Q 1. Prove that the determinant \begin{vmatrix}x&\sin\theta&\cos\theta\\-\sin\theta& x&1\\\cos\theta&1&x\end{vmatrix} is independent of \theta.
Solution:
Let \Delta = \begin{vmatrix}x&\sin\theta&\cos\theta\\-\sin\theta&-x&1\\\cos\theta&1&x\end{vmatrix}
To evaluate this, we will write the expansion of the determinant along the first row (R_1).
\Delta = x\begin{vmatrix}-x&1\\1&x\end{vmatrix} - \sin\theta\begin{vmatrix}-\sin\theta&1\\\cos\theta&x\end{vmatrix} + \cos\theta\begin{vmatrix}-\sin\theta&-x\\\cos\theta&1\end{vmatrix}
Now, calculate the second-order determinants:
\Delta = x(-x(x) - (1)(1)) - \sin\theta(-x\sin\theta - \cos\theta) + \cos\theta(-\sin\theta - (-x\cos\theta))
\Delta = x(-x^2 - 1) - \sin\theta(-x\sin\theta - \cos\theta) + \cos\theta(-\sin\theta + x\cos\theta)
Next, carefully multiply the terms into the brackets:
\Delta = -x^3 - x + x\sin^2\theta + \sin\theta\cos\theta - \sin\theta\cos\theta + x\cos^2\theta
Notice that the terms +\sin\theta\cos\theta and -\sin\theta\cos\theta cancel each other out.
We can then group the remaining terms containing x:
\Delta = -x^3 - x + x(\sin^2\theta + \cos^2\theta)
Using the fundamental trigonometric identity \sin^2\theta + \cos^2\theta = 1,
we substitute back into our equation:
\Delta = -x^3 - x + x(1)
\Delta = -x^3 - x + x
\Delta = -x^3
Final Answer: The final expression is -x^3, which does not contain the variable \theta. Therefore, the given determinant is entirely independent of \theta.
Q 2. Evaluate \begin{vmatrix}\cos\alpha\cos\beta&\cos\alpha\sin\beta&-\sin\alpha\\ \sin\beta&\cos\beta&0\\\sin\alpha\cos\beta&\sin\alpha\sin\beta&\cos\alpha\end{vmatrix}
Solution:
Let the determinant be \Delta = \begin{vmatrix}\cos\alpha\cos\beta&\cos\alpha\sin\beta&-\sin\alpha\\-\sin\beta&\cos\beta&0\\\sin\alpha\cos\beta&\sin\alpha\sin\beta&\cos\alpha\end{vmatrix}
Teacher’s Pro-Tip: For easier calculations, we shall expand the determinant along that row or column which contains the maximum number of zeros.
Since the second row (R_2) contains a zero, expanding along R_2 will save us a step!
\Delta = -(-\sin\beta)\begin{vmatrix}\cos\alpha\sin\beta&-\sin\alpha\\\sin\alpha\sin\beta&\cos\alpha\end{vmatrix} + \cos\beta\begin{vmatrix}\cos\alpha\cos\beta&-\sin\alpha\\\sin\alpha\cos\beta&\cos\alpha\end{vmatrix} - 0
Now, evaluate the 2\times 2 determinants carefully:
\Delta = \sin\beta(\cos^2\alpha\sin\beta - (-\sin^2\alpha\sin\beta)) + \cos\beta(\cos^2\alpha\cos\beta - (-\sin^2\alpha\cos\beta))
\Delta = \sin\beta(\cos^2\alpha\sin\beta + \sin^2\alpha\sin\beta) + \cos\beta(\cos^2\alpha\cos\beta + \sin^2\alpha\cos\beta)
Factor out the common terms from inside the brackets
(\sin\beta from the first bracket, and \cos\beta from the second):
\Delta = \sin\beta[\sin\beta(\cos^2\alpha + \sin^2\alpha)] + \cos\beta[\cos\beta(\cos^2\alpha + \sin^2\alpha)]
Apply the trigonometric identity \cos^2\alpha + \sin^2\alpha = 1:
\Delta = \sin\beta[\sin\beta(1)] + \cos\beta[\cos\beta(1)]
\Delta = \sin^2\beta + \cos^2\beta
\Delta = 1
Final Answer: The evaluated result of the determinant is 1.
