 2y - 4z = -5 \)
\( x + y - 2z = -3 \)
The system of equations can be written as \( AX = B \)
where, \( A = \begin{bmatrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{bmatrix} \); \( X = \begin{bmatrix} x \\ y \\ z \end{bmatrix} \); \( B = \begin{bmatrix} 11 \\ -5 \\ -3 \end{bmatrix} \);
Since \( A^{-1} \) exists. So, system of equations has unique solution given by \( X = A^{-1} B \)
\( \Rightarrow \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13 \end{bmatrix} \begin{bmatrix} 11 \\ -5 \\ -3 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} \)
\( \Rightarrow x = 1, y = 2, z = 3 \)

Question. If \( A = \begin{bmatrix} 1 & 2 & 5 \\ 1 & -1 & -1 \\ 2 & 3 & -1 \end{bmatrix} \), find \( A^{-1} \). Hence solve the following system of equations :
\( x + 2y + 5z = 10 \),
\( x - y - z = -2 \),
\( 2x + 3y - z = -11 \)
Answer: Here, \( A = \begin{bmatrix} 1 & 2 & 5 \\ 1 & -1 & -1 \\ 2 & 3 & -1 \end{bmatrix} \)
\( \Rightarrow |A| = \begin{vmatrix} 1 & 2 & 5 \\ 1 & -1 & -1 \\ 2 & 3 & -1 \end{vmatrix} = 1(1 + 3) - 2(-1 + 2) + 5(3 + 2) = 4 - 2 + 25 = 27 \neq 0 \)
\( \therefore A^{-1} \) exists.
Now, \( A_{11} = 4, A_{12} = -1, A_{13} = 5, A_{21} = 17, A_{22} = -11, A_{23} = 1, A_{31} = 3, A_{32} = 6, A_{33} = -3 \)
\( \dots \text{adj } A = \begin{bmatrix} 4 & 17 & 3 \\ -1 & -11 & 6 \\ 5 & 1 & -3 \end{bmatrix} \)
\( A^{-1} = \frac{1}{|A|} \text{adj } A = \frac{1}{27} \begin{bmatrix} 4 & 17 & 3 \\ -1 & -11 & 6 \\ 5 & 1 & -3 \end{bmatrix} \)
The given system of equations is
\( x + 2y + 5z = 10 \)
\( x - y - z = -2 \)
\( 2x + 3y - z = -11 \)
The given system of equations can be written as \( AX = B \)
where, \( A = \begin{bmatrix} 1 & 2 & 5 \\ 1 & -1 & -1 \\ 2 & 3 & -1 \end{bmatrix} \); \( X = \begin{bmatrix} x \\ y \\ z \end{bmatrix} \) and \( B = \begin{bmatrix} 10 \\ -2 \\ -11 \end{bmatrix} \)
\( \therefore A^{-1} \) exists. So, system of equations has a unique solution given by \( X = A^{-1} B \)
\( \Rightarrow \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \frac{1}{27} \begin{bmatrix} 4 & 17 & 3 \\ -1 & -11 & 6 \\ 5 & 1 & -3 \end{bmatrix} \begin{bmatrix} 10 \\ -2 \\ -11 \end{bmatrix} \)
\( \Rightarrow \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \frac{1}{27} \begin{bmatrix} -27 \\ -54 \\ 81 \end{bmatrix} = \begin{bmatrix} -1 \\ -2 \\ 3 \end{bmatrix} \)
\( \Rightarrow x = -1, y = -2, z = 3 \)

Question. Using matrix method, solve the following system of equations :
\( \frac{2}{x} + \frac{3}{y} + \frac{10}{z} = 4 \),
\( \frac{4}{x} - \frac{6}{y} + \frac{5}{z} = 1 \),
\( \frac{6}{x} + \frac{9}{y} - \frac{20}{z} = 2 \);
\( x, y, z \neq 0 \)
Answer: The given equations are
\( \frac{2}{x} + \frac{3}{y} + \frac{10}{z} = 4 \), \( \frac{4}{x} - \frac{6}{y} + \frac{5}{z} = 1 \), \( \frac{6}{x} + \frac{9}{y} - \frac{20}{z} = 2 \)
Given equations can be written as \( AX = B \)
where, \( A = \begin{bmatrix} 2 & 3 & 10 \\ 4 & -6 & 5 \\ 6 & 9 & -20 \end{bmatrix} \), \( X = \begin{bmatrix} 1/x \\ 1/y \\ 1/z \end{bmatrix} \) and \( B = \begin{bmatrix} 4 \\ 1 \\ 2 \end{bmatrix} \)
\( |A| = \begin{vmatrix} 2 & 3 & 10 \\ 4 & -6 & 5 \\ 6 & 9 & -20 \end{vmatrix} = [2(120 - 45) - 3(-80 - 30) + 10(36 + 36)] = [150 + 330 + 720] = 1200 \neq 0 \)
Since \( A^{-1} \) exists, therefore system of equations has a unique solution given by \( X = A^{-1} B \)
\( A_{11} = 75, A_{12} = 110, A_{13} = 72, A_{21} = 150, A_{22} = -100, A_{23} = 0, A_{31} = 75, A_{32} = 30, A_{33} = -24 \)
\( \therefore \text{adj } A = \begin{bmatrix} 75 & 150 & 75 \\ 110 & -100 & 30 \\ 72 & 0 & -24 \end{bmatrix} \)
\( A^{-1} = \frac{1}{|A|} (\text{adj } A) = \frac{1}{1200} \begin{bmatrix} 75 & 150 & 75 \\ 110 & -100 & 30 \\ 72 & 0 & -24 \end{bmatrix} \)
Now, \( X = A^{-1} B \).
\( \Rightarrow \begin{bmatrix} 1/x \\ 1/y \\ 1/z \end{bmatrix} = \frac{1}{1200} \begin{bmatrix} 75 & 150 & 75 \\ 110 & -100 & 30 \\ 72 & 0 & -24 \end{bmatrix} \begin{bmatrix} 4 \\ 1 \\ 2 \end{bmatrix} = \frac{1}{1200} \begin{bmatrix} 600 \\ 400 \\ 240 \end{bmatrix} = \begin{bmatrix} 1/2 \\ 1/3 \\ 1/5 \end{bmatrix} \)
\( \Rightarrow \frac{1}{x} = \frac{1}{2} \Rightarrow x = 2 \), \( \frac{1}{y} = \frac{1}{3} \Rightarrow y = 3 \), \( \frac{1}{z} = \frac{1}{5} \Rightarrow z = 5 \)

