GSEB Class 12 Maths Solutions Chapter 4 નિશ્ચાયક Exercise 4.1

Gujarat Board Textbook Solutions Class 12 Maths Chapter 4 નિશ્ચાયક Ex 4.1

પ્રશ્ન 1 અને 2 માં આપેલા નિશ્વારાકનું મૂલ્ય શોધો

Question 1. \( \left|\begin{array}{cc} 2 & 4 \\ -5 & -1 \end{array}\right| \)

Answer: The value of the determinant is found by multiplying the diagonal elements and subtracting. First, multiply 2 by -1, which gives -2. Next, multiply -5 by 4, which gives -20. Then, subtract the second result from the first: \(-2 - (-20) = -2 + 20 = 18\).
\[ \left|\begin{array}{cc} 2 & 4 \\ -5 & -1 \end{array}\right| = 2(-1) - (-5)(4) \] \[ = -2 + 20 \] \[ = 18 \]In simple words: To find the value, multiply the numbers on the main diagonal, then multiply the numbers on the other diagonal, and subtract the second product from the first.

🎯 Exam Tip: Remember the formula for a 2x2 determinant: \( \left|\begin{array}{cc} a & b \\ c & d \end{array}\right| = ad - bc \). Care with negative signs is crucial.

 

Question 2.

(i) \( \left|\begin{array}{cc} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right| \)
Answer: To find the value of this determinant, we multiply \(\cos \theta\) by \(\cos \theta\), which gives \(\cos^2 \theta\). Then, we multiply \(\sin \theta\) by \(-\sin \theta\), resulting in \(-\sin^2 \theta\). Subtracting the second product from the first yields \(\cos^2 \theta - (-\sin^2 \theta) = \cos^2 \theta + \sin^2 \theta\), which simplifies to 1 based on a trigonometric identity.
\[ \left|\begin{array}{cc} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right| \] \[ = (\cos \theta)(\cos \theta) - (-\sin \theta)(\sin \theta) \] \[ = \cos^2 \theta + \sin^2 \theta \] \[ = 1 \]In simple words: Multiply the top-left with bottom-right, then bottom-left with top-right, and subtract. The answer comes out to 1 because of a math rule about sin and cos.

🎯 Exam Tip: Familiarity with basic trigonometric identities like \( \sin^2 \theta + \cos^2 \theta = 1 \) is essential for quickly solving such determinants.

 

Question 2.

(ii) \( \left|\begin{array}{cc} x^2-x+1 & x-1 \\ x+1 & x+1 \end{array}\right| \)
Answer: We calculate the determinant by cross-multiplication. First, multiply \((x^2-x+1)\) by \((x+1)\). This product is \(x^3+1\). Next, multiply \((x-1)\) by \((x+1)\), which equals \(x^2-1\). Subtracting the second product from the first gives \((x^3+1) - (x^2-1)\). Simplifying this expression, we get \(x^3+1-x^2+1\), which further reduces to \(x^3-x^2+2\).
\[ \left|\begin{array}{cc} x^2-x+1 & x-1 \\ x+1 & x+1 \end{array}\right| \] \[ = (x+1)(x^2-x+1) - (x+1)(x-1) \] \[ = (x^3+1^3) - (x^2-1) \] \[ = x^3+1-x^2+1 \] \[ = x^3-x^2+2 \]In simple words: Multiply the elements diagonally and subtract. The result is a simple polynomial expression.

🎯 Exam Tip: Recognize algebraic identities like \( (a+b)(a^2-ab+b^2) = a^3+b^3 \) and \( (a+b)(a-b) = a^2-b^2 \) to simplify calculations efficiently.

 

Question 3. જો A = \( \left[\begin{array}{II} 1 & 2 \\ 4 & 2 \end{array}\right] \) હોય, તો સાબિત કરો કે \(|2A| = 4|A|\).

Answer: First, we find the determinant of matrix A. For \(A = \left[\begin{array}{cc} 1 & 2 \\ 4 & 2 \end{array}\right]\), \(|A| = (1)(2) - (2)(4) = 2 - 8 = -6\). Next, we calculate matrix 2A by multiplying each element of A by 2: \(2A = \left[\begin{array}{cc} 2(1) & 2(2) \\ 2(4) & 2(2) \end{array}\right] = \left[\begin{array}{cc} 2 & 4 \\ 8 & 4 \end{array}\right]\). Now, we find the determinant of 2A: \(|2A| = (2)(4) - (4)(8) = 8 - 32 = -24\). To verify the statement, we calculate \(4|A| = 4(-6) = -24\). Since \(|2A| = -24\) and \(4|A| = -24\), we have successfully proven that \(|2A| = 4|A|\).
\[ |A| = \left|\begin{array}{ll} 1 & 2 \\ 4 & 2 \end{array}\right| \] \[ = (2)(1) - (2)(4) \] \[ = 2-8 \] \[ = -6 \] \[ 2A = \left[\begin{array}{ll} 2 & 4 \\ 8 & 4 \end{array}\right] \] \[ |2A| = \left|\begin{array}{ll} 2 & 4 \\ 8 & 4 \end{array}\right| \] \[ = (2)(4) - (4)(8) \] \[ = 8 - 32 \] \[ = -24 \] \[ 4|A| = 4(-6) = -24 \]In simple words: First, find the determinant of matrix A. Then, multiply matrix A by 2 and find its determinant. Finally, check if the determinant of 2A is four times the determinant of A.

