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 }