CBSE Class 12 Mathematics HOTs Determinants Set 06

Question. Find \( \text{adj } A \), if \( A = \begin{bmatrix} 5 & 2 \\ 7 & 3 \end{bmatrix} \).
Answer: Here, \( A = \begin{bmatrix} 5 & 2 \\ 7 & 3 \end{bmatrix} \)
Cofactor of \( A = \begin{bmatrix} 3 & -7 \\ -2 & 5 \end{bmatrix} \)
\( \therefore \text{adj } A = \begin{bmatrix} 3 & -7 \\ -2 & 5 \end{bmatrix}^T = \begin{bmatrix} 3 & -2 \\ -7 & 5 \end{bmatrix} \).

Question. If \( A \) is a square matrix of order 3 such that \( |\text{adj } A| = 64 \), find \( |A| \).
Answer: \( A \) is a \( 3 \times 3 \) matrix s.t. \( |\text{adj } A| = 64 \).
We know that \( |\text{adj } A| = |A|^{n-1} \)
\( \Rightarrow 64 = |A|^{3-1} \Rightarrow |A|^2 = 64 \Rightarrow |A| = \pm 8 \).

Question. If \( A \) is an invertible square matrix of order 3 and \( |A| = 5 \), then find the value of \( |\text{adj } A| \).
Answer: Given, \( |A| = 5 \) and \( A \) is invertible matrix of order 3.
We know that \( |\text{adj } A| = |A|^{n-1} \)
\( \Rightarrow |\text{adj } A| = (5)^{3-1} \Rightarrow |\text{adj } A| = 25 \).

Question. For what value of \( x \), is the given matrix \( A = \begin{bmatrix} 3 - 2x & x + 1 \\ 2 & 4 \end{bmatrix} \) singular?
Answer: Let \( A = \begin{bmatrix} 3 - 2x & x + 1 \\ 2 & 4 \end{bmatrix} \).
Matrix \( A \) is singular, iff \( |A| = 0 \)
\( \Rightarrow \begin{vmatrix} 3 - 2x & x + 1 \\ 2 & 4 \end{vmatrix} = 0 \)
\( \Rightarrow 4(3 - 2x) - 2(x + 1) = 0 \)
\( \Rightarrow 12 - 8x - 2x - 2 = 0 \)
\( \Rightarrow -10x + 10 = 0 \Rightarrow x = 1 \).

Question. For what value of \( x \), the matrix \( \begin{bmatrix} 1 + x & 7 \\ 3 - x & 8 \end{bmatrix} \) is a singular matrix?
Answer: Let \( A = \begin{bmatrix} 1 + x & 7 \\ 3 - x & 8 \end{bmatrix} \).
For \( A \) to be singular, \( |A| = 0 \)
\( \Rightarrow \begin{vmatrix} 1 + x & 7 \\ 3 - x & 8 \end{vmatrix} = 0 \)
\( \Rightarrow 8 + 8x - 21 + 7x = 0 \Rightarrow 15x = 13 \Rightarrow x = \frac{13}{15} \).

Question. Write \( A^{-1} \) for \( A = \begin{bmatrix} 2 & 5 \\ 1 & 3 \end{bmatrix} \).
Answer: We have, \( A = \begin{bmatrix} 2 & 5 \\ 1 & 3 \end{bmatrix} \)
\( \therefore |A| = \begin{vmatrix} 2 & 5 \\ 1 & 3 \end{vmatrix} = 6 - 5 = 1 \neq 0 \), so \( A^{-1} \) exists.
\( \Rightarrow \text{adj } A = \begin{bmatrix} 3 & -5 \\ -1 & 2 \end{bmatrix} \)
Now, \( A^{-1} = \frac{\text{adj } A}{|A|} = \begin{bmatrix} 3 & -5 \\ -1 & 2 \end{bmatrix} \).

Question. For what value of \( x \), the matrix \( \begin{bmatrix} 5 - x & x + 1 \\ 2 & 4 \end{bmatrix} \) is singular?
Answer: Let \( A = \begin{bmatrix} 5 - x & x + 1 \\ 2 & 4 \end{bmatrix} \).
Matrix \( A \) is singular, \( \therefore |A| = 0 \)
\( \Rightarrow \begin{vmatrix} 5 - x & x + 1 \\ 2 & 4 \end{vmatrix} = 0 \)
\( \Rightarrow 20 - 4x - 2x - 2 = 0 \)
\( \Rightarrow -6x + 18 = 0 \Rightarrow x = 3 \).

Question. If \( A = \begin{bmatrix} 2 & 3 \\ 5 & -2 \end{bmatrix} \), then write \( A^{-1} \) in terms of \( A \).
Answer: We have, \( A = \begin{bmatrix} 2 & 3 \\ 5 & -2 \end{bmatrix} \)
\( \therefore |A| = \begin{vmatrix} 2 & 3 \\ 5 & -2 \end{vmatrix} = -4 - 15 = -19 \neq 0 \), so \( A^{-1} \) exists.
\( \text{adj } A = \begin{bmatrix} -2 & -3 \\ -5 & 2 \end{bmatrix} \)
Now, \( A^{-1} = \frac{\text{adj } A}{|A|} = \frac{1}{-19} \begin{bmatrix} -2 & -3 \\ -5 & 2 \end{bmatrix} = \frac{1}{19} \begin{bmatrix} 2 & 3 \\ 5 & -2 \end{bmatrix} \)
\( \Rightarrow A^{-1} = \frac{1}{19} A \).

Question. For what value of \( x \) is the matrix \( \begin{bmatrix} 6 - x & 4 \\ 3 - x & 1 \end{bmatrix} \) singular?
Answer: For the matrix \( A = \begin{bmatrix} 6 - x & 4 \\ 3 - x & 1 \end{bmatrix} \) to be singular, \( |A| = 0 \).
\( \Rightarrow \begin{vmatrix} 6 - x & 4 \\ 3 - x & 1 \end{vmatrix} = 0 \)
\( \Rightarrow (6 - x)(1) - 4(3 - x) = 0 \)
\( \Rightarrow 6 - x - 12 + 4x = 0 \)
\( \Rightarrow 3x - 6 = 0 \Rightarrow x = 2 \).