🎯 Exam Tip: For an n x n matrix A and any scalar k, \(|kA| = k^n|A|\). In this 2x2 case, n=2, so \(|2A| = 2^2|A| = 4|A|\).

 

Question 4. જો A = \( \left[\begin{array}{III} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{array}\right] \), હોય, તો સાબિત કરો કે \(|3A| = 27|A|\).

Answer: First, let's find the determinant of matrix A. Since A is an upper triangular matrix, its determinant is the product of its diagonal elements: \(|A| = 1 \times 1 \times 4 = 4\). Alternatively, expanding along the first column: \(|A| = 1 \left|\begin{array}{cc} 1 & 2 \\ 0 & 4 \end{array}\right| - 0 \left|\begin{array}{cc} 0 & 1 \\ 0 & 4 \end{array}\right| + 0 \left|\begin{array}{cc} 0 & 1 \\ 1 & 2 \end{array}\right| = 1(4-0) = 4\). Now, let's calculate \(27|A|\): \(27|A| = 27 \times 4 = 108\). Next, we find matrix 3A by multiplying each element of A by 3: \[ 3A = 3\left[\begin{array}{III} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{array}\right] = \left[\begin{array}{ccc} 3 & 0 & 3 \\ 0 & 3 & 6 \\ 0 & 0 & 12 \end{array}\right] \] Finally, we find the determinant of 3A. Again, it's an upper triangular matrix, so \(|3A| = 3 \times 3 \times 12 = 108\). Since \(|3A| = 108\) and \(27|A| = 108\), we have proven that \(|3A| = 27|A|\).
\[ |A| = \left|\begin{array}{III} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{array}\right| = 1(4-0) = 4 \] \[ 27|A| = 27(4) = 108 \] \[ 3A = \left[\begin{array}{ccc} 3 & 0 & 3 \\ 0 & 3 & 6 \\ 0 & 0 & 12 \end{array}\right] \] \[ |3A| = \left|\begin{array}{ccc} 3 & 0 & 3 \\ 0 & 3 & 6 \\ 0 & 0 & 12 \end{array}\right| = 3(3 \times 12 - 6 \times 0) - 0(0 \times 12 - 6 \times 0) + 3(0 \times 0 - 3 \times 0) \] \[ = 3(36 - 0) = 108 \] \[ |3A| = 27|A| = 108 \]In simple words: Calculate the determinant of A. Multiply A by 3 to get 3A, then find its determinant. Check if the determinant of 3A is 27 times the determinant of A.

🎯 Exam Tip: For an n x n matrix A and any scalar k, \(|kA| = k^n|A|\). Here, n=3 for a 3x3 matrix, so \(|3A| = 3^3|A| = 27|A|\). Recognizing this property can save time.

 

પ્રશ્ન 5. નીચે આપેલાં નિશ્ચાયકનાં મૂલ્યો શોધો :

Question 5.

(i) \( \left|\begin{array}{ccc} 3 & -1 & -2 \\ 0 & 0 & -1 \\ 3 & -5 & 0 \end{array}\right| \)
Answer: We expand the determinant along the first row. For the element 3, its minor determinant is \( \left|\begin{array}{cc} 0 & -1 \\ -5 & 0 \end{array}\right| = (0)(0) - (-1)(-5) = 0 - 5 = -5 \). For the element -1, its minor determinant is \( \left|\begin{array}{cc} 0 & -1 \\ 3 & 0 \end{array}\right| = (0)(0) - (-1)(3) = 0 - (-3) = 3 \). For the element -2, its minor determinant is \( \left|\begin{array}{cc} 0 & 0 \\ 3 & -5 \end{array}\right| = (0)(-5) - (0)(3) = 0 - 0 = 0 \). So, the determinant is \( 3(-5) - (-1)(3) + (-2)(0) = -15 + 3 - 0 = -12 \).
\[ \left|\begin{array}{ccc} 3 & -1 & -2 \\ 0 & 0 & -1 \\ 3 & -5 & 0 \end{array}\right| \] \[ = 3\left|\begin{array}{cc} 0 & -1 \\ -5 & 0 \end{array}\right| - (-1)\left|\begin{array}{cc} 0 & -1 \\ 3 & 0 \end{array}\right| + (-2)\left|\begin{array}{cc} 0 & 0 \\ 3 & -5 \end{array}\right| \] \[ = 3(0 - 5) + 1(0 + 3) - 2(0) \] \[ = -15 + 3 - 0 \] \[ = -12 \]In simple words: To find the value, pick a row or column, and for each number, multiply it by its smaller determinant (minor), adjusting signs. Then add or subtract these results.

