Have you solved Class 12 Maths Matrices and Determinants Test – 1 and are looking for a solutions to the questions? Here you’ll find complete, step-by-step NCERT-style solutions for every question. Designed for both students and teachers, these answers make board exam revision easier with clear explanations, formula highlights, and downloadable PDF options.
Class 12 Matrices and Determinants Test-1 2026 ⇒
Section A
Q1: Find the value of x if: \begin{vmatrix} 2 & 4 \\ 5 & 1 \end{vmatrix} = \begin{vmatrix} 2x & 4 \\ 6 & x \end{vmatrix}
(A) \sqrt{5}
(B) 9
(C) \sqrt{3}
(D) 25
Solution:
To solve for x , we must evaluate the determinants on both sides of the equation.
Step 1: Expand the Left Hand Side (LHS)
\text{LHS} = \begin{vmatrix} 2 & 4 \\ 5 & 1 \end{vmatrix}
\Rightarrow (2 \times 1) - (4 \times 5)
\Rightarrow 2 - 20
\Rightarrow -18
Step 2: Expand the Right Hand Side (RHS)
\text{RHS} = \begin{vmatrix} 2x & 4 \\ 6 & x \end{vmatrix}
\Rightarrow (2x \times x) - (4 \times 6)
\Rightarrow 2x^2 - 24
Step 3: Equate LHS and RHS
Now, we equate the calculated values:
-18 = 2x^2 - 24
Step 4: Solve for x
2x^2 = 24 - 18
2x^2 = 6
x^2 = 3
Taking the square root of both sides:
x = \pm \sqrt{3}
Correct Option: (C)
Q 2: If A=\begin{bmatrix}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix} , then the value of \text{Det}(A^n) is equal to:
(A) 3
(B) 1
(C) -1
(D) 2
Solution:
We need to find the determinant of A^n .
Recall the property of determinants: |A^n| = |A|^n .
Step 1: Find the determinant of matrix A
A=\begin{vmatrix}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{vmatrix}
|A| = (\cos \theta)(\cos \theta) - (\sin \theta)(-\sin \theta)
|A| = \cos^2 \theta + \sin^2 \theta
|A| = 1
Step 2: Calculate |A|^n
|A^n| = (1)^n = 1
Correct Option: (B)
Q 3: Let A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} be a square matrix such that \text{adj} A = A . Then (a+b+c+d) is equal to:
(A) 2a
(B) 2b
(C) 2c
(D) 0
Solution:
We are given that the Adjoint of A is equal to the matrix A itself.
Step 1: Find adj A
For a 2 \times 2 matrix A = \begin{bmatrix} a & b \ c & d \end{bmatrix} , the adjoint is found by swapping diagonal elements and changing the signs of non-diagonal elements:
\text{Adj}A = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}
Step 2: Equate adj A to A Given \text{adj} A = A :
A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} = \text{Adj}A = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}
Comparing corresponding elements:
d = a
-b = b \Rightarrow 2b = 0 \Rightarrow b = 0
-c = c \Rightarrow 2c = 0 \Rightarrow c = 0
Step 3: Calculate the sum (a + b + c + d) Substitute the values we found:
a+b+c+d = a+0+0+a
\text{Sum} = 2a
Correct Option: (A)
Q4: If the determinant is \Delta = \begin{vmatrix}-3 & 4 & 1 \\ 2 & 7 & 0 \\ 5 & 6 & -8\end{vmatrix} , find the value of a_{11}A_{11}+a_{21}A_{21}+a_{31}A_{31} , where A_{ij} is the cofactor of the element a_{ij} .
(A) 109
(B) 418
(C) 209
(D) 218
Solution:
The expression a_{11}A_{11}+a_{21}A_{21}+a_{31}A_{31} represents the expansion of the determinant along the first column.
Therefore, we simply need to calculate the value of the determinant.
Note: Determinant is essentially just the sum of the products of elements of any row (or column) with their corresponding cofactors. The question was testing this definition directly rather than asking for a blind calculation!