Question. For what value of \( x \) is the matrix \( \begin{bmatrix} 2x & 4 \\ x + 2 & 3 \end{bmatrix} \) singular?
Answer: Let \( A = \begin{bmatrix} 2x & 4 \\ x + 2 & 3 \end{bmatrix} \).
For the matrix \( A \) to be singular, \( |A| = 0 \)
\( \Rightarrow \begin{vmatrix} 2x & 4 \\ x + 2 & 3 \end{vmatrix} = 0 \)
\( \Rightarrow 6x - 4x - 8 = 0 \Rightarrow 2x = 8 \Rightarrow x = 4 \).

Question. For what value of \( x \) is the matrix \( \begin{bmatrix} 2x + 4 & 4 \\ x + 5 & 3 \end{bmatrix} \) a singular matrix?
Answer: For the matrix \( A = \begin{bmatrix} 2x + 4 & 4 \\ x + 5 & 3 \end{bmatrix} \) to be singular, \( |A| = 0 \).
\( \Rightarrow \begin{vmatrix} 2x + 4 & 4 \\ x + 5 & 3 \end{vmatrix} = 0 \)
\( \Rightarrow 3(2x + 4) - 4(x + 5) = 0 \)
\( \Rightarrow 6x + 12 - 4x - 20 = 0 \)
\( \Rightarrow 2x - 8 = 0 \Rightarrow x = 4 \).

Question. For what value of \( x \) is \( A = \begin{bmatrix} 2(x + 1) & 2x \\ x & x - 2 \end{bmatrix} \) a singular matrix?
Answer: For the given matrix \( A = \begin{bmatrix} 2(x + 1) & 2x \\ x & x - 2 \end{bmatrix} \) to be singular, \( |A| = 0 \).
\( \Rightarrow \begin{vmatrix} 2(x + 1) & 2x \\ x & x - 2 \end{vmatrix} = 0 \)
\( \Rightarrow 2(x + 1)(x - 2) - 2x^2 = 0 \)
\( \Rightarrow 2(x^2 - x - 2) - 2x^2 = 0 \Rightarrow 2(-x - 2) = 0 \)
\( \Rightarrow x = -2 \).

Question. If \( A \) is a square matrix of order 3 and \( |A| = 7 \), then write the value of \( |\text{adj } A| \).
Answer: We know that \( |\text{adj } A| = |A|^{n-1} \).
Given \( |A| = 7 \) and order \( n = 3 \),
\( \therefore |\text{adj } A| = 7^{3-1} = 7^2 = 49 \).

Question. Write the adjoint of the following matrix : \( \begin{bmatrix} 2 & -1 \\ 4 & 3 \end{bmatrix} \).
Answer: Let \( A = \begin{bmatrix} 2 & -1 \\ 4 & 3 \end{bmatrix} \).
The cofactor matrix of \( A = \begin{bmatrix} 3 & -4 \\ 1 & 2 \end{bmatrix} \).
\( \therefore \text{adj } A = \begin{bmatrix} 3 & -4 \\ 1 & 2 \end{bmatrix}^T = \begin{bmatrix} 3 & 1 \\ -4 & 2 \end{bmatrix} \).

Question. If \( A = \begin{bmatrix} 3 & 1 \\ 2 & -3 \end{bmatrix} \), then find \( |\text{adj } A| \).
Answer: We have \( A = \begin{bmatrix} 3 & 1 \\ 2 & -3 \end{bmatrix} \).
\( \therefore |A| = \begin{vmatrix} 3 & 1 \\ 2 & -3 \end{vmatrix} = -9 - 2 = -11 \)
Since \( |\text{adj } A| = |A|^{n-1} \), for order \( n = 2 \):
\( \dots |\text{adj } A| = (-11)^{2-1} = -11 \).

Question. If \( |A| = 2 \), where \( A \) is a \( 2 \times 2 \) matrix, find \( |\text{adj } A| \).
Answer: We know that \( |\text{adj } A| = |A|^{n-1} \).
Given \( |A| = 2 \) and order \( n = 2 \),
\( \therefore |\text{adj } A| = 2^{2-1} = 2 \).

Question. If \( A \) is a non-singular matrix of order 3 and \( |\text{adj } A| = |A|^k \), then write the value of \( k \).
Answer: We know that \( |\text{adj } A| = |A|^{n-1} \).
For order \( n = 3 \):
\( \Rightarrow |\text{adj } A| = |A|^{3-1} = |A|^2 = |A|^k \)
\( \Rightarrow k = 2 \).

Question. If \( A = \begin{bmatrix} 1 & -2 & 3 \\ 0 & -1 & 4 \\ -2 & 2 & 1 \end{bmatrix} \), find \( (A^T)^{-1} \).
Answer: We have, \( A = \begin{bmatrix} 1 & -2 & 3 \\ 0 & -1 & 4 \\ -2 & 2 & 1 \end{bmatrix} \)
\( \Rightarrow A^T = \begin{bmatrix} 1 & 0 & -2 \\ -2 & -1 & 2 \\ 3 & 4 & 1 \end{bmatrix} \)
\( |A^T| = \begin{vmatrix} 1 & 0 & -2 \\ -2 & -1 & 2 \\ 3 & 4 & 1 \end{vmatrix} = 1(-1 - 8) - 2(-8 + 3) = -9 + 10 = 1 \neq 0 \). So, \( (A^T)^{-1} \) exists.
Let the cofactors of elements \( a_{ij} \) in \( A^T \) be \( A_{ij} \):
\( A_{11} = -9, A_{12} = 8, A_{13} = -5, \)
\( A_{21} = -8, A_{22} = 7, A_{23} = -4, \)
\( A_{31} = -2, A_{32} = 2, A_{33} = -1 \)
\( \therefore \text{adj}(A^T) = \begin{bmatrix} -9 & -8 & -2 \\ 8 & 7 & 2 \\ -5 & -4 & -1 \end{bmatrix} \)
\( \therefore (A^T)^{-1} = \frac{\text{adj}(A^T)}{|A^T|} = \frac{1}{1} \begin{bmatrix} -9 & -8 & -2 \\ 8 & 7 & 2 \\ -5 & -4 & -1 \end{bmatrix} = \begin{bmatrix} -9 & -8 & -2 \\ 8 & 7 & 2 \\ -5 & -4 & -1 \end{bmatrix} \).

