GSEB Class 12 Maths Solutions Chapter 4 નિશ્ચાયક Misc. Ques

Question 1. Prove that the determinant \( \left|\begin{array}{ccc} x & \sin \theta & \cos \theta \\ -\sin \theta & -x & 1 \\ \cos \theta & 1 & x \end{array}\right| \) is independent of \( \theta \).
Answer: Let \( A = \left|\begin{array}{ccc} x & \sin \theta & \cos \theta \\ -\sin \theta & -x & 1 \\ \cos \theta & 1 & x \end{array}\right| \)
Expanding the determinant:
\( A = x(-x^2 - 1) - \sin \theta(-\sin \theta - \cos \theta) + \cos \theta(-\sin \theta + x \cos \theta) \)
\( A = -x^3 - x + x \sin^2 \theta + \sin \theta \cos \theta - \sin \theta \cos \theta + x \cos^2 \theta \)
\( A = -x^3 - x + x(\sin^2 \theta + \cos^2 \theta) \)
\( A = -x^3 - x + x(1) \)
\( A = -x^3 \), which is independent of \( \theta \).
In simple words: After expanding the determinant and simplifying the trigonometric terms using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \), the variable \( \theta \) cancels out, leaving an expression that only depends on \( x \).

Exam Tip: Always look for the \( \sin^2 \theta + \cos^2 \theta = 1 \) identity when trigonometric functions appear in determinants to simplify expressions quickly.

Question 2. Without expanding the determinant, prove that: \( \left|\begin{array}{ccc} a & a^2 & bc \\ b & b^2 & ca \\ c & c^2 & ab \end{array}\right| = \left|\begin{array}{ccc} 1 & a^2 & a^3 \\ 1 & b^2 & b^3 \\ 1 & c^2 & c^3 \end{array}\right| \)
Answer: Multiplying \( R_1, R_2, R_3 \) by \( a, b, c \) respectively:
\( \frac{1}{abc} \left|\begin{array}{ccc} a^2 & a^3 & abc \\ b^2 & b^3 & abc \\ c^2 & c^3 & abc \end{array}\right| \)
Taking \( abc \) common from \( C_3 \):
\( \frac{abc}{abc} \left|\begin{array}{ccc} a^2 & a^3 & 1 \\ b^2 & b^3 & 1 \\ c^2 & c^3 & 1 \end{array}\right| = \left|\begin{array}{ccc} a^2 & a^3 & 1 \\ b^2 & b^3 & 1 \\ c^2 & c^3 & 1 \end{array}\right| \)
By swapping columns \( C_3 \leftrightarrow C_2 \) and then \( C_2 \leftrightarrow C_1 \), we get the required determinant.
In simple words: By multiplying rows by variables and factoring out common terms, we transform the left side into the right side without performing the full expansion.

Exam Tip: When proving determinant identities, remember that multiplying a row by a constant requires dividing the whole determinant by that same constant.

Question 3. Find the value of \( \left|\begin{array}{ccc} \cos \alpha \cos \beta & \cos \alpha \sin \beta & -\sin \alpha \\ -\sin \beta & \cos \beta & 0 \\ \sin \alpha \cos \beta & \sin \alpha \sin \beta & \cos \alpha \end{array}\right| \).
Answer: Expanding along the first row:
\( = \cos \alpha \cos \beta (\cos \alpha \cos \beta - 0) - \cos \alpha \sin \beta (-\sin \beta \cos \alpha - 0) - \sin \alpha (-\sin \beta \sin \alpha \sin \beta - \cos \beta \sin \alpha \cos \beta) \)
Simplifying the terms:
\( = \cos^2 \alpha \cos^2 \beta + \cos^2 \alpha \sin^2 \beta + \sin^2 \alpha \sin^2 \beta + \sin^2 \alpha \cos^2 \beta \)
\( = \cos^2 \alpha (\cos^2 \beta + \sin^2 \beta) + \sin^2 \alpha (\sin^2 \beta + \cos^2 \beta) \)
\( = \cos^2 \alpha (1) + \sin^2 \alpha (1) = 1 \).
In simple words: Expand the determinant and group the terms using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \) to reduce the complex trigonometric expression to 1.

Exam Tip: Expansion is often the safest route for determinants with many trigonometric terms; just be careful with signs during multiplication.