Step 1:
Expand along Column 1
\Delta = a_{11}A_{11}+a_{21}A_{21}+a_{31}A_{31}
Element a_{11} = -3 . Cofactor: \begin{vmatrix} 7 & 0 \ 6 & -8 \end{vmatrix} = -56 - 0 = -56
Element a_{21} = 2 . Cofactor: - \begin{vmatrix} 4 & 1 \ 6 & -8 \end{vmatrix} = -(-32 - 6) = 38
Element a_{31} = 5 . Cofactor: \begin{vmatrix} 4 & 1 \ 7 & 0 \end{vmatrix} = 0 - 7 = -7
Step 2: Sum the products
\Delta = (-3)(-56) + (2)(38) + (5)(-7)
\Delta = 168 + 76 - 35
\Delta = 244 - 35
\Delta = 209
Correct Option: (C)
Q5: If A=\begin{bmatrix}1 & 4 & x \\ z & 2 & y \\ -3 & -1 & 3\end{bmatrix} is a symmetric matrix, then the value of x+y+z is:
(A) 10
(B) 6
(C) 8
(D) 0
Solution:
A matrix is symmetric if A = A^T (the matrix equals its transpose).
This implies that a_{ij} = a_{ji} for all i, j .
Step 1: Compare conjugate elements
a_{12} = a_{21} \Rightarrow 4 = z
a_{13} = a_{31} \Rightarrow x = -3
a_{23} = a_{32} \Rightarrow y = -1
Step 2: Find the sum
x + y + z = (-3) + (-1) + 4
x + y + z = -4 + 4 = 0
Correct Option: (D)
Q6: If A is a square matrix of order 3 and |A|=5 , then the value of |A \cdot (\text{adj} A)| is:
Solution:
We use the fundamental property: A \cdot (\text{adj} A) = |A| I .
Step 1: Express the product matrix
Since |A| = 5 and the order is 3:
A \cdot (\text{adj} A) = 5 \text{I}
A \cdot (\text{adj} A) = \begin{bmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{bmatrix}
Step 2: Calculating determinant
|A \cdot (\text{adj} A)| = ||A| I|
|A \cdot (\text{adj} A)| = (|A|)^n (where n is the order of the matrix)
|A \cdot (\text{adj} A)| = 5^3
|A \cdot (\text{adj} A)| = 125
Correct Option : (B)
Q7: If A=\begin{bmatrix}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix} and A+A^{T}=I , then the value of \theta is equal to:
(A) \frac{\pi}{3}
(B) \frac{\pi}{4}
(C) \frac{\pi}{5}
(D) \frac{\pi}{6}
Solution:
We are given the condition A + A^T = I .
Step 1: Write A and A^T
A=\begin{bmatrix}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}
A^T = \begin{bmatrix}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix} (Rows become columns)
Step 2: Add A and A^T
A + A^T=\begin{bmatrix}\cos \theta + \cos \theta & -\sin \theta + \sin \theta \\ \sin \theta -\sin \theta & \cos \theta + \cos \theta \end{bmatrix}
A + A^T = \begin{bmatrix} 2\cos \theta & 0 \\ 0 & 2\cos \theta \end{bmatrix}
Step 3: Equate to Identity Matrix I
\begin{bmatrix} 2\cos \theta & 0 \\ 0 & 2\cos \theta \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}
Comparing the corresponding elements:
2 \cos \theta = 1
\cos \theta = \frac{1}{2}
Step 4: Find \theta
The principal value for \cos \theta = 1/2 is 60^{\circ} or \frac{\pi}{3} .
Correct Option: (A)
Question 8: A and B are skew-symmetric matrices of the same order. AB is symmetric if:
(A) AB=0
(B) AB=-BA
(C) AB=BA
(D) BA=0
Solution:
We need to find the condition under which the product AB is symmetric.