Question. Find the adjoint of the matrix \( A = \begin{bmatrix} -1 & -2 & -2 \\ 2 & 1 & -2 \\ 2 & -2 & 1 \end{bmatrix} \) and hence show that \( A \cdot (\text{adj } A) = |A| I_3 \).
Answer: Here, \( A = \begin{bmatrix} -1 & -2 & -2 \\ 2 & 1 & -2 \\ 2 & -2 & 1 \end{bmatrix} \)
\( \Rightarrow |A| = \begin{vmatrix} -1 & -2 & -2 \\ 2 & 1 & -2 \\ 2 & -2 & 1 \end{vmatrix} = -1(1 - 4) - (-2)(2 + 4) - 2(-4 - 2) = 3 + 12 + 12 = 27 \)
Now, let's find the cofactors of elements of \( A \):
\( A_{11} = -3, A_{12} = -6, A_{13} = -6, \)
\( A_{21} = 6, A_{22} = 3, A_{23} = -6, \)
\( A_{31} = 6, A_{32} = -6, A_{33} = 3 \)
\( \therefore \text{adj } A = \begin{bmatrix} -3 & 6 & 6 \\ -6 & 3 & -6 \\ -6 & -6 & 3 \end{bmatrix} \)
\( \therefore A \cdot (\text{adj } A) = \begin{bmatrix} -1 & -2 & -2 \\ 2 & 1 & -2 \\ 2 & -2 & 1 \end{bmatrix} \begin{bmatrix} -3 & 6 & 6 \\ -6 & 3 & -6 \\ -6 & -6 & 3 \end{bmatrix} \)
\( = \begin{bmatrix} 27 & 0 & 0 \\ 0 & 27 & 0 \\ 0 & 0 & 27 \end{bmatrix} = 27 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = |A| I_3 \).

Question. If \( A = \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix} \) and \( I \) is the identity matrix of order 2, then show that \( A^2 = 4A - 3I \). Hence find \( A^{-1} \).
Answer: \( A = \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix} \)
\( \Rightarrow A^2 = \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 5 & -4 \\ -4 & 5 \end{bmatrix} \) ...(i)
Now, \( 4A - 3I = 4 \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix} - 3 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 5 & -4 \\ -4 & 5 \end{bmatrix} \) ...(ii)
From (i) and (ii), we get:
\( A^2 = 4A - 3I \)
Pre-multiplying by \( A^{-1} \) on both sides, we get:
\( A^{-1}(A^2) = 4A^{-1}A - 3A^{-1} \)
\( \Rightarrow A = 4I - 3A^{-1} \quad [\because A A^{-1} = I] \)
\( \Rightarrow 3A^{-1} = 4I - A \)
\( \Rightarrow A^{-1} = \frac{1}{3}(4I - A) = \frac{1}{3} \left( 4\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} - \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix} \right) = \begin{bmatrix} 2/3 & 1/3 \\ 1/3 & 2/3 \end{bmatrix} \).

Question. Find the inverse of the matrix \( A = \begin{bmatrix} 1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \end{bmatrix} \).
Answer: Given \( A = \begin{bmatrix} 1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \end{bmatrix} \).
Now, finding cofactors of the elements of \( A \):
\( A_{11} = 7, A_{12} = -1, A_{13} = -1 \),
\( A_{21} = -3, A_{22} = 1, A_{23} = 0 \),
\( A_{31} = -3, A_{32} = 0, A_{33} = 1 \)
\( \therefore \text{adj } A = \begin{bmatrix} 7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{bmatrix} \)
and \( |A| = 1(16 - 9) - 3(4 - 3) + 3(3 - 4) = 7 - 3 - 3 = 1 \neq 0 \).
So, \( A^{-1} \) exists and is given by:
\( A^{-1} = \frac{\text{adj } A}{|A|} = \frac{1}{1} \begin{bmatrix} 7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{bmatrix} \).

Question. If \( A = \begin{bmatrix} 2 & 3 \\ 1 & -4 \end{bmatrix} \), \( B = \begin{bmatrix} 1 & -2 \\ -1 & 3 \end{bmatrix} \), verify that \( (AB)^{-1} = B^{-1} A^{-1} \).
Answer: Here, \( A = \begin{bmatrix} 2 & 3 \\ 1 & -4 \end{bmatrix} \) and \( B = \begin{bmatrix} 1 & -2 \\ -1 & 3 \end{bmatrix} \)
\( \Rightarrow |A| = 2(-4) - 3(1) = -11 \), \( |B| = 1(3) - (-2)(-1) = 1 \)
\( \therefore A^{-1} = \frac{\text{adj } A}{|A|} = -\frac{1}{11} \begin{bmatrix} -4 & -3 \\ -1 & 2 \end{bmatrix} \) and \( B^{-1} = \begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix} \)
\( \therefore \text{R.H.S.} = B^{-1} A^{-1} = \begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix} \left( -\frac{1}{11} \begin{bmatrix} -4 & -3 \\ -1 & 2 \end{bmatrix} \right) = -\frac{1}{11} \begin{bmatrix} -14 & -5 \\ -5 & -1 \end{bmatrix} \) ...(i)
Next, \( AB = \begin{bmatrix} 2 & 3 \\ 1 & -4 \end{bmatrix} \begin{bmatrix} 1 & -2 \\ -1 & 3 \end{bmatrix} = \begin{bmatrix} -1 & 5 \\ 5 & -14 \end{bmatrix} \)
\( \Rightarrow |AB| = 14 - 25 = -11 \)
\( \dots \text{L.H.S.} = (AB)^{-1} = -\frac{1}{11} \begin{bmatrix} -14 & -5 \\ -5 & -1 \end{bmatrix} \) ...(ii)
From (i) and (ii), we get \( (AB)^{-1} = B^{-1} A^{-1} \).