Q 3: If A^{-1} = \begin{bmatrix}3&-1&1\\-15&6&-5\\5&-2&2\end{bmatrix} and B = \begin{bmatrix}1&2&-2\\-1&3&0\\0&-2&1\end{bmatrix}, find (AB)^{-1}.
Solution:
We know the property of inverses that (AB)^{-1} = B^{-1}A^{-1}.
Since we are already given A^{-1}, we just need to find B^{-1} and multiply them.
First, find the determinant of matrix B:
|B| = 1(3 - 0) - 2(-1 - 0) - 2(2 - 0)
|B| = 1(3) + 2 - 4 = 1
Since |B| \neq 0, B^{-1} exists.
Next, find the cofactor matrix of B:
- B_{11} = 3(1) - (-2)(0) = 3
- B_{12} = -(-1(1) - 0(0)) = 1
- B_{13} = -1(-2) - 0(3) = 2
- B_{21} = -(2(1) - (-2)(-2)) = -(2 - 4) = 2
- B_{22} = 1(1) - (-2)(0) = 1
- B_{23} = -(1(-2) - 0(2)) = -(-2) = 2
- B_{31} = 2(0) - 3(-2) = 6
- B_{32} = -(1(0) - (-1)(-2)) = -(-2) = 2
- B_{33} = 1(3) - (-1)(2) = 5
The Adjoint matrix adj B is the transpose of the cofactor matrix:
adj B = \begin{bmatrix}3&2&6\\1&1&2\\2&2&5\end{bmatrix}
Since B^{-1} = \frac{1}{|B|}adj B and |B| = 1:
B^{-1} = \begin{bmatrix}3&2&6\\1&1&2\\2&2&5\end{bmatrix}
Now, calculate (AB)^{-1} = B^{-1}A^{-1}:
(AB)^{-1} = \begin{bmatrix}3&2&6\\1&1&2\\2&2&5\end{bmatrix} \begin{bmatrix}3&-1&1\\-15&6&-5\\5&-2&2\end{bmatrix}
Multiply the matrices row by column:
(AB)^{-1} = \begin{bmatrix}9 - 30 + 30 & -3 + 12 - 12 & 3 - 10 + 12 \\ 3 - 15 + 10 & -1 + 6 - 4 & 1 - 5 + 4 \\ 6 - 30 + 25 & -2 + 12 - 10 & 2 - 10 + 10\end{bmatrix}
Final Answer: (AB)^{-1} = \begin{bmatrix}9&-3&5\\-2&1&0\\1&0&2\end{bmatrix}
Q 4: Let A = \begin{bmatrix}1&2&1\\2&3&1\\1&1&5\end{bmatrix}. Verify that:
(i) [adj A]^{-1} = adj(A^{-1})
(ii) (A^{-1})^{-1} = A
Solution:
Before we tackle parts (i) and (ii), we need to find the determinant (|A|), the adjoint (adj A), and the inverse (A^{-1}) of the given matrix.
Step 1: Find |A|
|A| = 1(15 - 1) - 2(10 - 1) + 1(2 - 3)
|A| = 1(14) - 2(9) + 1(-1)
|A| = 14 - 18 - 1 = -5
Since |A| = -5 \neq 0, the matrix is non-singular, and its inverse exists.
