Have you solved Class 12 Maths Matrices and Determinants Test – 2, 2027 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 Maths Matrices and Determinants Test – 2, 2027 ⇒
1 Mark Questions
Q 1. If A=\begin{bmatrix}-1&0&0\\ 0&1&0\\ 0&0&1\end{bmatrix}, then find A^{-1}.
Solution:
The given matrix A is a diagonal matrix.
The inverse of a diagonal matrix is simply a diagonal matrix with the reciprocals of the original diagonal elements.
A^{-1} = \begin{bmatrix}\frac{1}{-1}&0&0\\ 0&\frac{1}{1}&0\\ 0&0&\frac{1}{1}\end{bmatrix} = \begin{bmatrix}-1&0&0\\ 0&1&0\\ 0&0&1\end{bmatrix}
Notice that A^{-1} = A.
Final Answer: \boxed { \begin{bmatrix}-1&0&0\\ 0&1&0\\ 0&0&1\end{bmatrix} }
Q 2. If A is a square matrix of order 2 such that det(A)=4, then find det(4Adj~A).
Solution:
We know the properties of determinants:
- |kA| = k^n|A| where n is the order of the matrix.
- |Adj~A| = |A|^{n-1}
Here, order n = 2 and |A| = 4.
|4Adj~A| = 4^2 \times |Adj~A| = 16 \times |A|^{2-1} = 16 \times |A|^1 = 16 \times 4 = 64
Final Answer: \boxed{ 64 }
Q 3. Let A=\begin{bmatrix}1&-2&1\\ 0&4&-1\\ -3&2&1\end{bmatrix}, B=\begin{bmatrix}-2\\ -5\\ -7\end{bmatrix}, C=\begin{bmatrix}9&8&7\end{bmatrix}. Which of the following is defined?
Solution:
Let’s check the orders of the matrices:
Order of A = 3 \times 3
Order of B = 3 \times 1
Order of C = 1 \times 3
For matrix multiplication XY to be defined, the number of columns in X must equal the number of rows in Y.
- AB: (3 \times 3) \times (3 \times 1) \rightarrow Defined (Resulting order 3 \times 1)
- AC: (3 \times 3) \times (1 \times 3) \rightarrow Not defined
- BA: (3 \times 1) \times (3 \times 3) \rightarrow Not defined
Answer: a. Only AB
2 Marks Questions
Q 4. If \begin{bmatrix}x&4&1\end{bmatrix}\begin{bmatrix}2&1&2\\ 1&0&2\\ 0&2&-4\end{bmatrix}\begin{bmatrix}x\\ 4\\ -1\end{bmatrix}=0, find x.
Solution:
First, let’s multiply the first two matrices:
\begin{bmatrix}x(2) + 4(1) + 1(0) & x(1) + 4(0) + 1(2) & x(2) + 4(2) + 1(-4)\end{bmatrix} \begin{bmatrix}x\\ 4\\ -1\end{bmatrix} = 0
\begin{bmatrix}2x+4 & x+2 & 2x+4\end{bmatrix} \begin{bmatrix}x\\ 4\\ -1\end{bmatrix} = 0
Now, multiply the resulting row matrix by the column matrix:
(2x+4)(x) + (x+2)(4) + (2x+4)(-1) = 0
2x^2 + 4x + 4x + 8 - 2x - 4 = 0
2x^2 + 6x + 4 = 0
Dividing by 2:
x^2 + 3x + 2 = 0
(x+1)(x+2) = 0
Answer: \boxed{ x = -1 \text{ or } x = -2 }
Q 5. Evaluate \begin{vmatrix}a+ib&c+id\\ -c+id&a-ib\end{vmatrix}
Solution:
Cross-multiply to evaluate the 2 \times 2 determinant:
= (a+ib)(a-ib) - (c+id)(-c+id)
Using the identity (x+y)(x-y) = x^2 - y^2:
= (a^2 - i^2b^2) - (-c^2 + i^2d^2)
Since i^2 = -1:
= (a^2 - (-1)b^2) - (-c^2 + (-1)d^2)
= (a^2 + b^2) - (-c^2 - d^2)
= a^2 + b^2 + c^2 + d^2
Answer: \boxed{ a^2 + b^2 + c^2 + d^2 }
Q 6. Evaluate \begin{vmatrix}x+a&x&x\\ x&x+a&x\\ x&x&x+a\end{vmatrix}
Solution:
Apply column operation C_1 \rightarrow C_1 + C_2 + C_3:
\begin{vmatrix}3x+a&x&x\\ 3x+a&x+a&x\\ 3x+a&x&x+a\end{vmatrix}
Take (3x+a) common from C_1:
= (3x+a) \begin{vmatrix}1&x&x\\ 1&x+a&x\\ 1&x&x+a\end{vmatrix}
Apply row operations R_2 \rightarrow R_2 - R_1 and R_3 \rightarrow R_3 - R_1:
= (3x+a) \begin{vmatrix}1&x&x\\ 0&a&0\\ 0&0&a\end{vmatrix}
Expanding along C_1:
= (3x+a) [1(a^2 - 0)] = a^2(3x+a)
Answer: \boxed{ a^2(3x+a)}
Q 7. If A=\begin{bmatrix}0&0\\ 4&0\end{bmatrix}, find A^{16}.