Question. Using matrix method, solve the following system of equations:
\( x + 2y + z = 7 \),
\( x + 3z = 11 \),
\( 2x - 3y = 1 \)
Answer: The given equations are
\( x + 2y + z = 7 \)
\( x + 3z = 11 \)
\( 2x - 3y = 1 \)
The given system of equations can be written as \( AX = B \)
where, \( A = \begin{bmatrix} 1 & 2 & 1 \\ 1 & 0 & 3 \\ 2 & -3 & 0 \end{bmatrix} \), \( X = \begin{bmatrix} x \\ y \\ z \end{bmatrix} \), \( B = \begin{bmatrix} 7 \\ 11 \\ 1 \end{bmatrix} \)
Now, \( |A| = \begin{vmatrix} 1 & 2 & 1 \\ 1 & 0 & 3 \\ 2 & -3 & 0 \end{vmatrix} = 2(6 - 0) + 3(3 - 1) = 12 + 6 = 18 \neq 0 \)
\( \therefore A^{-1} \) exists. So, system of equations has a unique solution given by \( X = A^{-1} B \)
Now, \( A_{11} = 9, A_{12} = 6, A_{13} = -3, A_{21} = -3, A_{22} = -2, A_{23} = 7, A_{31} = 6, A_{32} = -2, A_{33} = -2 \)
\( \therefore \text{adj } A = \begin{bmatrix} 9 & -3 & 6 \\ 6 & -2 & -2 \\ -3 & 7 & -2 \end{bmatrix} \)
\( A^{-1} = \frac{1}{18} \text{adj } A = \frac{1}{18} \begin{bmatrix} 9 & -3 & 6 \\ 6 & -2 & -2 \\ -3 & 7 & -2 \end{bmatrix} \)
Now, \( X = A^{-1} B \)
\( \Rightarrow \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \frac{1}{18} \begin{bmatrix} 9 & -3 & 6 \\ 6 & -2 & -2 \\ -3 & 7 & -2 \end{bmatrix} \begin{bmatrix} 7 \\ 11 \\ 1 \end{bmatrix} = \frac{1}{18} \begin{bmatrix} 36 \\ 18 \\ 54 \end{bmatrix} = \begin{bmatrix} 2 \\ 1 \\ 3 \end{bmatrix} \)
\( \Rightarrow x = 2, y = 1, z = 3 \)

Question. Using matrices, solve the following system of equations :
\( 4x + 3y + 2z = 60 \),
\( x + 2y + 3z = 45 \),
\( 6x + 2y + 3z = 70 \)
Answer: The given system of equations are
\( 4x + 3y + 2z = 60 \),
\( x + 2y + 3z = 45 \),
\( 6x + 2y + 3z = 70 \)
The given system of equations can be written as \( AX = B \)
where, \( A = \begin{bmatrix} 4 & 3 & 2 \\ 1 & 2 & 3 \\ 6 & 2 & 3 \end{bmatrix} \), \( X = \begin{bmatrix} x \\ y \\ z \end{bmatrix} \), \( B = \begin{bmatrix} 60 \\ 45 \\ 70 \end{bmatrix} \)
Now, \( |A| = \begin{vmatrix} 4 & 3 & 2 \\ 1 & 2 & 3 \\ 6 & 2 & 3 \end{vmatrix} = 4(6 - 6) - 3(3 - 18) + 2(2 - 12) = 0 + 45 - 20 = 25 \neq 0 \)
\( \therefore A^{-1} \) exists. So system of equations has a unique solution \( X = A^{-1} B \)
Now,
\( A_{11} = 0, A_{12} = 15, A_{13} = -10, A_{21} = -5, A_{22} = 0, A_{23} = 10, A_{31} = 5, A_{32} = -10, A_{33} = 5 \)
\( \therefore \text{adj } A = \begin{bmatrix} 0 & -5 & 5 \\ 15 & 0 & -10 \\ -10 & 10 & 5 \end{bmatrix} \)
\( A^{-1} = \frac{1}{|A|} \text{adj } A = \frac{1}{25} \begin{bmatrix} 0 & -5 & 5 \\ 15 & 0 & -10 \\ -10 & 10 & 5 \end{bmatrix} \)
Now, \( X = A^{-1} B \)
\( \therefore X = \frac{1}{25} \begin{bmatrix} 0 & -5 & 5 \\ 15 & 0 & -10 \\ -10 & 10 & 5 \end{bmatrix} \begin{bmatrix} 60 \\ 45 \\ 70 \end{bmatrix} \)
\( \Rightarrow \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \frac{1}{25} \begin{bmatrix} 0 - 225 + 350 \\ 900 + 0 - 700 \\ -600 + 450 + 350 \end{bmatrix} = \frac{1}{25} \begin{bmatrix} 125 \\ 200 \\ 200 \end{bmatrix} = \begin{bmatrix} 5 \\ 8 \\ 8 \end{bmatrix} \)
\( \therefore x = 5, y = 8, z = 8 \)