Question. Show that the matrix \( A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \) satisfies the equation \( A^2 - 5A + 7I = O \). Hence, find \( A^{-1} \).
Answer: We have, \( A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \)
\( \text{L.H.S.} = A^2 - 5A + 7I = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} - 5\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} + 7\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)
\( = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix} - \begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix} + \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix} \)
\( = \begin{bmatrix} 8 - 15 + 7 & 5 - 5 + 0 \\ -5 + 5 + 0 & 3 - 10 + 7 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O = \text{R.H.S.} \)
\( \Rightarrow A^2 - 5A + 7I = O \)
Pre-multiplying by \( A^{-1} \) on both sides, we get:
\( A(A A^{-1}) - 5 A A^{-1} + 7 I A^{-1} = O \)
\( \Rightarrow A I - 5I + 7 A^{-1} = O \)
\( \Rightarrow A - 5I + 7 A^{-1} = O \Rightarrow 7 A^{-1} = 5I - A \)
\( \Rightarrow 7A^{-1} = 5 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} - \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \)
\( \Rightarrow 7A^{-1} = \begin{bmatrix} 2 & -1 \\ 1 & 3 \end{bmatrix} \Rightarrow A^{-1} = \frac{1}{7} \begin{bmatrix} 2 & -1 \\ 1 & 3 \end{bmatrix} \).

Question. If \( A = \begin{bmatrix} \cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{bmatrix} \), find \( \text{adj } A \) and verify that \( A(\text{adj } A) = (\text{adj } A)A = |A| I_3 \).
Answer: Given, \( A = \begin{bmatrix} \cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{bmatrix} \)
Now, let's find the cofactors of elements of \( A \):
\( A_{11} = \cos \alpha, A_{12} = -\sin \alpha, A_{13} = 0 \),
\( A_{21} = \sin \alpha, A_{22} = \cos \alpha, A_{23} = 0 \),
\( A_{31} = 0, A_{32} = 0, A_{33} = 1 \)
\( \therefore \text{adj } A = \begin{bmatrix} \cos \alpha & \sin \alpha & 0 \\ -\sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{bmatrix} \)
Now, verifying \( A(\text{adj } A) \):
\( A(\text{adj } A) = \begin{bmatrix} \cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} \cos \alpha & \sin \alpha & 0 \\ -\sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} \cos^2\alpha + \sin^2\alpha & 0 & 0 \\ 0 & \sin^2\alpha + \cos^2\alpha & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \) ...(i)
And, \( (\text{adj } A)A \):
\( (\text{adj } A)A = \begin{bmatrix} \cos \alpha & \sin \alpha & 0 \\ -\sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} \cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \) ...(ii)
Now, let's find \( |A| \):
\( |A| = \begin{vmatrix} \cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{vmatrix} = \cos \alpha(\cos \alpha - 0) + \sin \alpha(\sin \alpha - 0) + 0 = \cos^2\alpha + \sin^2\alpha = 1 \) ...(iii)
From (i), (ii) and (iii), we get:
\( A(\text{adj } A) = (\text{adj } A)A = |A| I_3 \).

Applications of Determinants and Matrices

Question. The monthly incomes of Aryan and Babban are in the ratio 3 : 4 and their monthly expenditures are in the ratio 5 : 7. If each saves ₹ 15,000 per month, find their monthly incomes using matrix method. This problem reflects which value?
Answer: Let the monthly income of Aryan be ₹ \( 3x \) and that of Babban be ₹ \( 4x \).
Also, let the monthly expenditure of Aryan be ₹ \( 5y \) and that of Babban be ₹ \( 7y \).
According to the question,
\( 3x - 5y = 15000 \)
\( 4x - 7y = 15000 \)
These equations can be written as:
\( AX = B \)
where, \( A = \begin{bmatrix} 3 & -5 \\ 4 & -7 \end{bmatrix} \), \( X = \begin{bmatrix} x \\ y \end{bmatrix} \), and \( B = \begin{bmatrix} 15000 \\ 15000 \end{bmatrix} \).
\( |A| = \begin{vmatrix} 3 & -5 \\ 4 & -7 \end{vmatrix} = -21 - (-20) = -1 \neq 0 \).
Thus, \( A^{-1} \) exists. So, the system of equations has a unique solution given by \( X = A^{-1} B \).
Now, \( \text{adj}(A) = \begin{bmatrix} -7 & 5 \\ -4 & 3 \end{bmatrix} \)
\( \therefore A^{-1} = \frac{\text{adj}(A)}{|A|} = \frac{1}{-1} \begin{bmatrix} -7 & 5 \\ -4 & 3 \end{bmatrix} = \begin{bmatrix} 7 & -5 \\ 4 & -3 \end{bmatrix} \)
Now, \( X = A^{-1}B \Rightarrow \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 7 & -5 \\ 4 & -3 \end{bmatrix} \begin{bmatrix} 15000 \\ 15000 \end{bmatrix} \)
\( \Rightarrow \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 30000 \\ 15000 \end{bmatrix} \)
\( \Rightarrow x = 30000 \) and \( y = 15000 \).
So, monthly income of Aryan = ( 3 \times 30000 = ₹ 90,000 \)
Monthly income of Babban = ( 4 \times 30000 = ₹ 1,20,000 \).
Value: From this question we are encouraged to save a part of money every month.