Step 1: Use properties of skew-symmetric matrices
Since A and B are skew-symmetric:
A^T = -A
B^T = -B
Step 2: Apply definition of symmetric matrix to AB
For AB to be symmetric, its transpose must equal itself:
(AB)^T = AB
Step 3: Expand the transpose
Using the reversal law of transposes (AB)^T = B^T A^T :
B^T A^T= AB :
Substitute the skew-symmetric values:
(-B)(-A) = AB
BA = AB
Thus, the matrices A and B must commute.
Correct Option: (C)
Question 9: If A=\begin{bmatrix}2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 5\end{bmatrix} , then A^{-1} is:
(A) \begin{bmatrix}\frac{1}{2} & 0 & 0 \\ 0 & \frac{1}{3} & 0 \\ 0 & 0 & \frac{1}{5}\end{bmatrix}
(B) \frac{1}{30} \begin{bmatrix}2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 5\end{bmatrix}
(C) …
(D) …
Solution:
Matrix A is a diagonal matrix.
Step 1: Property of Diagonal Matrices
The inverse of a diagonal matrix \text{diag}(d_1, d_2, d_3) is simply \text{diag}(\frac{1}{d_1}, \frac{1}{d_2}, \frac{1}{d_3}) .
Step 2: Calculate the Inverse
A^{-1} = \begin{bmatrix} \frac{1}{2} & 0 & 0 \\ 0 & \frac{1}{3} \\ 0 & 0 & \frac{1}{5} \end{bmatrix}
Correct Option: (A)
Question 10: If A=[a_{ij}] is an identity matrix, then which of the following is true?
(A) a_{ij} = \begin{cases} 0, & \text{if } i=j \\ 1, & \text{if } i \neq j \end{cases}
(B) a_{ij} = 1 for all i, j
(C) a_{ij} = 0 for all i, j
(D) a_{ij} = \begin{cases} 1, & \text{if } i=j \\ 0, & \text{if } i \neq j \end{cases}
Solution:
This is the standard definition of an Identity Matrix.
An identity matrix has 1s on the main diagonal (where row index equals column index, i=j ) and 0s everywhere else (where i \neq j ).
Correct Option: (D)
Q 11: Given A \cdot (\text{adj} A)=\begin{bmatrix}3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3\end{bmatrix} , find the value of |A| + |\text{adj} A| .
Solution:
We need to use the fundamental properties of the adjoint matrix and determinants.
Step 1: Identify the value of |A|
Recall the property: A \cdot (\text{adj} A) = |A| I .
The given matrix can be rewritten as:
A \cdot (\text{adj} A)=3 \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}
A \cdot (\text{adj} A)=3 I
Comparing this with the property, we get:
|A|I=3 I
|A| = 3
Step 2: Identify the order of the matrix
Since the given matrix has 3 rows and 3 columns, the order n = 3 .
Step 3: Calculate |\text{adj} A|
We use the property: |\text{adj} A| = |A|^{n-1} Substitute |A|=3 and n=3 :
|\text{adj} A| = 3^{3-1}
|\text{adj} A| = 3^2 = 9
Step 4: Find the final sum
|A| + |\text{adj} A| = 3 + 9 = 12
Final Answer: \boxed{ |A| + |\text{adj} A| = 12 }
Q12: If the area of a triangle is 35 sq units with vertices (2, -6), (5, 4), and (k, 4), then find the value of k.
Solution:
The area of a triangle with vertices (x_1, y_1), (x_2, y_2), (x_3, y_3) is given by the determinant formula.
Step 1: Set up the determinant equation
\text{Area} = \frac{1}{2} \left| \begin{matrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{matrix} \right|
Since the area is 35, the value of the determinant can be positive or negative:
\text{Area} = \frac{1}{2} \begin{vmatrix} 2 & -6 & 1 \\ 5 & 4 & 1 \\ k & 4 & 1 \end{vmatrix} = \pm 35
\begin{vmatrix} 2 & -6 & 1 \\ 5 & 4 & 1 \\ k & 4 & 1 \end{vmatrix} = \pm 70
Step 2: Expand the determinant
Expanding along the first column (or any convenient row/column):
2(4-4) - (-6)(5-k) + 1(20-4k) = \pm 70
2(0) + 6(5 - k) + 20 - 4k = \pm 70
30 - 6k + 20 - 4k = \pm 70
50 - 10k = \pm 70
Step 3: Solve for both cases
Case 1 (Positive 70):
50 - 10k = 70
- 10k = 20
k = -2
Case 2 (Negative 70):
50 - 10k = -70
-10k = -120
k = 12
Final Answer: \boxed{ k = 12, -2 }
Teacher’s Note: It is crucial to take \pm 35 .