Question. Using matrices, solve the following system of equations :
\( x + 2y - 3z = -4 \),
\( 2x + 3y + 2z = 2 \),
\( 3x - 3y - 4z = 11 \)
Answer: Refer to answer 136.

Question. If \( A = \begin{bmatrix} 3 & -4 & 2 \\ 2 & 3 & 5 \\ 1 & 0 & 1 \end{bmatrix} \), find \( A^{-1} \) and hence solve the following system of equations :
\( 3x - 4y + 2z = -1 \),
\( 2x + 3y + 5z = 7 \) and
\( x + z = 2 \)
Answer: Here, \( A = \begin{bmatrix} 3 & -4 & 2 \\ 2 & 3 & 5 \\ 1 & 0 & 1 \end{bmatrix} \)
\( \Rightarrow |A| = \begin{vmatrix} 3 & -4 & 2 \\ 2 & 3 & 5 \\ 1 & 0 & 1 \end{vmatrix} = 3(3 - 0) + 4(2 - 5) + 2(0 - 3) = 9 - 12 - 6 = -9 \neq 0 \)
\( \therefore A^{-1} \) exists. So, system of equations has a unique solution \( X = A^{-1} B \)
Now, \( A_{11} = 3, A_{12} = 3, A_{13} = -3, A_{21} = 4, A_{22} = 1, A_{23} = -4, A_{31} = -26, A_{32} = -11, A_{33} = 17 \)
\( \therefore \text{adj } A = \begin{bmatrix} 3 & 4 & -26 \\ 3 & 1 & -11 \\ -3 & -4 & 17 \end{bmatrix} \)
\( A^{-1} = \frac{1}{|A|} \text{adj } A = -\frac{1}{9} \begin{bmatrix} 3 & 4 & -26 \\ 3 & 1 & -11 \\ -3 & -4 & 17 \end{bmatrix} \)
The given system of equations is
\( 3x - 4y + 2z = -1 \),
\( 2x + 3y + 5z = 7 \),
\( x + z = 2 \)
The given system of equations can be written as \( AX = B \)
where, \( A = \begin{bmatrix} 3 & -4 & 2 \\ 2 & 3 & 5 \\ 1 & 0 & 1 \end{bmatrix} \), \( X = \begin{bmatrix} x \\ y \\ z \end{bmatrix} \), \( B = \begin{bmatrix} -1 \\ 7 \\ 2 \end{bmatrix} \)
Since \( A^{-1} \) exists, so, system of equations has a unique solution given by \( X = A^{-1} B \)
\( \Rightarrow \begin{bmatrix} x \\ y \\ z \end{bmatrix} = -\frac{1}{9} \begin{bmatrix} 3 & 4 & -26 \\ 3 & 1 & -11 \\ -3 & -4 & 17 \end{bmatrix} \begin{bmatrix} -1 \\ 7 \\ 2 \end{bmatrix} = -\frac{1}{9} \begin{bmatrix} -27 \\ -18 \\ 9 \end{bmatrix} = \begin{bmatrix} 3 \\ 2 \\ -1 \end{bmatrix} \)
\( \Rightarrow x = 3, y = 2, z = -1 \)