Solution:
First, let’s find A^2:
A^2 = \begin{bmatrix}0&0\\ 4&0\end{bmatrix} \begin{bmatrix}0&0\\ 4&0\end{bmatrix} = \begin{bmatrix}0(0)+0(4) & 0(0)+0(0) \\ 4(0)+0(4) & 4(0)+0(0)\end{bmatrix} = \begin{bmatrix}0&0\\ 0&0\end{bmatrix} = O
Since A^2 is a zero matrix, any higher power of A will also be a zero matrix.
Therefore, A^{16} = O = \begin{bmatrix}0&0\\ 0&0\end{bmatrix}.
Final Answer: \boxed{ \begin{bmatrix}0&0\\ 0&0\end{bmatrix} }
Q 8. Express the matrix A=\begin{bmatrix}3&-4\\ 1&-1\end{bmatrix} as the sum of a symmetric and a skew-symmetric matrix.
Solution:
Any square matrix A can be expressed as A = P + Q,
where P = \frac{1}{2}(A+A') is symmetric and
Q = \frac{1}{2}(A-A') is skew-symmetric.
A' = \begin{bmatrix}3&1\\ -4&-1\end{bmatrix}
P = \frac{1}{2}(A+A') = \frac{1}{2} \left( \begin{bmatrix}3&-4\\ 1&-1\end{bmatrix} + \begin{bmatrix}3&1\\ -4&-1\end{bmatrix} \right) = \frac{1}{2} \begin{bmatrix}6&-3\\ -3&-2\end{bmatrix} = \begin{bmatrix}3&-3/2\\ -3/2&-1\end{bmatrix}
Q = \frac{1}{2}(A-A') = \frac{1}{2} \left( \begin{bmatrix}3&-4\\ 1&-1\end{bmatrix} - \begin{bmatrix}3&1\\ -4&-1\end{bmatrix} \right) = \frac{1}{2} \begin{bmatrix}0&-5\\ 5&0\end{bmatrix} = \begin{bmatrix}0&-5/2\\ 5/2&0\end{bmatrix}
Thus, the expression is:
Final Answer: \boxed{ \begin{bmatrix}3&-4\\ 1&-1\end{bmatrix} = \begin{bmatrix}3&-3/2\\ -3/2&-1\end{bmatrix} + \begin{bmatrix}0&-5/2\\ 5/2&0\end{bmatrix} }
3 Marks Questions
Q 9. If \begin{bmatrix}xy&4\\ z+6&x+y\end{bmatrix}=\begin{bmatrix}8&w\\ 0&6\end{bmatrix}, then find the value of x, y, z and w.
Solution:
By equating corresponding elements of the matrices:
- xy = 8
- w = 4
- z + 6 = 0 \Rightarrow z = -6
- x + y = 6 \Rightarrow y = 6 - x
Substitute y = 6 - x into the first equation:
x(6 - x) = 8
6x - x^2 = 8
x^2 - 6x + 8 = 0
(x-4)(x-2) = 0
So, x = 4 or x = 2.
If x = 4, then y = 2.
If x = 2, then y = 4.