Step 2: Find adj A
Calculate the cofactors of matrix A:
- A_{11} = 15 - 1 = 14
- A_{12} = -(10 - 1) = -9
- A_{13} = 2 - 3 = -1
- A_{21} = -(10 - 1) = -9
- A_{22} = 5 - 1 = 4
- A_{23} = -(1 - 2) = 1
- A_{31} = 2 - 3 = -1
- A_{32} = -(1 - 2) = 1
- A_{33} = 3 - 4 = -1
Writing the cofactor matrix and taking its transpose gives us adj A:
adj A = \begin{bmatrix}14&-9&-1\\-9&4&1\\-1&1&-1\end{bmatrix}
Step 3: Find A^{-1}
A^{-1} = \frac{1}{|A|}(adj A) = -\frac{1}{5}\begin{bmatrix}14&-9&-1\\-9&4&1\\-1&1&-1\end{bmatrix} = \begin{bmatrix}-14/5 & 9/5 & 1/5 \\ 9/5 & -4/5 & -1/5 \\ 1/5 & -1/5 & 1/5\end{bmatrix}
Part (i): Verify [adj A]^{-1} = adj(A^{-1})
Calculating LHS: [adj A]^{-1}
Let B = adj A = \begin{bmatrix}14&-9&-1\\-9&4&1\\-1&1&-1\end{bmatrix}. We need to find B^{-1}.
First, find |B|:
|B| = 14(-4 - 1) - (-9)(9 - (-1)) + (-1)(-9 + 4)
|B| = 14(-5) + 9(10) - 1(-5) = -70 + 90 + 5 = 25
Now, find the cofactors of B to construct adj B:
- B_{11} = -4 - 1 = -5
- B_{12} = -(9 + 1) = -10
- B_{13} = -9 + 4 = -5
- B_{21} = -(9 + 1) = -10
- B_{22} = -14 - 1 = -15
- B_{23} = -(14 - 9) = -5
- B_{31} = -9 + 4 = -5
- B_{32} = -(14 - 9) = -5
- B_{33} = 56 - 81 = -25
Taking the transpose gives us adj B:
adj B = \begin{bmatrix}-5&-10&-5\\-10&-15&-5\\-5&-5&-25\end{bmatrix}
So, [adj A]^{-1} = B^{-1} = \frac{1}{|B|}(adj B) = \frac{1}{25}\begin{bmatrix}-5&-10&-5\\-10&-15&-5\\-5&-5&-25\end{bmatrix}
Dividing every element by 5 gives:
LHS: [adj A]^{-1} = \frac{1}{5}\begin{bmatrix}-1&-2&-1\\-2&-3&-1\\-1&-1&-5\end{bmatrix}
Calculating RHS: adj(A^{-1})
We need to find the cofactors of A^{-1} = \begin{bmatrix}-14/5 & 9/5 & 1/5 \\ 9/5 & -4/5 & -1/5 \\ 1/5 & -1/5 & 1/5\end{bmatrix}:
- C_{11} = \left(-\frac{4}{5}\right)\left(\frac{1}{5}\right) - \left(-\frac{1}{5}\right)\left(-\frac{1}{5}\right) = -\frac{4}{25} - \frac{1}{25} = -\frac{5}{25} = -\frac{1}{5}
- C_{12} = -\left(\frac{9}{25} - \left(-\frac{1}{25}\right)\right) = -\frac{10}{25} = -\frac{2}{5}
- C_{13} = -\frac{9}{25} - \left(-\frac{4}{25}\right) = -\frac{5}{25} = -\frac{1}{5}
- C_{21} = -\left(\frac{9}{25} - \left(-\frac{1}{25}\right)\right) = -\frac{10}{25} = -\frac{2}{5}
- C_{22} = -\frac{14}{25} - \frac{1}{25} = -\frac{15}{25} = -\frac{3}{5}
- C_{23} = -\left(\frac{14}{25} - \frac{9}{25}\right) = -\frac{5}{25} = -\frac{1}{5}
- C_{31} = -\frac{9}{25} - \left(-\frac{4}{25}\right) = -\frac{5}{25} = -\frac{1}{5}
- C_{32} = -\left(\frac{14}{25} - \frac{9}{25}\right) = -\frac{5}{25} = -\frac{1}{5}
- C_{33} = \frac{56}{25} - \frac{81}{25} = -\frac{25}{25} = -1 = -\frac{5}{5}
Taking the transpose gives us adj(A^{-1}):
adj(A^{-1}) = \begin{bmatrix}-1/5&-2/5&-1/5\\-2/5&-3/5&-1/5\\-1/5&-1/5&-5/5\end{bmatrix} = \frac{1}{5}\begin{bmatrix}-1&-2&-1\\-2&-3&-1\\-1&-1&-5\end{bmatrix}
Final Answer for (i): Since LHS = RHS, [adj A]^{-1} = adj(A^{-1}) is verified.