Question. If \( A = \begin{bmatrix} 1 & -2 & 0 \\ 2 & 1 & 3 \\ 0 & -2 & 1 \end{bmatrix} \) and \( B = \begin{bmatrix} 7 & 2 & -6 \\ -2 & 1 & -3 \\ -4 & 2 & 5 \end{bmatrix} \), find \( AB \). Hence solve the system of equations :
\( x - 2y = 10 \),
\( 2x + y + 3z = 8 \) and
\( -2y + z = 7 \)
Answer: Here, \( A = \begin{bmatrix} 1 & -2 & 0 \\ 2 & 1 & 3 \\ 0 & -2 & 1 \end{bmatrix} \); \( B = \begin{bmatrix} 7 & 2 & -6 \\ -2 & 1 & -3 \\ -4 & 2 & 5 \end{bmatrix} \)
\( \therefore AB = \begin{bmatrix} 1 & -2 & 0 \\ 2 & 1 & 3 \\ 0 & -2 & 1 \end{bmatrix} \begin{bmatrix} 7 & 2 & -6 \\ -2 & 1 & -3 \\ -4 & 2 & 5 \end{bmatrix} = \begin{bmatrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{bmatrix} = 11 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = 11 I \)
\( \Rightarrow A \left(\frac{1}{11} B\right) = I \Rightarrow A^{-1} = \frac{1}{11} B \)
Now the given system of equations is
\( x - 2y = 10 \)
\( 2x + y + 3z = 8 \)
\( -2y + z = 7 \)
The system of equations can be written as \( AX = P \)
where, \( A = \begin{bmatrix} 1 & -2 & 0 \\ 2 & 1 & 3 \\ 0 & -2 & 1 \end{bmatrix} \), \( X = \begin{bmatrix} x \\ y \\ z \end{bmatrix} \), \( P = \begin{bmatrix} 10 \\ 8 \\ 7 \end{bmatrix} \)
\( \therefore A^{-1} \) exists, so given system of equations has a unique solution given by \( X = A^{-1} P \)
\( \Rightarrow \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \frac{1}{11} BP = \frac{1}{11} \begin{bmatrix} 7 & 2 & -6 \\ -2 & 1 & -3 \\ -4 & 2 & 5 \end{bmatrix} \begin{bmatrix} 10 \\ 8 \\ 7 \end{bmatrix} = \frac{1}{11} \begin{bmatrix} 44 \\ -33 \\ 11 \end{bmatrix} = \begin{bmatrix} 4 \\ -3 \\ 1 \end{bmatrix} \)
\( \Rightarrow x = 4, y = -3, z = 1 \)

Question. If \( A = \begin{bmatrix} 1 & -2 & 1 \\ 0 & -1 & 1 \\ 2 & 0 & -3 \end{bmatrix} \), find \( A^{-1} \) and hence solve the system of equations :
\( x - 2y + z = 0 \),
\( -y + z = -2 \),
\( 2x - 3z = 10 \)
Answer: Here, \( A = \begin{bmatrix} 1 & -2 & 1 \\ 0 & -1 & 1 \\ 2 & 0 & -3 \end{bmatrix} \)
\( \Rightarrow |A| = \begin{vmatrix} 1 & -2 & 1 \\ 0 & -1 & 1 \\ 2 & 0 & -3 \end{vmatrix} = 1(3 - 0) + 2(-2 + 1) = 1 \neq 0 \)
\( \therefore A^{-1} \) exists.
Now, \( A_{11} = 3, A_{12} = 2, A_{13} = 2, A_{21} = -6, A_{22} = -5, A_{23} = -4, A_{31} = -1, A_{32} = -1, A_{33} = -1 \)
\( \therefore \text{adj } A = \begin{bmatrix} 3 & -6 & -1 \\ 2 & -5 & -1 \\ 2 & -4 & -1 \end{bmatrix} \)
\( A^{-1} = \frac{1}{|A|} \text{adj } A = \begin{bmatrix} 3 & -6 & -1 \\ 2 & -5 & -1 \\ 2 & -4 & -1 \end{bmatrix} \)
Now the given linear equations are
\( x - 2y + z = 0 \)
\( -y + z = -2 \)
\( 2x - 3z = 10 \)
The system of equations can be written as \( AX = B \)
where, \( A = \begin{bmatrix} 1 & -2 & 1 \\ 0 & -1 & 1 \\ 2 & 0 & -3 \end{bmatrix} \), \( X = \begin{bmatrix} x \\ y \\ z \end{bmatrix} \), \( B = \begin{bmatrix} 0 \\ -2 \\ 10 \end{bmatrix} \)
Since \( A^{-1} \) exists, so given system of equations has a unique solution given by \( X = A^{-1} B \)
\( \Rightarrow \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 3 & -6 & -1 \\ 2 & -5 & -1 \\ 2 & -4 & -1 \end{bmatrix} \begin{bmatrix} 0 \\ -2 \\ 10 \end{bmatrix} = \begin{bmatrix} 2 \\ 0 \\ -2 \end{bmatrix} \)
\( \Rightarrow x = 2, y = 0, z = -2 \)