Question. A trust invested some money in two type of bonds. The first bond pays 10% interest and second bond pays 12% interest. The trust received ₹ 2,800 as interest. However, if trust had interchanged money in bonds they would have got ₹ 100 less as interest. Using matrix method, find the amount invested by the trust. Interest received on this amount will be given to Helpage India as donation. Which value is reflected in this question?
Answer: Let ₹ \( x \) be invested in the first bond and ₹ \( y \) be invested in the second bond.
According to the question,
\( \frac{10}{100}x + \frac{12}{100}y = 2800 \Rightarrow 10x + 12y = 280000 \) ...(i)
If the rate of interest had been interchanged, then the total interest earned is ₹ 100 less than the previous interest, i.e., ₹ 2,700.
\( \therefore \frac{12}{100}x + \frac{10}{100}y = 2700 \Rightarrow 12x + 10y = 270000 \) ...(ii)
The system of equations (i) and (ii) can be represented as:
\( AX = B \)
where, \( A = \begin{bmatrix} 10 & 12 \\ 12 & 10 \end{bmatrix} \), \( X = \begin{bmatrix} x \\ y \end{bmatrix} \), and \( B = \begin{bmatrix} 280000 \\ 270000 \end{bmatrix} \).
\( |A| = \begin{vmatrix} 10 & 12 \\ 12 & 10 \end{vmatrix} = 100 - 144 = -44 \neq 0 \).
Thus, \( A^{-1} \) exists. So, the system of equations has a unique solution given by \( X = A^{-1} B \).
Now, \( \text{adj } A = \begin{bmatrix} 10 & -12 \\ -12 & 10 \end{bmatrix} \).
Now, \( X = A^{-1}B \Rightarrow X = \frac{\text{adj } A}{|A|} B \)
\( \Rightarrow \begin{bmatrix} x \\ y \end{bmatrix} = \frac{1}{-44} \begin{bmatrix} 10 & -12 \\ -12 & 10 \end{bmatrix} \begin{bmatrix} 280000 \\ 270000 \end{bmatrix} \)
\( \Rightarrow \begin{bmatrix} x \\ y \end{bmatrix} = \frac{1}{-44} \begin{bmatrix} 10(280000) - 12(270000) \\ -12(280000) + 10(270000) \end{bmatrix} = \frac{1}{-44} \begin{bmatrix} -440000 \\ -660000 \end{bmatrix} = \begin{bmatrix} 10000 \\ 15000 \end{bmatrix} \)
\( \Rightarrow x = 10000 \) and \( y = 15000 \).
Therefore, ₹ 10,000 is invested in the first bond and ₹ 15,000 is invested in the second bond. Thus, the total amount invested by the trust = ( 10,000 + 15,000 = ₹ 25,000 ).
Value: The interest received will be given to Helpage India as donation, which reflects the helping and caring nature of the trust.

Question. A coaching institute of English (Subject) conducts classes in two batches I and II and fees for rich and poor children are different. In batch I, it has 20 poor and 5 rich children and total monthly collection is ₹ 9,000, whereas in batch II, it has 5 poor and 25 rich children and total monthly collection is ₹ 26,000. Using matrix method, find monthly fees paid by each child of two types. What values the coaching institute is inculcating in the society?
Answer: Let the monthly fees paid by poor and rich children be ₹ \( x \) and ₹ \( y \), respectively.
For batch I :
\( 20x + 5y = 9000 \) ...(i)
For batch II :
\( 5x + 25y = 26000 \) ...(ii)
The system of equations (i) and (ii) can be written as:
\( AX = B \)
where, \( A = \begin{bmatrix} 20 & 5 \\ 5 & 25 \end{bmatrix} \), \( X = \begin{bmatrix} x \\ y \end{bmatrix} \), and \( B = \begin{bmatrix} 9000 \\ 26000 \end{bmatrix} \).
First, find the determinant of \( A \):
\( |A| = 20(25) - 5(5) = 500 - 25 = 475 \neq 0 \).
Since \( |A| \neq 0 \), \( A^{-1} \) exists and the system has a unique solution given by \( X = A^{-1}B \).
Now, \( \text{adj } A = \begin{bmatrix} 25 & -5 \\ -5 & 20 \end{bmatrix} \).
\( \therefore A^{-1} = \frac{1}{475} \begin{bmatrix} 25 & -5 \\ -5 & 20 \end{bmatrix} \)
Now, \( X = A^{-1}B \Rightarrow \begin{bmatrix} x \\ y \end{bmatrix} = \frac{1}{475} \begin{bmatrix} 25 & -5 \\ -5 & 20 \end{bmatrix} \begin{bmatrix} 9000 \\ 26000 \end{bmatrix} \)
\( \Rightarrow \begin{bmatrix} x \\ y \end{bmatrix} = \frac{1}{475} \begin{bmatrix} 25(9000) - 5(26000) \\ -5(9000) + 20(26000) \end{bmatrix} \)
\( \Rightarrow \begin{bmatrix} x \\ y \end{bmatrix} = \frac{1}{475} \begin{bmatrix} 225000 - 130000 \\ -45000 + 520000 \end{bmatrix} \)
\( \Rightarrow \begin{bmatrix} x \\ y \end{bmatrix} = \frac{1}{475} \begin{bmatrix} 95000 \\ 475000 \end{bmatrix} = \begin{bmatrix} 200 \\ 1000 \end{bmatrix} \)
\( \Rightarrow x = 200 \) and \( y = 1000 \).
Therefore, the monthly fees paid by a poor child is ₹ 200 and by a rich child is ₹ 1,000.
Value: The coaching institute is promoting social equity, empathy, and supporting under-privileged children by making education accessible to them.

Question. Two schools A and B decided to award prizes to their students for three values, team spirit, truthfulness and tolerance at the rate of ₹ x, ₹ y and ₹ z per student respectively. School A, decided to award a total of ₹ 1,100 for the three values to 3, 1 and 2 students respectively while school B decided to award ₹ 1,400 for the three values to 1, 2 and 3 students respectively. If one prize for all the three values together amount to ₹ 600 then
(i) Represent the above situation by a matrix equation after forming linear equations.
(ii) Is it possible to solve the system of equations so obtained using matrices?
(iii) Which value you prefer to be rewarded most and why?
Answer: Let ₹ \( x \), ₹ \( y \), and ₹ \( z \) be the values of prizes awarded for team spirit, truthfulness, and tolerance respectively.
Based on the given information, we form the following linear equations:
For School A: \( 3x + y + 2z = 1100 \)
For School B: \( x + 2y + 3z = 1400 \)
For the total of one prize of each type: \( x + y + z = 600 \)
(i) The above system of linear equations can be represented as a matrix equation:
\( PX = Q \)
where, \( P = \begin{bmatrix} 3 & 1 & 2 \\ 1 & 2 & 3 \\ 1 & 1 & 1 \end{bmatrix} \), \( X = \begin{bmatrix} x \\ y \\ z \end{bmatrix} \), and \( Q = \begin{bmatrix} 1100 \\ 1400 \\ 600 \end{bmatrix} \).
(ii) To check if the system of equations is solvable, we evaluate the determinant of \( P \):
\( |P| = \begin{vmatrix} 3 & 1 & 2 \\ 1 & 2 & 3 \\ 1 & 1 & 1 \end{vmatrix} = 3(2(1) - 3(1)) - 1(1(1) - 3(1)) + 2(1(1) - 2(1)) \)
\( = 3(2 - 3) - 1(1 - 3) + 2(1 - 2) \)
\( = 3(-1) - 1(-2) + 2(-1) = -3 + 2 - 2 = -3 \neq 0 \).
Since \( |P| \neq 0 \), matrix \( P \) is non-singular and \( P^{-1} \) exists. Therefore, it is possible to solve the system of equations using matrices.
(iii) I prefer Truthfulness to be rewarded most because honesty and truthfulness lead to a strong moral character, build mutual trust, and help establish a highly ethical and harmonious society.