Part (ii): Verify (A^{-1})^{-1} = A
Calculating LHS: (A^{-1})^{-1}
To find the inverse of A^{-1}, we use the formula:
(A^{-1})^{-1} = \frac{1}{|A^{-1}|} adj(A^{-1})
We know from the properties of determinants that |A^{-1}| = \frac{1}{|A|}.
Since |A| = -5, we have |A^{-1}| = -\frac{1}{5}.
From Part (i), we already know that:
adj(A^{-1}) = \frac{1}{5}\begin{bmatrix}-1&-2&-1\\-2&-3&-1\\-1&-1&-5\end{bmatrix}
Substituting these values into our inverse formula:
(A^{-1})^{-1} = \frac{1}{-1/5} \left( \frac{1}{5}\begin{bmatrix}-1&-2&-1\\-2&-3&-1\\-1&-1&-5\end{bmatrix} \right)
(A^{-1})^{-1} = -5 \left( \frac{1}{5}\begin{bmatrix}-1&-2&-1\\-2&-3&-1\\-1&-1&-5\end{bmatrix} \right)
(A^{-1})^{-1} = -1 \begin{bmatrix}-1&-2&-1\\-2&-3&-1\\-1&-1&-5\end{bmatrix}
(A^{-1})^{-1} = \begin{bmatrix}1&2&1\\2&3&1\\1&1&5\end{bmatrix}
If we look at the matrix we just calculated, it is exactly the original matrix A!
Final Answer for (ii): LHS = RHS, therefore (A^{-1})^{-1} = A is verified.
Teacher’s Pro-Tip: While standard manual verification is expected in board exams, keep in mind for competitive exams like JEE that these are universal properties! For any invertible matrix A, (A^{-1})^{-1} will always return you right back to matrix A, and |adj A| is always equal to |A|^{n-1}.
Q 5: Evaluate the determinant \begin{vmatrix}x&y&x+y\\y&x+y&x\\x+y&x&y\end{vmatrix}
Solution:
Let \Delta = \begin{vmatrix}x&y&x+y\\y&x+y&x\\x+y&x&y\end{vmatrix}
Apply the column operation C_1 \rightarrow C_1 + C_2 + C_3:
\Delta = \begin{vmatrix}2(x+y)&y&x+y\\2(x+y)&x+y&x\\2(x+y)&x&y\end{vmatrix}
Take 2(x+y) common from the first column C_1:
\Delta = 2(x+y)\begin{vmatrix}1&y&x+y\\1&x+y&x\\1&x&y\end{vmatrix}
Apply row operations R_2 \rightarrow R_2 - R_1 and R_3 \rightarrow R_3 - R_1 to create zeros in the first column:
\Delta = 2(x+y)\begin{vmatrix}1&y&x+y\\0&x&-y\\0&x-y&-x\end{vmatrix}
Now, expand the determinant along the first column (C_1):
\Delta = 2(x+y) \left[ 1 \begin{vmatrix}x&-y\\x-y&-x\end{vmatrix} - 0 + 0 \right]
Evaluate the 2\times 2 determinant:
\Delta = 2(x+y) [ (x)(-x) - (-y)(x-y) ]
\Delta = 2(x+y) [ -x^2 + y(x-y) ]
\Delta = 2(x+y) [ -x^2 + xy - y^2 ]
Take a negative sign common from the bracket:
\Delta = -2(x+y)(x^2 - xy + y^2)
Using the algebraic identity (a+b)(a^2 - ab + b^2) = a^3 + b^3:
\Delta = -2(x^3 + y^3)
Final Answer: The evaluated value of the determinant is -2(x^3 + y^3).
Q 6: Evaluate the determinant \begin{vmatrix}1&x&y\\1&x+y&y\\1&x&x+y\end{vmatrix}
Solution:
Let \Delta = \begin{vmatrix}1&x&y\\1&x+y&y\\1&x&x+y\end{vmatrix}
To make evaluation simple, create zeros in the first column using row operations.