A common error in board exams is taking only the positive value, which leads to finding only one value for k instead of two, resulting in a loss of marks.
Q13: If the matrix A=\begin{bmatrix}6 & x & 2 \\ 2 & -1 & 2 \\ -10 & 5 & 2\end{bmatrix} is a singular matrix, find the value of x.
Solution:
A square matrix is called a singular matrix if its determinant is equal to zero (i.e., |A| = 0 ).
Step 1: Set up the determinant equation
A=\begin{bmatrix}6 & x & 2 \\ 2 & -1 & 2 \\ -10 & 5 & 2\end{bmatrix} = 0
Step 2: Expand along Row 1
6 \begin{vmatrix} -1 & 2 \\ 5 & 2 \end{vmatrix} - x \begin{vmatrix} 2 & 2 \\ -10 & 2 \end{vmatrix} + \begin{vmatrix} 2 & -1 \\ -10 & 5 \end{vmatrix} = 0
Step 3: Calculate the 2×2 determinants
- Term 1: 6[(-1)(2) - (2)(5)] = 6[-2 - 10] = 6(-12) = -72
- Term 2: -x[(2)(2) - (2)(-10)] = -x[4 - (-20)] = -x(24) = -24x
- Term 3: 2[(2)(5) - (-1)(-10)] = 2[10 - 10] = 2(0) = 0
Step 4: Solve for x
Combine the terms:
-72 -24x + 0 = 0
-24x = 72
x = \frac{72}{-24}
x = -3
Final Answer: \boxed{ x = -3 }
Q 14: If A=\begin{bmatrix}3 & 1 \\ -1 & 2\end{bmatrix} , show that A^{2}-5A+7I=0 . Hence find A^{-1} .
Solution:
This is a two-part question: first prove the equation, then use it to find the inverse.
Part 1: Show A^2 - 5A + 7I = 0
Calculate A^2 :
A^2 = A \cdot A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 2 & 1 \\ -1 & 2 \end{bmatrix}
A^2 = \begin{bmatrix} (3)(3)+(1)(-1) & (3)(1)+(1)(2) \ (-1)(3)+(2)(-1) & (-1)(1)+(2)(2) \end{bmatrix}
A^2 = \begin{bmatrix} 9-1 & 3+2 \\ -3-2 & -1+4 \end{bmatrix} = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix}
Substitute into the expression:
\text{LHS} = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix} - 5\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} + 7\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}
\text{LHS} = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix} - \begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix} + \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}
\text{LHS} = \begin{bmatrix} 8 - 15 + 7 & 5-5+0 \\ -5-(-5)+0 & 3-10+7 \end{bmatrix}
\text{LHS} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = \text{RHS}
Hence Proved.
Part 2: Find A^{-1} using the equation
Multiply the equation by A^{-1} :
A^2 - 5A + 7I = 0
Multiply by A^{-1} :
A^{-1}A^2 - 5A^{-1}A + 7A^{-1}I = A^{-1}0
A - 5I + 7A^{-1} = 0
7A^{-1} = 5I - A
7A^{-1} = 5\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} - \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}
7A^{-1} = \begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix} - \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}
7A^{-1} = \begin{bmatrix} 2 & -1 \\ 1 & 3 \end{bmatrix}
A^{-1} = \frac{1}{7} \begin{bmatrix} 2 & -1 \\ 1 & 3 \end{bmatrix}
Final Answer: A^{-1} = \boxed{ \frac{1}{7} \begin{bmatrix} 2 & -1 \\ 1 & 3 \end{bmatrix} }
Teacher’s Note: In Board exams, explicitly verify the “Hence find A^{-1}” instruction. Many students calculate the inverse using the determinant/adjoint method instead of deriving it from the equation. This often leads to a deduction of marks because the question specifically tests the algebraic manipulation of the matrix polynomial.