Question. If \( A = \begin{bmatrix} 2 & -1 & 1 \\ 3 & 0 & -1 \\ 2 & 6 & 0 \end{bmatrix} \), find \( A^{-1} \). Using \( A^{-1} \), solve the following system of equations:
\( 2x - y + z = -3 \),
\( 3x - z = 0 \),
\( 2x + 6y - 2 = 0 \)
Answer: Here, \( A = \begin{bmatrix} 2 & -1 & 1 \\ 3 & 0 & -1 \\ 2 & 6 & 0 \end{bmatrix} \)
\( \Rightarrow |A| = \begin{vmatrix} 2 & -1 & 1 \\ 3 & 0 & -1 \\ 2 & 6 & 0 \end{vmatrix} = 2(0 + 6) + 1(0 + 2) + 1(18 - 0) = 12 + 2 + 18 = 32 \neq 0 \)
\( \therefore A^{-1} \) exists.
Now, \( A_{11} = 6, A_{12} = -2, A_{13} = 18, A_{21} = 6, A_{22} = -2, A_{23} = -14, A_{31} = 1, A_{32} = 5, A_{33} = 3 \)
\( \therefore \text{adj } A = \begin{bmatrix} 6 & 6 & 1 \\ -2 & -2 & 5 \\ 18 & -14 & 3 \end{bmatrix} \)
\( A^{-1} = \frac{1}{|A|} \text{adj } A = \frac{1}{32} \begin{bmatrix} 6 & 6 & 1 \\ -2 & -2 & 5 \\ 18 & -14 & 3 \end{bmatrix} \)
Now the given equations are
\( 2x - y + z = -3 \),
\( 3x - z = 0 \),
\( 2x + 6y - 2 = 0 \)
The given system of equations can be written as \( AX = B \)
where, \( A = \begin{bmatrix} 2 & -1 & 1 \\ 3 & 0 & -1 \\ 2 & 6 & 0 \end{bmatrix} \), \( X = \begin{bmatrix} x \\ y \\ z \end{bmatrix} \), \( B = \begin{bmatrix} -3 \\ 0 \\ 2 \end{bmatrix} \)
\( \therefore A^{-1} \) exists, so, the system has a unique solution given by \( X = A^{-1} B \)
\( \Rightarrow \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \frac{1}{32} \begin{bmatrix} 6 & 6 & 1 \\ -2 & -2 & 5 \\ 18 & -14 & 3 \end{bmatrix} \begin{bmatrix} -3 \\ 0 \\ 2 \end{bmatrix} = \frac{1}{32} \begin{bmatrix} -16 \\ 16 \\ -48 \end{bmatrix} = \begin{bmatrix} -1/2 \\ 1/2 \\ -3/2 \end{bmatrix} \)
\( \Rightarrow x = -\frac{1}{2}, y = \frac{1}{2}, z = -\frac{3}{2} \)

Question. If \( A = \begin{bmatrix} 3 & 2 & 1 \\ 4 & -1 & 2 \\ 7 & 3 & -3 \end{bmatrix} \), then find \( A^{-1} \). Hence, solve the following system of equations:
\( 3x + 2y + z = 6 \),
\( 4x - y + 2z = 5 \),
\( 7x + 3y - 3z = 7 \)
Answer: Here, \( A = \begin{bmatrix} 3 & 2 & 1 \\ 4 & -1 & 2 \\ 7 & 3 & -3 \end{bmatrix} \)
\( \therefore |A| = \begin{vmatrix} 3 & 2 & 1 \\ 4 & -1 & 2 \\ 7 & 3 & -3 \end{vmatrix} = 3(3 - 6) - 2(-12 - 14) + 1(12 + 7) = -9 + 52 + 19 = 62 \neq 0 \therefore A^{-1} \text{ exists} \)
Now,
\( A_{11} = -3, A_{12} = 26, A_{13} = 19, A_{21} = 9, A_{22} = -16, A_{23} = 5, A_{31} = 5, A_{32} = -2, A_{33} = -11 \)
\( \dots A^{-1} = \frac{1}{|A|} \text{adj } A = \frac{1}{62} \begin{bmatrix} -3 & 9 & 5 \\ 26 & -16 & -2 \\ 19 & 5 & -11 \end{bmatrix} \)
Now, the given equations are
\( 3x + 2y + z = 6 \)
\( 4x - y + 2z = 5 \)
\( 7x + 3y - 3z = 7 \)
Given equations can be written as
\( AX = B \)
where, \( A = \begin{bmatrix} 3 & 2 & 1 \\ 4 & -1 & 2 \\ 7 & 3 & -3 \end{bmatrix} \), \( X = \begin{bmatrix} x \\ y \\ z \end{bmatrix} \), \( B = \begin{bmatrix} 6 \\ 5 \\ 7 \end{bmatrix} \)
Since \( A^{-1} \) exists. So, system has a unique solution given by \( X = A^{-1} B \)
\( \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \frac{1}{62} \begin{bmatrix} -3 & 9 & 5 \\ 26 & -16 & -2 \\ 19 & 5 & -11 \end{bmatrix} \begin{bmatrix} 6 \\ 5 \\ 7 \end{bmatrix} = \frac{1}{62} \begin{bmatrix} -18 + 45 + 35 \\ 156 - 80 - 14 \\ 114 + 25 - 77 \end{bmatrix} = \frac{1}{62} \begin{bmatrix} 62 \\ 62 \\ 62 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} \)
\( \therefore x = 1, y = 1, z = 1 \)