Question 4. If \( a, b, c \) are real numbers and \( \Delta = \left|\begin{array}{ccc} b+c & c+a & a+b \\ c+a & a+b & b+c \\ a+b & b+c & c+a \end{array}\right| = 0 \), then prove that \( a+b+c=0 \) or \( a=b=c \).
Answer: Applying \( C_1 \rightarrow C_1 + C_2 + C_3 \):
\( \Delta = 2(a+b+c) \left|\begin{array}{ccc} 1 & c+a & a+b \\ 1 & a+b & b+c \\ 1 & b+c & c+a \end{array}\right| \)
Applying \( R_2 \rightarrow R_2 - R_1 \) and \( R_3 \rightarrow R_3 - R_1 \):
\( \Delta = 2(a+b+c) \left|\begin{array}{ccc} 1 & c+a & a+b \\ 0 & b-c & c-a \\ 0 & b-a & c-b \end{array}\right| \)
Expanding gives \( -2(a+b+c) [(a-b)^2 + (b-c)^2 + (c-a)^2] = 0 \).
Thus, \( a+b+c=0 \) or \( a=b=c \).
In simple words: By adding all columns and simplifying, we factor the determinant into a product that equals zero, which implies either the sum is zero or all variables are equal.

Exam Tip: Whenever you see cyclic sums like \( a+b, b+c, c+a \) in a determinant, adding all rows or columns is almost always the first step.

Question 5. Solve the equation \( \left|\begin{array}{ccc} x+a & x & x \\ x & x+a & x \\ x & x & x+a \end{array}\right| = 0 \) for non-zero \( a \).
Answer: Applying \( C_1 \rightarrow C_1 + C_2 + C_3 \):
\( (3x+a) \left|\begin{array}{ccc} 1 & x & x \\ 1 & x+a & x \\ 1 & x & x+a \end{array}\right| = 0 \)
Applying \( R_2 \rightarrow R_2 - R_1 \) and \( R_3 \rightarrow R_3 - R_1 \):
\( (3x+a) \left|\begin{array}{ccc} 1 & x & x \\ 0 & a & 0 \\ 0 & 0 & a \end{array}\right| = 0 \)
\( (3x+a)(a^2) = 0 \). Since \( a \neq 0 \), \( 3x+a = 0 \), so \( x = -\frac{a}{3} \).
In simple words: Add all columns to get a common factor, then use row operations to create zeros, making the determinant easy to solve for \( x \).

Exam Tip: When a determinant has a symmetric structure like this, adding columns usually reveals a factor that simplifies the entire equation.

Question 6. Prove that \( \left|\begin{array}{ccc} a^2 & bc & ac+c^2 \\ a^2+ab & b^2 & ac \\ ab & b^2+bc & c^2 \end{array}\right| = 4a^2b^2c^2 \).
Answer: Taking \( a, b, c \) common from \( R_1, R_2, R_3 \):
\( abc \left|\begin{array}{ccc} a & c & a+c \\ a+b & b & a \\ b & b+c & c \end{array}\right| \)
Applying \( C_3 \rightarrow C_3 - C_1 - C_2 \):
\( abc \left|\begin{array}{ccc} a & c & -2b \\ a+b & b & -2b \\ b & b+c & -2b \end{array}\right| \)
Factoring out \( -2b \) and simplifying further leads to \( 4a^2b^2c^2 \).
In simple words: Factor out variables from each row to simplify the determinant, then use column operations to create zeros and solve.

Exam Tip: Always check if you can factor out variables from rows or columns before starting complex operations; it makes the numbers much smaller.

Question 7. If \( A = \left[\begin{array}{ccc} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{array}\right] \) and \( B = \left[\begin{array}{ccc} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{array}\right] \), find \( (AB)^{-1} \).
Answer: First, calculate \( |B| = 1(3-0) - 2(-1-0) - 2(2-0) = 3+2-4 = 1 \).
Find the adjoint of \( B \):
\( \text{adj } B = \left[\begin{array}{ccc} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{array}\right] \)
\( B^{-1} = \frac{1}{|B|} \text{adj } B = \left[\begin{array}{ccc} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{array}\right] \).
Since \( A=B \), \( (AB)^{-1} = B^{-1} A^{-1} = B^{-1} B^{-1} \).
Multiplying the matrices gives the result.
In simple words: Find the inverse of the matrix, then multiply it by itself to get the inverse of the product.

Exam Tip: Remember the property \( (AB)^{-1} = B^{-1} A^{-1} \); it is much faster than multiplying matrices first and then finding the inverse.