Q 15: If A, B are square matrices of the same order, then prove that \text{adj}(AB) = \text{adj}(B) \cdot \text{adj}(A) .
Solution:
We know that for any invertible square matrix X, X^{-1} = \frac{1}{|X|} \text{adj}(X) ,
which implies \text{adj}(X) = |X| X^{-1} .
Step 1: Apply this definition to matrix AB
\text{adj}(AB) = |AB| (AB)^{-1}
Step 2: Use the Reversal Law of Inverses
Recall that (AB)^{-1} = B^{-1} A^{-1} .
\text{adj}(AB) = |AB| (B^{-1} A^{-1})
Step 3: Use the determinant property
Recall that |AB| = |A||B| .
\text{adj}(AB) = |A||B| (B^{-1} A^{-1})
Step 4: Rearrange the scalar terms
Since determinants are scalars, we can associate them with their respective matrices:
\text{adj}(AB) = (|B| B^{-1}) \cdot (|A|A^{-1})
Step 5: Substitute back the definition of Adjoint
Since |B| B^{-1} = \text{adj}(B) and |A| A^{-1} = \text{adj}(A) :
\text{adj}(AB) = \text{adj}(B) \cdot \text{adj}(A)
Hence Proved
Q16: Find the value of K, such that the following points are collinear: A(-3,7) , B(7,k) and C(2,1) .
Solution:
Three points are collinear if the area of the triangle formed by them is zero.
Step 1: Set the area determinant to zero
\frac{1}{2} \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} = 0
\begin{vmatrix} -3 & 7 & 1 \\ 7 & k & 1 \\ 2 & 1 & 1 \end{vmatrix} = 0
Step 2: Expand the determinant
Expand along Row 1:
-3( k - 1) - 7(7 - 2) + 1(7 - 2k) = 0
Step 3: Simplify the equation
-3k + 3 - 7(5) + 7 - 2k = 0
-3k + 3 - 35 + 7 - 2k = 0
-3k -2k + 3 - 35 + 7 = 0
-5k - 25 = 0
Step 4: Solve for K
-5k = 25
k = -5
Final Answer: \boxed {k = -5}
Question 17: Given A = \begin{bmatrix}3 & 4 & 2 \\ 0 & 2 & -3 \\ 1 & -2 & 6\end{bmatrix} , find A^{-1} . Hence solve the system of equations:
3x + 4y + 2z = 8
2y - 3z = 3
x - 2y + 6z = -2
Solution:
Part 1: Find A^{-1}
To find the inverse, we calculate the determinant and the adjoint matrix.
Step 1: Calculate Determinant |A|
Expanding along Row 1:
|A| = 3(12 - 6) - 4(0 - (-3)) + 2 (0 - 2)
|A| = 3(6) - 4(3) + 2(-2)
|A| = 18 - 12 -4
|A| = 2
Since |A| \neq 0 , the inverse exists.
Step 2: Find the Cofactors
C_{11} = +(12 - 6) = 6
C_{12} = -(0 + 3) = -3
C_{13} = +(0 - 2) = -2
C_{21} = -(24 + 4) = -28
C_{22} = +(18 - 2) = 16
C_{23} = -(-6 - 4) = 10
C_{31} = +(-12 - 4) = -16
C_{33} = -(-9 - 0) = 9
C_{33} = +(6 - 0) = 6
Step 3: Write the Adjoint Matrix (Transpose of Cofactors)
\text{adj} A = \begin{bmatrix} 6 & -28 & -16 \\ -3 & 16 & 9 \\ -2 & 10 & 6 \end{bmatrix}
Step 4: Write the Inverse
A^{-1} = \frac{1}{|A|} (\text{adj} A)
A^{-1} = \frac{1}{2} \begin{bmatrix} 6 & -28 & -16 \\ -3 & 16 & 9 \\ -2 & 10 & 6 \end{bmatrix}
Part 2: Solve the System of Equations
The system can be written as AX = B , where:
A = \begin{bmatrix}3 & 4 & 2 \\ 0 & 2 & -3 \\ 1 & -2 & 6\end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad B = \begin{bmatrix} 8 \ 3 \ -2 \end{bmatrix}
The solution is given by X = A^{-1} B .