Question. If \( A = \begin{bmatrix} 8 & -4 & 1 \\ 10 & 0 & 6 \\ 8 & 1 & 6 \end{bmatrix} \), find \( A^{-1} \). Hence solve the following system of equations :
\( 8x - 4y + z = 5 \),
\( 10x + 6z = 4 \),
\( 8x + y + 6z = \frac{5}{2} \)
Answer: We have, \( A = \begin{bmatrix} 8 & -4 & 1 \\ 10 & 0 & 6 \\ 8 & 1 & 6 \end{bmatrix} \)
\( \Rightarrow |A| = \begin{vmatrix} 8 & -4 & 1 \\ 10 & 0 & 6 \\ 8 & 1 & 6 \end{vmatrix} = 8(0 - 6) + 4(60 - 48) + 1(10 - 0) = -48 + 48 + 10 = 10 \neq 0 \)
\( \therefore A^{-1} \) exists.
Now, \( A_{11} = -6, A_{12} = -12, A_{13} = 10, A_{21} = 25, A_{22} = 40, A_{23} = -40, A_{31} = -24, A_{32} = -38, A_{33} = 40 \)
\( \therefore \text{adj } A = \begin{bmatrix} -6 & 25 & -24 \\ -12 & 40 & -38 \\ 10 & -40 & 40 \end{bmatrix} \)
\( A^{-1} = \frac{1}{|A|} \text{adj } A = \frac{1}{10} \begin{bmatrix} -6 & 25 & -24 \\ -12 & 40 & -38 \\ 10 & -40 & 40 \end{bmatrix} \)
Now the given equations are
\( 8x - 4y + z = 5 \)
\( 10x + 6z = 4 \)
\( 8x + y + 6z = \frac{5}{2} \)
Given system of equations can be written as \( AX = B \)
where, \( A = \begin{bmatrix} 8 & -4 & 1 \\ 10 & 0 & 6 \\ 8 & 1 & 6 \end{bmatrix} \), \( X = \begin{bmatrix} x \\ y \\ z \end{bmatrix} \), \( B = \begin{bmatrix} 5 \\ 4 \\ 5/2 \end{bmatrix} \)
Since \( A^{-1} \) exists. So, system of equations has a unique solution given by \( X = A^{-1} B \)
\( \Rightarrow \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \frac{1}{10} \begin{bmatrix} -6 & 25 & -24 \\ -12 & 40 & -38 \\ 10 & -40 & 40 \end{bmatrix} \begin{bmatrix} 5 \\ 4 \\ 5/2 \end{bmatrix} = \frac{1}{10} \begin{bmatrix} 10 \\ 5 \\ -10 \end{bmatrix} = \begin{bmatrix} 1 \\ 1/2 \\ -1 \end{bmatrix} \)
\( \Rightarrow x = 1, y = \frac{1}{2}, z = -1 \)

Question. Using matrices, solve the following system of equations :
\( 2x - 3y + 5z = 11 \),
\( 3x + 2y - 4z = -5 \),
\( x + y - 2z = -3 \)
Answer: Refer to answer 138.

Question. Using matrices, solve the following system of equations :
\( x + y + z = 6 \),
\( x + 2z = 7 \),
\( 3x + y + z = 12 \)
Answer: Given equations can be written as \( AX = B \Rightarrow X = A^{-1} B \)
where, \( A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 0 & 2 \\ 3 & 1 & 1 \end{bmatrix} \), \( X = \begin{bmatrix} x \\ y \\ z \end{bmatrix} \), \( B = \begin{bmatrix} 6 \\ 7 \\ 12 \end{bmatrix} \)
\( |A| = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 0 & 2 \\ 3 & 1 & 1 \end{vmatrix} = 1(0 - 2) - 1(1 - 6) + 1(1 - 0) = -2 + 5 + 1 = 4 \neq 0 \)
\( \therefore A^{-1} \) exists
Now, \( A_{11} = -2, A_{12} = 5, A_{13} = 1, A_{21} = 0, A_{22} = -2, A_{23} = 2, A_{31} = 2, A_{32} = -1, A_{33} = -1 \)
\( \therefore \text{adj } A = \begin{bmatrix} -2 & 0 & 2 \\ 5 & -2 & -1 \\ 1 & 2 & -1 \end{bmatrix} \)
\( A^{-1} = \frac{1}{|A|} \text{adj } A = \frac{1}{4} \begin{bmatrix} -2 & 0 & 2 \\ 5 & -2 & -1 \\ 1 & 2 & -1 \end{bmatrix} \)
Since, \( A^{-1} \) exists, so, system of equations has a unique solution given by \( X = A^{-1} B \)
\( \Rightarrow \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \frac{1}{4} \begin{bmatrix} -2 & 0 & 2 \\ 5 & -2 & -1 \\ 1 & 2 & -1 \end{bmatrix} \begin{bmatrix} 6 \\ 7 \\ 12 \end{bmatrix} = \frac{1}{4} \begin{bmatrix} -12 + 0 + 24 \\ 30 - 14 - 12 \\ 6 + 14 - 12 \end{bmatrix} = \frac{1}{4} \begin{bmatrix} 12 \\ 4 \\ 8 \end{bmatrix} = \begin{bmatrix} 3 \\ 1 \\ 2 \end{bmatrix} \)
\( \therefore x = 3, y = 1, z = 2. \)