Question 8. For \( A = \left[\begin{array}{ccc} 1 & -2 & 1 \\ -2 & 3 & 1 \\ 1 & 1 & 5 \end{array}\right] \), verify that (i) \( [\text{adj } A]^{-1} = \text{adj } (A^{-1}) \) and (ii) \( (A^{-1})^{-1} = A \).
Answer: Calculate \( |A| = 1(15-1) + 2(-10-1) + 1(-2-3) = 14 - 22 - 5 = -13 \).
Find the cofactor matrix and its transpose to get \( \text{adj } A \).
Then find \( A^{-1} = \frac{1}{|A|} \text{adj } A \).
Perform the required operations to verify both identities.
In simple words: Calculate the inverse and adjoint step-by-step, then compare the results to confirm the matrix properties hold true.

Exam Tip: These verification questions are long; keep your cofactor calculations organized in a table to avoid simple arithmetic errors.

Question 9. \left|\begin{array}{ccc} x & y & x+y \\ y & x+y & x \\ x+y & x & y \end{array}\right| નું મૂલ્ય શોધો.
Answer: Applying \( C_1 \rightarrow C_1 + C_2 + C_3 \), we get:
\( = 2(x+y) \left|\begin{array}{ccc} 1 & y & x+y \\ 1 & x+y & x \\ 1 & x & y \end{array}\right| \)
Applying \( R_2 \rightarrow R_2 - R_1 \) and \( R_3 \rightarrow R_3 - R_1 \):
\( = 2(x+y) \left|\begin{array}{ccc} 1 & y & x+y \\ 0 & x & -y \\ 0 & x-y & -x \end{array}\right| \)
\( = 2(x+y) [x(-x) - (-y)(x-y)] \)
\( = 2(x+y) [-x^2 - xy + y^2] \)
\( = -2(x+y)(x^2 + xy - y^2) \)
\( = -2(x^3 + y^3) \)
In simple words: By adding all columns together, we can factor out 2(x+y). Then, by subtracting rows, we simplify the determinant to solve it easily.

Exam Tip: Always look for row or column operations that create zeros, as they make expanding the determinant much faster.

Question 10. \left|\begin{array}{ccc} 1 & x & y \\ 1 & x+y & y \\ 1 & x & x+y \end{array}\right| નું મૂલ્ય શોધો.
Answer: Applying \( R_2 \rightarrow R_2 - R_1 \) and \( R_3 \rightarrow R_3 - R_1 \):
\( = \left|\begin{array}{ccc} 1 & x & y \\ 0 & y & 0 \\ 0 & 0 & x \end{array}\right| \)
Expanding along the first column:
\( = 1(xy - 0) = xy \)
In simple words: Subtracting the first row from the others creates a simple triangular matrix, which is very easy to calculate.

Exam Tip: When you see a column of 1s, subtracting rows is the most efficient way to simplify the determinant.

Question 11. \left|\begin{array}{lll} \alpha & \alpha^2 & \beta+\gamma \\ \beta & \beta^2 & \gamma+\alpha \\ \gamma & \gamma^2 & \alpha+\beta \end{array}\right| = (\beta - \gamma) (\gamma - \alpha) (\alpha - \beta) (\alpha + \beta + \gamma)
Answer: Applying \( C_3 \rightarrow C_3 + C_1 \):
\( = \left|\begin{array}{lll} \alpha & \alpha^2 & \alpha+\beta+\gamma \\ \beta & \beta^2 & \alpha+\beta+\gamma \\ \gamma & \gamma^2 & \alpha+\beta+\gamma \end{array}\right| \)
Taking \( (\alpha+\beta+\gamma) \) common from \( C_3 \):
\( = (\alpha+\beta+\gamma) \left|\begin{array}{lll} \alpha & \alpha^2 & 1 \\ \beta & \beta^2 & 1 \\ \gamma & \gamma^2 & 1 \end{array}\right| \)
Applying \( R_1 \rightarrow R_1 - R_2 \) and \( R_2 \rightarrow R_2 - R_3 \):
\( = (\alpha+\beta+\gamma) (\alpha-\beta)(\beta-\gamma) \left|\begin{array}{lll} 1 & \alpha+\beta & 0 \\ 1 & \beta+\gamma & 0 \\ \gamma & \gamma^2 & 1 \end{array}\right| \)
Expanding gives \( (\alpha+\beta+\gamma)(\alpha-\beta)(\beta-\gamma)(\gamma-\alpha) \).
In simple words: By adding the first column to the third, we get a common factor. Then we subtract rows to get zeros and solve the remaining part.

Exam Tip: Factoring out common terms early significantly reduces the complexity of the algebraic expansion.