Question. Using elementary transformations, find the inverse of the matrix \( A = \begin{bmatrix} 8 & 4 & 3 \\ 2 & 1 & 1 \\ 1 & 2 & 2 \end{bmatrix} \) and use it to solve the following system of linear equations:
\( 8x + 4y + 3z = 19; \quad 2x + y + z = 5; \quad x + 2y + 2z = 7 \)
Answer: Given matrix \( A = \begin{bmatrix} 8 & 4 & 3 \\ 2 & 1 & 1 \\ 1 & 2 & 2 \end{bmatrix} \).
Since, \( A A^{-1} = I \)
\( \Rightarrow \begin{bmatrix} 8 & 4 & 3 \\ 2 & 1 & 1 \\ 1 & 2 & 2 \end{bmatrix} A^{-1} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \)
Applying \( R_1 \leftrightarrow R_3 \), we get
\( \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 1 \\ 8 & 4 & 3 \end{bmatrix} A^{-1} = \begin{bmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{bmatrix} \)
Applying \( R_2 \rightarrow R_2 - 2R_1 \) and \( R_3 \rightarrow R_3 - 8R_1 \), we get
\( \begin{bmatrix} 1 & 2 & 2 \\ 0 & -3 & -3 \\ 0 & -12 & -13 \end{bmatrix} A^{-1} = \begin{bmatrix} 0 & 0 & 1 \\ 0 & 1 & -2 \\ 1 & 0 & -8 \end{bmatrix} \)
Applying \( R_3 \rightarrow (-1)R_3 \), we get
\( \begin{bmatrix} 1 & 2 & 2 \\ 0 & -3 & -3 \\ 0 & 12 & 13 \end{bmatrix} A^{-1} = \begin{bmatrix} 0 & 0 & 1 \\ 0 & 1 & -2 \\ -1 & 0 & 8 \end{bmatrix} \)
Applying \( R_3 \rightarrow R_3 + 4R_2 \), we get
\( \begin{bmatrix} 1 & 2 & 2 \\ 0 & -3 & -3 \\ 0 & 0 & 1 \end{bmatrix} A^{-1} = \begin{bmatrix} 0 & 0 & 1 \\ 0 & 1 & -2 \\ -1 & 4 & 0 \end{bmatrix} \)
Applying \( R_2 \rightarrow -\frac{1}{3}R_2 \), we get
\( \begin{bmatrix} 1 & 2 & 2 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix} A^{-1} = \begin{bmatrix} 0 & 0 & 1 \\ 0 & -1/3 & 2/3 \\ -1 & 4 & 0 \end{bmatrix} \)
Applying \( R_2 \rightarrow R_2 - R_3 \) and \( R_1 \rightarrow R_1 - 2R_3 \), we get
\( \begin{bmatrix} 1 & 2 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} A^{-1} = \begin{bmatrix} 2 & -8 & 1 \\ 1 & -13/3 & 2/3 \\ -1 & 4 & 0 \end{bmatrix} \)
Applying \( R_1 \rightarrow R_1 - 2R_2 \), we get
\( \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} A^{-1} = \begin{bmatrix} 0 & 2/3 & -1/3 \\ 1 & -13/3 & 2/3 \\ -1 & 4 & 0 \end{bmatrix} \)
So, \( A^{-1} = \begin{bmatrix} 0 & 2/3 & -1/3 \\ 1 & -13/3 & 2/3 \\ -1 & 4 & 0 \end{bmatrix} \) ...(i)
The given system of linear equations can be written as \( AX = B \), where
\( A = \begin{bmatrix} 8 & 4 & 3 \\ 2 & 1 & 1 \\ 1 & 2 & 2 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad B = \begin{bmatrix} 19 \\ 5 \\ 7 \end{bmatrix} \)
\(\therefore\) The solution of the above equation is \( X = A^{-1}B \)
\( \Rightarrow \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 & 2/3 & -1/3 \\ 1 & -13/3 & 2/3 \\ -1 & 4 & 0 \end{bmatrix} \begin{bmatrix} 19 \\ 5 \\ 7 \end{bmatrix} \) [From (i)]
\( = \begin{bmatrix} 0 + 10/3 - 7/3 \\ 19 - 65/3 + 14/3 \\ -19 + 20 + 0 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix} \Rightarrow x = 1, y = 2, z = 1 \)

Question. A shopkeeper has 3 varieties of pens ‘A’, ‘B’ and ‘C’. Meenu purchased 1 pen of each variety for a total of ₹ 21. Jeevan purchased 4 pens of ‘A’ variety, 3 pens of ‘B’ variety and 2 pens of ‘C’ variety for ₹ 60. While Shikha purchased 6 pens of ‘A’ variety, 2 pens of ‘B’ variety and 3 pens of ‘C’ variety for ₹ 70. Using matrix method, find cost of each variety of pen.
Answer: Let one pen of variety ‘A’ costs ₹ \( x \), one pen of variety ‘B’ costs ₹ \( y \) and one pen of variety ‘C’ costs ₹ \( z \).
According to question,
\( x + y + z = 21 \) (For Meenu)
\( 4x + 3y + 2z = 60 \) (For Jeevan)
\( 6x + 2y + 3z = 70 \) (For Shikha)
The given system of equations can be written as \( AX = B \)
where, \( A = \begin{bmatrix} 1 & 1 & 1 \\ 4 & 3 & 2 \\ 6 & 2 & 3 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix} \) and \( B = \begin{bmatrix} 21 \\ 60 \\ 70 \end{bmatrix} \)
\( |A| = \begin{vmatrix} 1 & 1 & 1 \\ 4 & 3 & 2 \\ 6 & 2 & 3 \end{vmatrix} = 1(9 - 4) - 1(12 - 12) + 1(8 - 18) = -5 \neq 0 \)
\(\therefore A^{-1} \) exists and system of equations has a unique solution given by \( X = A^{-1}B \).
Now, \( A_{11} = 5, A_{12} = 0, A_{13} = -10 \),
\( A_{21} = -1, A_{22} = -3, A_{23} = 4 \),
\( A_{31} = -1, A_{32} = 2, A_{33} = -1 \)
\(\therefore \text{adj } A = \begin{bmatrix} 5 & -1 & -1 \\ 0 & -3 & 2 \\ -10 & 4 & -1 \end{bmatrix} \)
\( A^{-1} = \frac{1}{|A|}(\text{adj } A) = \frac{1}{-5} \begin{bmatrix} 5 & -1 & -1 \\ 0 & -3 & 2 \\ -10 & 4 & -1 \end{bmatrix} \)
Now, \( X = A^{-1}B \)
\( \Rightarrow \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \frac{1}{-5} \begin{bmatrix} 5 & -1 & -1 \\ 0 & -3 & 2 \\ -10 & 4 & -1 \end{bmatrix} \begin{bmatrix} 21 \\ 60 \\ 70 \end{bmatrix} \)
\( \Rightarrow \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \frac{1}{-5} \begin{bmatrix} -25 \\ -40 \\ -40 \end{bmatrix} \)
\( \Rightarrow \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 5 \\ 8 \\ 8 \end{bmatrix} \Rightarrow x = 5, y = 8, z = 8 \)
\(\dots\) Cost of 1 pen of variety ‘A’ = ₹ 5
Cost of 1 pen of variety ‘B’ = ₹ 8
Cost of 1 pen of variety ‘C’ = ₹ 8