X = \frac{1}{2} \begin{bmatrix} 6 & -28 & -16 \\ -3 & 16 & 9 \\ -2 & 10 & 6 \end{bmatrix} \begin{bmatrix} 8 \ 3 \ -2 \end{bmatrix}
X = \frac{1}{2} \begin{bmatrix} (6)(8) + (-28)(3) + (-16)(-2) \\ (-3)(8) + (16)(3) + (9)(-2) \\ (-2)(8) + (10)(3) + (6)(-2) \end{bmatrix}
X = \frac{1}{2} \begin{bmatrix} 48 - 84 + 32 \\ -24 + 48 - 18 \\ -16 + 30 - 12 \end{bmatrix}
X = \frac{1}{2} \begin{bmatrix} -4 \\ 6 \\ 2 \end{bmatrix}
\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} -2 \\ 3 \\ 1 \end{bmatrix}
Final Answer: \boxed{ x = -2, y = 3, z = 1}
Pro-Tip for Students: After finding x=-2, y=3, z=1 , strictly advise students to perform a Mental Check.
Plug these values into the simplest equation, e.g., 2y - 3z = 3 . Check: 2(3) - 3(1) = 6 - 3 = 3 . It matches! This 10-second check can save them 5 marks.
Q18: If A=\begin{bmatrix}1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3\end{bmatrix} , then show that A^{3}-6A^{2}+7A+2I=0 .
Solution:
Step 1: Calculate A^2
A^2 = A \cdot A = \begin{bmatrix}1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3\end{bmatrix} \begin{bmatrix}1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3\end{bmatrix}
A^2 = \begin{bmatrix} 1+0+4 & 0+0+0 & 2+0+6 \\ 0+0+2 & 0+4+0 & 0+2+3 \\ 2+0+6 & 0+0+0 & 4+0+9 \end{bmatrix}
A^2 = \begin{bmatrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{bmatrix}
Step 2: Calculate A^3
A^3 = \begin{bmatrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{bmatrix} \begin{bmatrix}1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3\end{bmatrix}
A^3 = \begin{bmatrix} 5+0+16 & 0+0+0 & 10+0+24 \\ 2+0+10 & 0+8+0 & 4+4+15 \\ 8+0+26 & 0+0+0 & 16+0+39 \end{bmatrix}
A^3 = \begin{bmatrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{bmatrix}
Step 3: Substitute into the LHS equation
\text{LHS} = A^3 - 6A^2 + 7A + 2I
\text{LHS} = \begin{bmatrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{bmatrix} - 6\begin{bmatrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{bmatrix} + 7\begin{bmatrix}1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3\end{bmatrix} + 2\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}
Step 4: Add corresponding elements
\text{LHS} = \begin{bmatrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{bmatrix} - \begin{bmatrix} 30 & 0 & 48 \\ 12 & 24 & 30 \\ 48 & 0 & 78 \end{bmatrix} + \begin{bmatrix}7 & 0 & 14 \\ 0 & 14 & 7 \\ 14 & 0 & 21\end{bmatrix} + \begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix}
\text{LHS} = \begin{bmatrix} 21 - 30 + 7 + 2 & 0 + 0 +0 + 0 & 34 - 48 + 14 + 0 \\ 12 -12 + 0 + 0 & 8 - 24 + 14 + 2 & 23 - 30 + 7 + 0 \\ 34 - 48 + 14 + 0 & 0 + 0 + 0 + 0 & 55 - 78 + 21 + 0 \end{bmatrix}
\text{LHS} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}
\text{LHS} = 0
\text{LHS} = \text{RHS}
Hence Proved.