Question. Using matrices, solve the following system of equations :
\( 2x + y + z = 7 \),
\( x - y - z = -4 \),
\( 3x + 2y + z = 10 \)
Answer: Given equations can be written as \( AX = B \)
where, \( A = \begin{bmatrix} 2 & 1 & 1 \\ 1 & -1 & -1 \\ 3 & 2 & 1 \end{bmatrix} \), \( X = \begin{bmatrix} x \\ y \\ z \end{bmatrix} \), \( B = \begin{bmatrix} 7 \\ -4 \\ 10 \end{bmatrix} \)
\( |A| = \begin{vmatrix} 2 & 1 & 1 \\ 1 & -1 & -1 \\ 3 & 2 & 1 \end{vmatrix} = 2(-1 + 2) - 1(1 + 3) + 1(2 + 3) = 2 - 4 + 5 = 3 \neq 0 \)
\( \therefore A^{-1} \) exists.
Now, \( A_{11} = 1, A_{12} = -4, A_{13} = 5, A_{21} = 1, A_{22} = -1, A_{23} = -1, A_{31} = 0, A_{32} = 3, A_{33} = -3 \)
\( \therefore \text{adj } A = \begin{bmatrix} 1 & 1 & 0 \\ -4 & -1 & 3 \\ 5 & -1 & -3 \end{bmatrix} \)
\( A^{-1} = \frac{1}{|A|} \text{adj } A = \frac{1}{3} \begin{bmatrix} 1 & 1 & 0 \\ -4 & -1 & 3 \\ 5 & -1 & -3 \end{bmatrix} \)
Since \( A^{-1} \) exists. So, system of equations has a unique solution given by \( X = A^{-1} B \)
\( \Rightarrow \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \frac{1}{3} \begin{bmatrix} 1 & 1 & 0 \\ -4 & -1 & 3 \\ 5 & -1 & -3 \end{bmatrix} \begin{bmatrix} 7 \\ -4 \\ 10 \end{bmatrix} = \frac{1}{3} \begin{bmatrix} 3 \\ 6 \\ 9 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} \)
\( \Rightarrow x = 1, y = 2, z = 3. \)

Question. Using matrices, solve the following system of equations: \( x + y + z = 6 \), \( x - y + z = 2 \), \( 2x + y - z = 1 \)
Answer: The given equations can be written as \( AX = B \Rightarrow X = A^{-1} B \)
where, \( A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & -1 & 1 \\ 2 & 1 & -1 \end{bmatrix} \), \( X = \begin{bmatrix} x \\ y \\ z \end{bmatrix} \), \( B = \begin{bmatrix} 6 \\ 2 \\ 1 \end{bmatrix} \)
\( |A| = \begin{vmatrix} 1 & 1 & 1 \\ 1 & -1 & 1 \\ 2 & 1 & -1 \end{vmatrix} = 1(1 - 1) - 1(-1 - 2) + 1(1 + 2) = 3 + 3 = 6 \neq 0 \)
\( \therefore A^{-1} \) exists.
Now, \( A_{11} = 0, A_{12} = 3, A_{13} = 3, A_{21} = 2, A_{22} = -3, A_{23} = 1, A_{31} = 2, A_{32} = 0, A_{33} = -2 \)
\( \text{adj } A = \begin{bmatrix} 0 & 2 & 2 \\ 3 & -3 & 0 \\ 3 & 1 & -2 \end{bmatrix} \)
\( \therefore A^{-1} = \frac{1}{|A|}(\text{adj } A) = \frac{1}{6} \begin{bmatrix} 0 & 2 & 2 \\ 3 & -3 & 0 \\ 3 & 1 & -2 \end{bmatrix} \)
Since \( A^{-1} \) exists. So, system of equations has a unique solution given by \( X = A^{-1} B \)
\( \Rightarrow \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \frac{1}{6} \begin{bmatrix} 0 & 2 & 2 \\ 3 & -3 & 0 \\ 3 & 1 & -2 \end{bmatrix} \begin{bmatrix} 6 \\ 2 \\ 1 \end{bmatrix} \)
\( \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \frac{1}{6} \begin{bmatrix} 0 + 4 + 2 \\ 18 - 6 + 0 \\ 18 + 2 - 2 \end{bmatrix} = \frac{1}{6} \begin{bmatrix} 6 \\ 12 \\ 18 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} \)
\( \therefore x = 1, y = 2, z = 3. \)

Question. Using matrices, solve the following system of equations for \( x, y \) and \( z \):
\( x + 2y - 3z = 6 \),
\( 3x + 2y - 2z = 3 \),
\( 2x - y + z = 2 \)
Answer: The given equations can be written as \( AX = B \Rightarrow X = A^{-1} B \)
where, \( A = \begin{bmatrix} 1 & 2 & -3 \\ 3 & 2 & -2 \\ 2 & -1 & 1 \end{bmatrix} \), \( X = \begin{bmatrix} x \\ y \\ z \end{bmatrix} \), \( B = \begin{bmatrix} 6 \\ 3 \\ 2 \end{bmatrix} \)
\( |A| = \begin{vmatrix} 1 & 2 & -3 \\ 3 & 2 & -2 \\ 2 & -1 & 1 \end{vmatrix} = 1(2 - 2) - 2(3 + 4) - 3(-3 - 4) = -14 + 21 = 7 \neq 0 \)
\( \dots A^{-1} \) exists.
Now, \( A_{11} = 0, A_{12} = -7, A_{13} = -7, A_{21} = 1, A_{22} = 7, A_{23} = 5, A_{31} = 2, A_{32} = -7, A_{33} = -4 \)
\( \therefore \text{adj } A = \begin{bmatrix} 0 & 1 & 2 \\ -7 & 7 & -7 \\ -7 & 5 & -4 \end{bmatrix} \)
\( A^{-1} = \frac{1}{|A|}(\text{adj } A) = \frac{1}{7} \begin{bmatrix} 0 & 1 & 2 \\ -7 & 7 & -7 \\ -7 & 5 & -4 \end{bmatrix} \)
Since \( A^{-1} \) exists. So, system of equations has a unique solution given by \( X = A^{-1} B \)
\( \Rightarrow \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \frac{1}{7} \begin{bmatrix} 0 & 1 & 2 \\ -7 & 7 & -7 \\ -7 & 5 & -4 \end{bmatrix} \begin{bmatrix} 6 \\ 3 \\ 2 \end{bmatrix} \)
\( \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \frac{1}{7} \begin{bmatrix} 0 + 3 + 4 \\ -42 + 21 - 14 \\ -42 + 15 - 8 \end{bmatrix} = \frac{1}{7} \begin{bmatrix} 7 \\ -35 \\ -35 \end{bmatrix} = \begin{bmatrix} 1 \\ -5 \\ -5 \end{bmatrix} \)
\( \therefore x = 1, y = -5, z = -5. \)