Question. Two schools P and Q want to award their selected students on the values of discipline, politeness and punctuality. The school P wants to award ₹ \( x \) each, ₹ \( y \) each and ₹ \( z \) each for the three respective values to its 3, 2 and 1 students with a total award money of ₹ 1,000. School Q wants to spend ₹ 1,500 to award its 4, 1 and 3 students on the respective values (by giving the same award money for the three values as before). If the total amount of awards for one prize on each value is ₹ 600, using matrices, find the award money for each value. Apart from the above three values, suggest one more value for awards.
Answer: According to question, we have,
\( 3x + 2y + z = 1000 \) ...(i)
\( 4x + y + 3z = 1500 \) ...(ii)
\( x + y + z = 600 \) ...(iii)
The given system of equations can be written as \( AX = B \)
where, \( A = \begin{bmatrix} 3 & 2 & 1 \\ 4 & 1 & 3 \\ 1 & 1 & 1 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix} \) and \( B = \begin{bmatrix} 1000 \\ 1500 \\ 600 \end{bmatrix} \)
\( |A| = \begin{vmatrix} 3 & 2 & 1 \\ 4 & 1 & 3 \\ 1 & 1 & 1 \end{vmatrix} = -5 \neq 0 \)
\(\therefore A \) is invertible and system of equations has a unique solution given by \( X = A^{-1} B \)
Now, \( A_{11} = -2, A_{12} = -1, A_{13} = 3 \),
\( A_{21} = -1, A_{22} = 2, A_{23} = -1 \),
\( A_{31} = 5, A_{32} = -5, A_{33} = -5 \)
\(\therefore \text{adj } A = \begin{bmatrix} -2 & -1 & 5 \\ -1 & 2 & -5 \\ 3 & -1 & -5 \end{bmatrix} \)
\(\therefore A^{-1} = \frac{\text{adj}(A)}{|A|} = \frac{-1}{5} \begin{bmatrix} -2 & -1 & 5 \\ -1 & 2 & -5 \\ 3 & -1 & -5 \end{bmatrix} \)
Now, \( X = A^{-1}B \)
\( \Rightarrow \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \frac{-1}{5} \begin{bmatrix} -2 & -1 & 5 \\ -1 & 2 & -5 \\ 3 & -1 & -5 \end{bmatrix} \begin{bmatrix} 1000 \\ 1500 \\ 600 \end{bmatrix} \)
\( = \frac{-1}{5} \begin{bmatrix} -500 \\ -1000 \\ -1500 \end{bmatrix} \)
\( \Rightarrow \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 100 \\ 200 \\ 300 \end{bmatrix} \Rightarrow x = 100, y = 200, z = 300 \)
Hence the money awarded for discipline, politeness and punctuality are ₹ 100, ₹ 200 and ₹ 300 respectively.
Apart from the above three values schools can award children for sincerity.

Question. A total amount of ₹ 7,000 is deposited in three different savings bank accounts with annual interest rates of 5%, 8% and \( 8\frac{1}{2}\% \) respectively. The total annual interest from these three accounts is ₹ 550. Equal amounts have been deposited in the 5% and 8% savings accounts. Find the amount deposited in each of the three accounts, with the help of matrices.
Answer: Let ₹ \( x \), ₹ \( y \) and ₹ \( z \) be deposited at the rates of interest 5%, 8% and \( 8\frac{1}{2}\% \) respectively.
According to question,
\( x + y + z = 7000 \)
\( x - y = 0 \)
\( x \cdot \frac{5}{100} + y \cdot \frac{8}{100} + z \cdot \frac{17}{2 \cdot 100} = 550 \)
\( \Rightarrow 10x + 16y + 17z = 110000 \)
The system of equations can be written as \( AX = B \)
where, \( A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & -1 & 0 \\ 10 & 16 & 17 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix} \) and \( B = \begin{bmatrix} 7000 \\ 0 \\ 110000 \end{bmatrix} \)
\( |A| = \begin{vmatrix} 1 & 1 & 1 \\ 1 & -1 & 0 \\ 10 & 16 & 17 \end{vmatrix} = 1(-17) - 1(17) + 1(16 + 10) = -8 \neq 0 \)
\(\therefore A^{-1} \) exists. So, system of equations has a unique solution and it is given by \( X = A^{-1}B \)
Now, \( A_{11} = -17, A_{12} = -17, A_{13} = 26 \),
\( A_{21} = -1, A_{22} = 7, A_{23} = -6 \),
\( A_{31} = 1, A_{32} = 1, A_{33} = -2 \)
\(\therefore \text{adj } A = \begin{bmatrix} -17 & -1 & 1 \\ -17 & 7 & 1 \\ 26 & -6 & -2 \end{bmatrix} \)
and \( A^{-1} = \frac{1}{|A|} \text{adj } A = \frac{1}{-8} \begin{bmatrix} -17 & -1 & 1 \\ -17 & 7 & 1 \\ 26 & -6 & -2 \end{bmatrix} \)
Now, \( X = A^{-1}B \)
\( \Rightarrow \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \frac{1}{-8} \begin{bmatrix} -17 & -1 & 1 \\ -17 & 7 & 1 \\ 26 & -6 & -2 \end{bmatrix} \begin{bmatrix} 7000 \\ 0 \\ 110000 \end{bmatrix} \)
\( = \frac{1}{-8} \begin{bmatrix} -9000 \\ -9000 \\ -38000 \end{bmatrix} = \begin{bmatrix} 1125 \\ 1125 \\ 4750 \end{bmatrix} \)
\( \Rightarrow x = 1125, y = 1125, z = 4750 \)