Case Study:
Q19 A scholarship is a sum of money provided to a student to help him or her pay for education. Some students are granted scholarships based on their academic achievements, while others are rewarded based on their financial needs.
Every year a school offers scholarships to girl children and meritorious achievers based on certain criteria. In the session 2022–23, the school offered monthly scholarships of ₹3,000 each to some girl students and ₹4,000 each to meritorious achievers in academics as well as sports.
In all, 50 students were given the scholarships, and the monthly expenditure incurred by the school on scholarships was ₹1,80,000.
Based on the above information, answer the following questions:
(i) Express the given information algebraically using matrices.
(ii) Check whether the system of matrix equations so obtained is consistent or not.
(iii) (A) Find the number of scholarships of each kind given by the school using matrices.
OR
(B). Had the amount of scholarship given to each girl child and meritorious student been interchanged, what would be the monthly expenditure incurred by the school?
Solution:
Given Information:
Scholarship for Girl Students: ₹3,000 per month.
Scholarship for Meritorious Achievers: ₹4,000 per month.
Total number of students: 50.
Total monthly expenditure: ₹1,80,000.
Let x denote the number of girl students and y denote the number of meritorious achievers.
From the given data, we can form the following linear equations:
Total students: x + y = 50
Total expenditure: 3000x + 4000y = 1,80,000
3x + 4y = 180
(i): Express the given information algebraically using matrices.
Solution: The system of linear equations can be written in the form AX = B , where:
A = \begin{bmatrix} 1 & 1 \\ 3 & 4 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \ \end{bmatrix}, \quad B = \begin{bmatrix} 50 \\ 180 \ \end{bmatrix}
\begin{bmatrix} 1 & 1 \\ 3 & 4 \end{bmatrix} \begin{bmatrix} x \\ y \ \end{bmatrix}= \begin{bmatrix} 50 \\ 180 \ \end{bmatrix}
(ii): Check whether the system of matrix equations so obtained is consistent or not.
Solution: To check consistency, we find the determinant of matrix A ( |A| ).
|A| = \begin{vmatrix} 1 & 1 \\ 3 & 4 \end{vmatrix}
|A| = (1)(4) - (1)(3)
|A| = 4 - 3
|A| = 1
Since |A| \neq 0 , the matrix is non-singular.
Final Answer: The system has a unique solution and is consistent.
Part (iii) (A): Find the number of scholarships of each kind given by the school using matrices.
Solution:
We use the formula X = A^{-1}B .
Step 1: Find A^{-1}
A^{-1} = \frac{1}{|A|} \text{adj}(A)
\text{adj}(A) = \begin{bmatrix} 4 & -1 \\ -3 & 1 \end{bmatrix}
Since |A| = 1 :
A^{-1} = \begin{bmatrix} 4 & -1 \\ -3 & 1 \end{bmatrix}
Step 2: Calculate X
X = \begin{bmatrix} 4 & -1 \\ -3 & 1 \end{bmatrix} \begin{bmatrix} 50 \\ 180 \end{bmatrix}
\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} (4)(50) + (-1)(180) \\ (-3)(50) + (1)(180) \end{bmatrix}
Final Answer: \boxed{ \text{Number of Girl Students Scholarships: } = 20}, \boxed{\text{Number of Meritorious Achievers Scholarships: } = 30}
(iii) (B): Had the amount of scholarship given to each girl child and meritorious student been interchanged, what would be the monthly expenditure incurred by the school?
Solution:
New amount for Girl Students: ₹4,000
New amount for Meritorious Achievers: ₹3,000
Number of students (calculated above): x = 20 , y = 30
\text{New Expenditure} = (4000 \times 20) + (3000 \times 30)
\text{New Expenditure} = 80,000 + 90,000
\text{New Expenditure} = 1,70,000
Final Answer: The new monthly expenditure would be ₹1,70,000.