Question. Using matrices, solve the following system of equations:
\( 9x - 5y - 11z = 12 \),
\( x - 3y + z = 1 \),
\( 2x + 3y - 7z = 2 \)
Answer: We have,
\( 9x - 5y - 11z = 12 \)
\( x - 3y + z = 1 \)
\( 2x + 3y - 7z = 2 \)
Given equations can be written as \( AX = B \)
where, \( A = \begin{bmatrix} 9 & -5 & -11 \\ 1 & -3 & 1 \\ 2 & 3 & -7 \end{bmatrix} \), \( X = \begin{bmatrix} x \\ y \\ z \end{bmatrix} \), \( B = \begin{bmatrix} 12 \\ 1 \\ 2 \end{bmatrix} \)
\( |A| = \begin{vmatrix} 9 & -5 & -11 \\ 1 & -3 & 1 \\ 2 & 3 & -7 \end{vmatrix} = 9(21 - 3) + 5(-7 - 2) - 11(3 + 6) = 162 - 45 - 99 = 18 \neq 0 \)
\( \therefore A^{-1} \) exists.
Now, \( A_{11} = 18, A_{12} = 9, A_{13} = 9, A_{21} = -68, A_{22} = -41, A_{23} = -37, A_{31} = -38, A_{32} = -20, A_{33} = -22 \)
\( \therefore \text{adj } A = \begin{bmatrix} 18 & -68 & -38 \\ 9 & -41 & -20 \\ 9 & -37 & -22 \end{bmatrix} \)
\( A^{-1} = \frac{1}{|A|}(\text{adj } A) = \frac{1}{18} \begin{bmatrix} 18 & -68 & -38 \\ 9 & -41 & -20 \\ 9 & -37 & -22 \end{bmatrix} \)
Since, \( A^{-1} \) exists, therefore, system of equations has unique solution given by \( X = A^{-1} B \)
\( \Rightarrow \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \frac{1}{18} \begin{bmatrix} 18 & -68 & -38 \\ 9 & -41 & -20 \\ 9 & -37 & -22 \end{bmatrix} \begin{bmatrix} 12 \\ 1 \\ 2 \end{bmatrix} \)
\( \Rightarrow \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \frac{1}{18} \begin{bmatrix} 72 \\ 27 \\ 27 \end{bmatrix} = \begin{bmatrix} 4 \\ 3/2 \\ 3/2 \end{bmatrix} \)
\( \therefore x = 4, y = 3/2, z = 3/2. \)

Question. Using matrices, solve the following system of equations :
\( 2x - y + z = 2 \),
\( 3x - z = 2 \),
\( x + 2y = 3 \)
Answer: Given equations are
\( 2x - y + z = 2 \)
\( 3x - z = 2 \)
\( x + 2y = 3 \)
Given equations can be written as \( AX = B \)
where, \( A = \begin{bmatrix} 2 & -1 & 1 \\ 3 & 0 & -1 \\ 1 & 2 & 0 \end{bmatrix} \), \( X = \begin{bmatrix} x \\ y \\ z \end{bmatrix} \), \( B = \begin{bmatrix} 2 \\ 2 \\ 3 \end{bmatrix} \)
\( |A| = \begin{vmatrix} 2 & -1 & 1 \\ 3 & 0 & -1 \\ 1 & 2 & 0 \end{vmatrix} = 2(0 + 2) + 1(0 + 1) + 1(6 - 0) = 4 + 1 + 6 = 11 \neq 0 \)
\( \therefore A^{-1} \) exists.
Now, \( A_{11} = 2, A_{12} = -1, A_{13} = 6, A_{21} = 2, A_{22} = -1, A_{23} = -5, A_{31} = 1, A_{32} = 5, A_{33} = 3 \)
\( \therefore \text{adj } A = \begin{bmatrix} 2 & 2 & 1 \\ -1 & -1 & 5 \\ 6 & -5 & 3 \end{bmatrix} \)
\( A^{-1} = \frac{1}{|A|}\text{adj}(A) = \frac{1}{11} \begin{bmatrix} 2 & 2 & 1 \\ -1 & -1 & 5 \\ 6 & -5 & 3 \end{bmatrix} \)
Since, \( A^{-1} \) exists, therefore, system of equations has a unique solution given by \( X = A^{-1} B \)
\( \Rightarrow \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \frac{1}{11} \begin{bmatrix} 2 & 2 & 1 \\ -1 & -1 & 5 \\ 6 & -5 & 3 \end{bmatrix} \begin{bmatrix} 2 \\ 2 \\ 3 \end{bmatrix} = \frac{1}{11} \begin{bmatrix} 11 \\ 11 \\ 11 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} \)
\( \Rightarrow x = 1, y = 1, z = 1. \)