Question. Two schools, P and Q, want to award their selected students for the values of sincerity, truthfulness and hard work at the rate of ₹ \( x \), ₹ \( y \) and ₹ \( z \) for each respective value per student. School P awards its 2, 3 and 4 students on the above respective values with a total prize money of ₹ 4,600. School Q wants to award its 3, 2 and 3 students on the respective values with a total award money of ₹ 4,100. If the total amount of award money for one prize on each value is ₹ 1,500, using matrices find the award money for each value. Suggest one other value which the school can consider for awarding the students.
Answer: According to question, we have
\( x + y + z = 1500 \)
\( 2x + 3y + 4z = 4600 \)
\( 3x + 2y + 3z = 4100 \)
The system of equations can be written as \( AX = B \)
where, \( A = \begin{bmatrix} 1 & 1 & 1 \\ 2 & 3 & 4 \\ 3 & 2 & 3 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix} \) and \( B = \begin{bmatrix} 1500 \\ 4600 \\ 4100 \end{bmatrix} \)
\( |A| = \begin{vmatrix} 1 & 1 & 1 \\ 2 & 3 & 4 \\ 3 & 2 & 3 \end{vmatrix} = 1(9 - 8) - 1(6 - 12) + 1(4 - 9) = 1 + 6 - 5 = 2 \neq 0 \)
\(\therefore A^{-1} \) exists and so, system of equations has a unique solution given by \( X = A^{-1}B \)
Now, \( A_{11} = 1, A_{12} = 6, A_{13} = -5 \),
\( A_{21} = -1, A_{22} = 0, A_{23} = 1 \),
\( A_{31} = 1, A_{32} = -2, A_{33} = 1 \)
\(\therefore \text{adj } A = \begin{bmatrix} 1 & -1 & 1 \\ 6 & 0 & -2 \\ -5 & 1 & 1 \end{bmatrix} \)
\( A^{-1} = \frac{1}{|A|} \cdot \text{adj } A = \frac{1}{2} \begin{bmatrix} 1 & -1 & 1 \\ 6 & 0 & -2 \\ -5 & 1 & 1 \end{bmatrix} \)
Now, \( X = A^{-1} B \)
\( \Rightarrow \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \frac{1}{2} \begin{bmatrix} 1 & -1 & 1 \\ 6 & 0 & -2 \\ -5 & 1 & 1 \end{bmatrix} \begin{bmatrix} 1500 \\ 4600 \\ 4100 \end{bmatrix} \)
\( = \frac{1}{2} \begin{bmatrix} 1000 \\ 800 \\ 1200 \end{bmatrix} = \begin{bmatrix} 500 \\ 400 \\ 600 \end{bmatrix} \)
\( \Rightarrow x = 500; y = 400; z = 600. \)
Apart from sincerity, truthfulness and hard work, the schools can include an award for regularity.

Question. A school wants to award its students for the value of honesty, regularity and hard work with a total cash award of ₹ 6,000. Three times the award money for hard work added to that given for honesty amounts to ₹ 11,000. The award money given for honesty and hard work together is double the one given for regularity. Represent the above situation algebraically and find the award money for each value, using matrix method. Apart from these values, namely, honesty, regularity and hard work, suggest one more value which the school must include for awards.
Answer: Let the award money for honesty = ₹ \( x \)
Let the award money for regularity = ₹ \( y \)
Let the award money for hard work = ₹ \( z \)
According to question, we have
\( x + y + z = 6000 \)
\( 3z + x = 11000 \)
\( x + z - 2y = 0 \)
The system of equations can be written as \( AX = B \)
where, \( A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 0 & 3 \\ 1 & -2 & 1 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad B = \begin{bmatrix} 6000 \\ 11000 \\ 0 \end{bmatrix} \)
Now, \( |A| = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 0 & 3 \\ 1 & -2 & 1 \end{vmatrix} = 1(0 + 6) - 1(1 - 3) + 1(-2 - 0) = 6 - (-2) - 2 = 6 \neq 0 \)
\(\therefore A \) is invertible. So, the given system has a unique solution given by \( X = A^{-1}B \)
Now, \( A_{11} = 6, A_{12} = 2, A_{13} = -2 \)
\( A_{21} = -3, A_{22} = 0, A_{23} = 3 \)
\( A_{31} = 3, A_{32} = -2, A_{33} = -1 \)
\(\therefore \text{adj}(A) = \begin{bmatrix} 6 & -3 & 3 \\ 2 & 0 & -2 \\ -2 & 3 & -1 \end{bmatrix} \)
\( A^{-1} = \frac{1}{|A|} \text{adj}(A) = \frac{1}{6} \begin{bmatrix} 6 & -3 & 3 \\ 2 & 0 & -2 \\ -2 & 3 & -1 \end{bmatrix} \)
Now, \( X = A^{-1}B \)
\( \Rightarrow \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \frac{1}{6} \begin{bmatrix} 6 & -3 & 3 \\ 2 & 0 & -2 \\ -2 & 3 & -1 \end{bmatrix} \begin{bmatrix} 6000 \\ 11000 \\ 0 \end{bmatrix} \)
\( = \frac{1}{6} \begin{bmatrix} 3000 \\ 12000 \\ 21000 \end{bmatrix} = \begin{bmatrix} 500 \\ 2000 \\ 3500 \end{bmatrix} \)
\( \Rightarrow x = 500, y = 2000, z = 3500 \)
One more value which the school can include for awards is discipline.