Answers: w = 4, z = -6. The values of (x,y) are (4,2) or (2,4).
Q 10. Find the matrix A such that \begin{bmatrix}2&-1\\ 1&0\\ -3&4\end{bmatrix} A = \begin{bmatrix}-1&-8&-10\\ 1&-2&-5\\ 9&22&15\end{bmatrix}
Solution:
Let the given matrices be X_{3 \times 2} \times A_{m \times n} = Y_{3 \times 3}.
For multiplication to be possible and yield a 3 \times 3 matrix, A must be a 2 \times 3 matrix.
Let A = \begin{bmatrix}a&b&c\\ d&e&f\end{bmatrix}.
\begin{bmatrix}2&-1\\ 1&0\\ -3&4\end{bmatrix} \begin{bmatrix}a&b&c\\ d&e&f\end{bmatrix} = \begin{bmatrix}2a-d & 2b-e & 2c-f \\ a & b & c \\ -3a+4d & -3b+4e & -3c+4f\end{bmatrix}
Equating this to \begin{bmatrix}-1&-8&-10\\ 1&-2&-5\\ 9&22&15\end{bmatrix}:
From the second row, we directly get:
a = 1, b = -2, c = -5.
Substitute these into the first row equations:
2(1) - d = -1 \Rightarrow d = 3
2(-2) - e = -8 \Rightarrow e = 4
2(-5) - f = -10 \Rightarrow f = 0
Therefore, the matrix is:
A = \begin{bmatrix}1&-2&-5\\ 3&4&0\end{bmatrix}
Final Answer: \boxed{ \begin{bmatrix}1&-2&-5\\ 3&4&0\end{bmatrix}}
Q 11. If P(x)=\begin{bmatrix}cos~x&sin~x\\ -sin~x&cos~x\end{bmatrix}, then show that P(x)P(y)=P(x+y)=P(y)P(x).
Solution:
LHS
P(x)P(y) = \begin{bmatrix}\cos x&\sin x\\ -\sin x&\cos x\end{bmatrix} \begin{bmatrix}\cos y&\sin y\\ -\sin y&\cos y\end{bmatrix}
= \begin{bmatrix}\cos x \cos y - \sin x \sin y & \cos x \sin y + \sin x \cos y \\ -\sin x \cos y - \cos x \sin y & -\sin x \sin y + \cos x \cos y\end{bmatrix}
Using trigonometric addition formulas:
= \begin{bmatrix}\cos(x+y) & \sin(x+y) \\ -\sin(x+y) & \cos(x+y)\end{bmatrix} = P(x+y)
LHS = MS (Middle Side)
Similarly, calculating P(y)P(x) yields the same matrix.
Hence proved.
Q 12. If the points (x,-2), (5, 2) and (8, 8) are collinear, find x using determinant.
Solution:
For three points (x_1, y_1), (x_2, y_2), (x_3, y_3) to be collinear, the area of the triangle formed by them is 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}x&-2&1\\ 5&2&1\\ 8&8&1\end{vmatrix} = 0
Expanding along R_1:
x(2 - 8) - (-2)(5 - 8) + 1(40 - 16) = 0
-6x + 2(-3) + 24 = 0
-6x - 6 + 24 = 0
6x = 18 \Rightarrow x = 3
Final Answer: \boxed { x = 3}
5 Marks Questions
Q 13. For the matrix A=\begin{bmatrix}1&1&1\\ 1&2&-3\\ 2&-1&3\end{bmatrix}, show that A^3-6A^2+5A+11I_3=0. Hence find A^{-1}.