Question 12. \left|\begin{array}{ccc} x & x^2 & 1+px^3 \\ y & y^2 & 1+py^3 \\ z & z^2 & 1+pz^3 \end{array}\right| = (1 + pxyz) (x - y)(y - z) (z - x)
Answer: Splitting the determinant into two:
\( = \left|\begin{array}{ccc} x & x^2 & 1 \\ y & y^2 & 1 \\ z & z^2 & 1 \end{array}\right| + \left|\begin{array}{ccc} x & x^2 & px^3 \\ y & y^2 & py^3 \\ z & z^2 & pz^3 \end{array}\right| \)
\( = \left|\begin{array}{ccc} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{array}\right| + pxyz \left|\begin{array}{ccc} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{array}\right| \)
\( = (1 + pxyz) \left|\begin{array}{ccc} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{array}\right| \)
\( = (1 + pxyz) (x-y)(y-z)(z-x) \)
In simple words: We split the determinant into two parts. One part has a common factor of pxyz, allowing us to combine them and factorize the result.

Exam Tip: Splitting a determinant is a powerful technique when one column contains a sum of two terms.

Question 13. \left|\begin{array}{ccc} 3a & -a+b & -a+c \\ -b+a & 3b & -b+c \\ -c+a & -c+b & 3c \end{array}\right| = 3(a + b + c)(ab + bc + ca)
Answer: Applying \( C_1 \rightarrow C_1 + C_2 + C_3 \):
\( = \left|\begin{array}{ccc} a+b+c & -a+b & -a+c \\ a+b+c & 3b & -b+c \\ a+b+c & -c+b & 3c \end{array}\right| \)
Taking \( (a+b+c) \) common from \( C_1 \):
\( = (a+b+c) \left|\begin{array}{ccc} 1 & -a+b & -a+c \\ 1 & 3b & -b+c \\ 1 & -c+b & 3c \end{array}\right| \)
Applying \( R_2 \rightarrow R_2 - R_1 \) and \( R_3 \rightarrow R_3 - R_1 \):
\( = (a+b+c) \left|\begin{array}{ccc} 1 & -a+b & -a+c \\ 0 & 2b+a & a-b \\ 0 & a-c & a+2c \end{array}\right| \)
Expanding gives \( 3(a+b+c)(ab+bc+ca) \).
In simple words: Adding all columns creates a common factor (a+b+c). Then we subtract rows to simplify the matrix and expand it.

Exam Tip: Always check if adding all rows or columns creates a common factor before starting row operations.

Question 14. \left|\begin{array}{ccc} 1 & 1+p & 1+p+q \\ 2 & 3+2p & 4+3p+2q \\ 3 & 6+3p & 10+6p+3q \end{array}\right| = 1
Answer: Applying \( R_2 \rightarrow R_2 - 2R_1 \) and \( R_3 \rightarrow R_3 - 3R_1 \):
\( = \left|\begin{array}{ccc} 1 & 1+p & 1+p+q \\ 0 & 1 & 2+p \\ 0 & 3 & 7+3p \end{array}\right| \)
Expanding along the first column:
\( = 1[(7+3p) - 3(2+p)] \)
\( = 7 + 3p - 6 - 3p = 1 \)
In simple words: By subtracting multiples of the first row from the others, we get zeros in the first column, making the final calculation very simple.

Exam Tip: Row reduction is the fastest way to solve determinants with numerical coefficients in the first column.

Question 15. \left|\begin{array}{lll} \sin \alpha & \cos \alpha & \cos (\alpha+\delta) \\ \sin \beta & \cos \beta & \cos (\beta+\delta) \\ \sin \gamma & \cos \gamma & \cos (\gamma+\delta) \end{array}\right| = 0
Answer: Using the formula \( \cos(A+B) = \cos A \cos B - \sin A \sin B \):
The third column becomes \( \cos \alpha \cos \delta - \sin \alpha \sin \delta \), etc.
Applying \( C_3 \rightarrow C_3 + (\sin \delta)C_1 - (\cos \delta)C_2 \):
All elements in the third column become zero.
Since one column is zero, the determinant value is 0.
In simple words: By using trigonometric identities, we can transform the third column into zeros, which automatically makes the whole determinant zero.

Exam Tip: If a determinant involves trigonometric addition formulas, try to use column operations to eliminate the third column.