Apply R_2 \rightarrow R_2 - R_1 and R_3 \rightarrow R_3 - R_1:
\Delta = \begin{vmatrix}1&x&y\\0&y&0\\0&0&x\end{vmatrix}
Now, expand the determinant along the first column (C_1):
\Delta = 1\begin{vmatrix}y&0\\0&x\end{vmatrix} - 0 + 0
Evaluate the remaining 2\times 2 determinant:
\Delta = 1((y)(x) - (0)(0))
\Delta = xy
Final Answer: The evaluated value of the determinant is xy.
Q 7: Solve the 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
Solution:
Let \frac{1}{x} = u, \frac{1}{y} = v, and \frac{1}{z} = w.
The system of equations transforms into a linear format:
2u + 3v + 10w = 4
4u - 6v + 5w = 1
6u + 9v - 20w = 2
Express this in matrix form AX = B:
A = \begin{bmatrix}2&3&10\\4&-6&5\\6&9&-20\end{bmatrix}, X = \begin{bmatrix}u\\v\\w\end{bmatrix}, B = \begin{bmatrix}4\\1\\2\end{bmatrix}
Find |A|:
|A| = 2(120 - 45) - 3(-80 - 30) + 10(36 - (-36))
|A| = 2(75) - 3(-110) + 10(72) = 150 + 330 + 720 = 1200
Since |A| = 1200 \neq 0, A^{-1} exists.
Calculate the cofactors to find adj A:
- A_{11} = 120 - 45 = 75
- A_{12} = -(-80 - 30) = 110
- A_{13} = 36 + 36 = 72
- A_{21} = -(-60 - 90) = 150
- A_{22} = -40 - 60 = -100
- A_{23} = -(18 - 18) = 0
- A_{31} = 15 + 60 = 75
- A_{32} = -(10 - 40) = 30
- A_{33} = -12 - 12 = -24
adj A = \begin{bmatrix}75&150&75\\110&-100&30\\72&0&-24\end{bmatrix}
A^{-1} = \frac{1}{1200} \begin{bmatrix}75&150&75\\110&-100&30\\72&0&-24\end{bmatrix}
Solve for X using X = A^{-1}B:
X = \frac{1}{1200} \begin{bmatrix}75&150&75\\110&-100&30\\72&0&-24\end{bmatrix} \begin{bmatrix}4\\1\\2\end{bmatrix}
X = \frac{1}{1200} \begin{bmatrix}300 + 150 + 150 \\ 440 - 100 + 60 \\ 288 + 0 - 48\end{bmatrix}
X = \frac{1}{1200} \begin{bmatrix}600\\400\\240\end{bmatrix} = \begin{bmatrix}\frac{1}{2}\\\frac{1}{3}\\\frac{1}{5}\end{bmatrix}
So, u = \frac{1}{2}, v = \frac{1}{3}, and w = \frac{1}{5}.
Substitute back x = \frac{1}{u}, y = \frac{1}{v}, z = \frac{1}{w}:
x = 2, y = 3, z = 5.
Final Answer: The solution to the system is x = 2, y = 3, z = 5.
Q 8: If x, y, z are nonzero real numbers, then the inverse of matrix A=\begin{bmatrix}x&0&0\\0&y&0\\0&0&z\end{bmatrix} is
(A) \begin{bmatrix}x^{-1}&0&0\\0&y^{-1}&0\\0&0&z^{-1}\end{bmatrix}
(B) xyz\begin{bmatrix}x^{-1}&0&0\\0&y^{-1}&0\\0&0&z^{-1}\end{bmatrix}
(C) \frac{1}{xyz}\begin{bmatrix}x&0&0\\0&y&0\\0&0&z\end{bmatrix}
(D) \frac{1}{xyz}\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}
Solution:
Let the given matrix be A = \begin{bmatrix}x&0&0\\0&y&0\\0&0&z\end{bmatrix}.