Solution:
First, calculate A^2:
A^2 = A \cdot A = \begin{bmatrix}1&1&1\\ 1&2&-3\\ 2&-1&3\end{bmatrix} \begin{bmatrix}1&1&1\\ 1&2&-3\\ 2&-1&3\end{bmatrix} = \begin{bmatrix}4&2&1\\ -3&8&-14\\ 7&-3&14\end{bmatrix}
Next, calculate A^3:
A^3 = A^2 \cdot A = \begin{bmatrix}4&2&1\\ -3&8&-14\\ 7&-3&14\end{bmatrix} \begin{bmatrix}1&1&1\\ 1&2&-3\\ 2&-1&3\end{bmatrix} = \begin{bmatrix}8&7&1\\ -23&27&-69\\ 32&-13&58\end{bmatrix}
Now, substitute into the equation A^3 - 6A^2 + 5A + 11I:
= \begin{bmatrix}8&7&1\\ -23&27&-69\\ 32&-13&58\end{bmatrix} - \begin{bmatrix}24&12&6\\ -18&48&-84\\ 42&-18&84\end{bmatrix} + \begin{bmatrix}5&5&5\\ 5&10&-15\\ 10&-5&15\end{bmatrix} + \begin{bmatrix}11&0&0\\ 0&11&0\\ 0&0&11\end{bmatrix}
Adding corresponding elements yields the zero matrix \begin{bmatrix}0&0&0\\ 0&0&0\\ 0&0&0\end{bmatrix} = O.
Hence shown.
To find A^{-1}:
Multiply the entire equation by A^{-1}:
A^{-1}(A^3 - 6A^2 + 5A + 11I) = A^{-1}(O)
A^2 - 6A + 5I + 11A^{-1} = O
11A^{-1} = -A^2 + 6A - 5I
A^{-1} = \frac{1}{11} \left( 6\begin{bmatrix}1&1&1\\ 1&2&-3\\ 2&-1&3\end{bmatrix} - \begin{bmatrix}4&2&1\\ -3&8&-14\\ 7&-3&14\end{bmatrix} - \begin{bmatrix}5&0&0\\ 0&5&0\\ 0&0&5\end{bmatrix} \right)
A^{-1} = \frac{1}{11} \begin{bmatrix}-3&4&5\\ 9&-1&-4\\ 5&-3&-1\end{bmatrix}
Final Answer: \boxed{ A^{-1} = \frac{1}{11} \begin{bmatrix}-3&4&5\\ 9&-1&-4\\ 5&-3&-1\end{bmatrix} }
Q 14. If A=\begin{bmatrix}2&3&1\\ 1&2&2\\ -3&1&-1\end{bmatrix}, find A^{-1} and hence solve the system of equations:
2x+y-3z=13
3x+2y+z=4
x+2y-z=8
Solution:
First, note that the coefficients of the given system of equations form the transpose of matrix A (A').
The system is A'X = B, where B = \begin{bmatrix}13\\ 4\\ 8\end{bmatrix}.
The solution will be X = (A')^{-1}B = (A^{-1})'B.
Let’s find A^{-1}.
Step 1: Find |A|
|A| = 2(-2 - 2) - 3(-1 + 6) + 1(1 + 6) = 2(-4) - 3(5) + 7 = -8 - 15 + 7 = -16
Since |A| \neq 0, A^{-1} exists.
Step 2: Find the Cofactors of A
C_{11} = -4, C_{12} = -5, C_{13} = 7
C_{21} = 4, C_{22} = 1, C_{23} = -11
C_{31} = 4, C_{32} = -3, C_{33} = 1
Step 3: Adjoint and Inverse
Adj~A = \begin{bmatrix}-4&4&4\\ -5&1&-3\\ 7&-11&1\end{bmatrix}
A^{-1} = \frac{1}{|A|} Adj~A = -\frac{1}{16} \begin{bmatrix}-4&4&4\\ -5&1&-3\\ 7&-11&1\end{bmatrix}
Step 4: Solve the equations
X = (A^{-1})' B
\begin{bmatrix}x\\ y\\ z\end{bmatrix} = -\frac{1}{16} \begin{bmatrix}-4&-5&7\\ 4&1&-11\\ 4&-3&1\end{bmatrix} \begin{bmatrix}13\\ 4\\ 8\end{bmatrix}
\begin{bmatrix}x\\ y\\ z\end{bmatrix} = -\frac{1}{16} \begin{bmatrix}-52 - 20 + 56 \\ 52 + 4 - 88 \\ 52 - 12 + 8\end{bmatrix}
\begin{bmatrix}x\\ y\\ z\end{bmatrix} = -\frac{1}{16} \begin{bmatrix}-16\\ -32\\ 48\end{bmatrix} = \begin{bmatrix}1\\ 2\\ -3\end{bmatrix}
Answer: \boxed{ x = 1, y = 2, z = -3 }