Question 16. નીચેની સમીકરણ સંહતિનો ઉકેલ મેળવો : \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
Answer: Let \( u = \frac{1}{x}, v = \frac{1}{y}, w = \frac{1}{z} \). The system is \( 2u+3v+10w=4, 4u-6v+5w=1, 6u+9v-20w=2 \).
Using matrix method \( AX=B \), where \( |A| = 1200 \neq 0 \).
Calculating the inverse \( A^{-1} \) and multiplying by \( B \):
\( \left[\begin{array}{l} u \\ v \\ w \end{array}\right] = \left[\begin{array}{l} 1/2 \\ 1/3 \\ 1/5 \end{array}\right] \)
Thus, \( x=2, y=3, z=5 \).
In simple words: We replace the fractions with variables, solve the system using matrices, and then flip the results to find x, y, and z.

Exam Tip: Always define new variables for reciprocal terms to simplify the system into a standard linear form.

Question 17. જો a, b, c સમાંતર શ્રેણીમાં હોય, તો નિશ્ચાયક \left|\begin{array}{lll} x+2 & x+3 & x+2a \\ x+3 & x+4 & x+2b \\ x+4 & x+5 & x+2c \end{array}\right| = _____________
Answer: (A) 0
In simple words: Since a, b, and c are in arithmetic progression, 2b = a + c. Using this property, the rows become dependent, making the determinant zero.

Exam Tip: Whenever you see terms in arithmetic progression in a determinant, use the property \( 2b = a+c \) to simplify the rows.

Question 18. જો x, y, z શૂન્યેતર વાસ્તવિક સંખ્યાઓ હોય, તો A = \left[\begin{array}{lll} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{array}\right] નો વ્યસ્ત શ્રેણિક _____________
Answer: The inverse of a diagonal matrix is found by taking the reciprocal of each diagonal element.
\( A^{-1} = \left[\begin{array}{lll} 1/x & 0 & 0 \\ 0 & 1/y & 0 \\ 0 & 0 & 1/z \end{array}\right] \)
In simple words: For a diagonal matrix, you just flip the numbers on the main diagonal to get the inverse.

Exam Tip: Remember that the inverse of a diagonal matrix is always another diagonal matrix with reciprocal entries.

Question 19. જો 0 \leq \theta \leq 2 માટે A = \left[\begin{array}{ccc} 1 & \sin \theta & 1 \\ -\sin \theta & 1 & \sin \theta \\ -1 & -\sin \theta & 1 \end{array}\right] હોય, તો
Answer: (D) det(A) \in [2, 4]
In simple words: By calculating the determinant, we get 2(1 + \sin^2 \theta). Since \sin^2 \theta ranges from 0 to 1, the determinant ranges from 2 to 4.

Exam Tip: Always determine the range of trigonometric functions involved to find the bounds of the determinant value.

Question 18. If x, y, z are non-zero real numbers, then the inverse matrix of A = \( \begin{bmatrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{bmatrix} \) is
Answer: The determinant \( |A| = xyz \). The cofactor matrix is calculated, and the adjoint matrix is \( \text{adj } A = \begin{bmatrix} yz & 0 & 0 \\ 0 & xz & 0 \\ 0 & 0 & xy \end{bmatrix} \). Thus, \( A^{-1} = \frac{1}{|A|} \text{adj } A = \frac{1}{xyz} \begin{bmatrix} yz & 0 & 0 \\ 0 & xz & 0 \\ 0 & 0 & xy \end{bmatrix} = \begin{bmatrix} x^{-1} & 0 & 0 \\ 0 & y^{-1} & 0 \\ 0 & 0 & z^{-1} \end{bmatrix} \).
In simple words: To find the inverse of a diagonal matrix, you simply replace each number on the main diagonal with its reciprocal (1 divided by the number).

Exam Tip: For a diagonal matrix, the inverse is always another diagonal matrix containing the reciprocals of the original diagonal elements.

Question 19. If \( 0 \le \theta \le 2 \) for A = \( \begin{bmatrix} 1 & \sin \theta & 1 \\ -\sin \theta & 1 & \sin \theta \\ -1 & -\sin \theta & 1 \end{bmatrix} \), then
(a) det(A) = 0
(b) det(A) ∈ (2, ∞)
(c) det(A) ∈ (2, 4)
(d) det(A) ∈ [2, 4]
Answer: (d) det(A) ∈ [2, 4]
In simple words: By calculating the determinant, we get \( 2(1 + \sin^2 \theta) \). Since the value of \( \sin^2 \theta \) ranges from 0 to 1, the total value must fall between 2 and 4.

Exam Tip: Always simplify the determinant expression first; here, it reduces to a simple function of \( \sin^2 \theta \), making the range easy to identify.