To find A^{-1}, we first find the determinant |A|:
|A| = x(yz - 0) - 0 + 0 = xyz
Since x, y, z are nonzero real numbers, |A| = xyz \neq 0. Thus, A^{-1} exists.
Next, we find the cofactor of each element to form the Adjoint matrix:
- A_{11} = yz - 0 = yz
- A_{12} = -(0 - 0) = 0
- A_{13} = 0 - 0 = 0
- A_{21} = -(0 - 0) = 0
- A_{22} = xz - 0 = xz
- A_{23} = -(0 - 0) = 0
- A_{31} = 0 - 0 = 0
- A_{32} = -(0 - 0) = 0
- A_{33} = xy - 0 = xy
Writing these cofactors into a matrix and taking the transpose gives us adj A:
adj A = \begin{bmatrix}yz&0&0\\0&xz&0\\0&0&xy\end{bmatrix}
Now, substitute this into the inverse formula A^{-1} = \frac{1}{|A|}(adj A):
A^{-1} = \frac{1}{xyz}\begin{bmatrix}yz&0&0\\0&xz&0\\0&0&xy\end{bmatrix}
Multiply the scalar \frac{1}{xyz} inside the matrix:
A^{-1} = \begin{bmatrix}\frac{yz}{xyz}&0&0\\0&\frac{xz}{xyz}&0\\0&0&\frac{xy}{xyz}\end{bmatrix} = \begin{bmatrix}\frac{1}{x}&0&0\\0&\frac{1}{y}&0\\0&0&\frac{1}{z}\end{bmatrix}
Using negative exponents, this can be written as:
A^{-1} = \begin{bmatrix}x^{-1}&0&0\\0&y^{-1}&0\\0&0&z^{-1}\end{bmatrix}
Final Answer: The correct option is (A).
Teacher’s Pro-Tip: For any purely diagonal matrix, the inverse is simply a new diagonal matrix where every diagonal element is the reciprocal of the original diagonal elements! You can use this trick to save a lot of time in competitive exams like JEE.
Q 9: Let A=\begin{bmatrix}1&\sin\theta&1\\-\sin\theta&1&\sin\theta\\-1&-\sin\theta&1\end{bmatrix}, where 0\le\theta\le2\pi. Then
(A) Det(A)=0
(B) Det(A)\in(2,\infty)
(C) Det(A)\in(2,4)
(D) Det(A)\in[2,4]
Solution:
First, we need to find the determinant of matrix A:
|A| = \begin{vmatrix}1&\sin\theta&1\\-\sin\theta&1&\sin\theta\\-1&-\sin\theta&1\end{vmatrix}
Expanding along the first row (R_1):
|A| = 1(1 - (-\sin^2\theta)) - \sin\theta(-\sin\theta - (-\sin\theta)) + 1(\sin^2\theta - (-1))
|A| = 1(1 + \sin^2\theta) - \sin\theta(-\sin\theta + \sin\theta) + 1(\sin^2\theta + 1)
|A| = (1 + \sin^2\theta) - \sin\theta(0) + (\sin^2\theta + 1)
|A| = 1 + \sin^2\theta + \sin^2\theta + 1
|A| = 2 + 2\sin^2\theta
|A| = 2(1 + \sin^2\theta)
Now, we must find the range of this expression given the domain 0\le\theta\le2\pi.
We know that for any value of \theta, the value of \sin\theta lies between -1 and 1:
-1 \le \sin\theta \le 1
Therefore, the square of \sin\theta must lie between 0 and 1:
0 \le \sin^2\theta \le 1
Now, we build our determinant expression step-by-step from this inequality:
Multiply all parts by 2:
0 \le 2\sin^2\theta \le 2
Add 2 to all parts:
2 + 0 \le 2 + 2\sin^2\theta \le 2 + 2
2 \le 2(1 + \sin^2\theta) \le 4
Since Det(A) = 2(1 + \sin^2\theta), we can say:
2 \le Det(A) \le 4
In interval notation, this is written as Det(A) \in [2, 4].
(Notice we use square brackets because 2 and 4 are included in the range).
Final Answer: The correct option is (D).