🎯 Exam Tip: Choose the row or column with the most zeros to minimize calculations. In this case, the second row has two zeros, making expansion along it easier.

 

Question 5.

(ii) \( \left|\begin{array}{ccc} 3 & -4 & 5 \\ 1 & 1 & -2 \\ 2 & 3 & 1 \end{array}\right| \)
Answer: We expand the determinant along the first row. For 3, the minor is \( \left|\begin{array}{cc} 1 & -2 \\ 3 & 1 \end{array}\right| = (1)(1) - (-2)(3) = 1 - (-6) = 1 + 6 = 7 \). For -4, the minor is \( \left|\begin{array}{cc} 1 & -2 \\ 2 & 1 \end{array}\right| = (1)(1) - (-2)(2) = 1 - (-4) = 1 + 4 = 5 \). For 5, the minor is \( \left|\begin{array}{cc} 1 & 1 \\ 2 & 3 \end{array}\right| = (1)(3) - (1)(2) = 3 - 2 = 1 \). So, the determinant is \( 3(7) - (-4)(5) + 5(1) = 21 + 20 + 5 = 46 \).
\[ \left|\begin{array}{ccc} 3 & -4 & 5 \\ 1 & 1 & -2 \\ 2 & 3 & 1 \end{array}\right| \] \[ = 3(1 \times 1 - (-2) \times 3) - (-4)(1 \times 1 - (-2) \times 2) + 5(1 \times 3 - 1 \times 2) \] \[ = 3(1+6) + 4(1+4) + 5(3-2) \] \[ = 3(7) + 4(5) + 5(1) \] \[ = 21 + 20 + 5 \] \[ = 46 \]In simple words: Multiply each number in the first row by its 2x2 determinant, remembering to change the sign for the middle term. Then add the results to get the final answer.

🎯 Exam Tip: Pay close attention to the signs when expanding a 3x3 determinant: \( + \quad - \quad + \).

 

Question 5.

(iii) \( \left|\begin{array}{ccc} 0 & 1 & 2 \\ -1 & 0 & -3 \\ -2 & 3 & 0 \end{array}\right| \)
Answer: We expand the determinant along the first row. For 0, its minor determinant is \( \left|\begin{array}{cc} 0 & -3 \\ 3 & 0 \end{array}\right| = (0)(0) - (-3)(3) = 0 - (-9) = 9 \). So, \(0 \times 9 = 0\). For 1, its minor determinant is \( \left|\begin{array}{cc} -1 & -3 \\ -2 & 0 \end{array}\right| = (-1)(0) - (-3)(-2) = 0 - 6 = -6 \). For 2, its minor determinant is \( \left|\begin{array}{cc} -1 & 0 \\ -2 & 3 \end{array}\right| = (-1)(3) - (0)(-2) = -3 - 0 = -3 \). So, the determinant is \( 0(9) - 1(-6) + 2(-3) = 0 + 6 - 6 = 0 \).
\[ \left|\begin{array}{ccc} 0 & 1 & 2 \\ -1 & 0 & -3 \\ -2 & 3 & 0 \end{array}\right| \] \[ = 0\left|\begin{array}{cc} 0 & -3 \\ 3 & 0 \end{array}\right| - 1\left|\begin{array}{cc} -1 & -3 \\ -2 & 0 \end{array}\right| + 2\left|\begin{array}{cc} -1 & 0 \\ -2 & 3 \end{array}\right| \] \[ = 0(0 - (-9)) - 1(0 - (-6)) + 2(-3 - 0) \] \[ = 0 - 1(6) + 2(-3) \] \[ = 0 - 6 - 6 \] \[ = 0 \]In simple words: Expand the determinant by the first row. The first term is zero. Calculate for the other two terms carefully with their signs. The final value is zero.

🎯 Exam Tip: If an entire row or column consists of zeros, the determinant is 0. While not the case here, having a zero as the first element can simplify calculation if expanded along that row/column.

 

Question 5.

(iv) \( \left|\begin{array}{ccc} 2 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0 \end{array}\right| \)
Answer: We expand the determinant along the first row. For 2, its minor determinant is \( \left|\begin{array}{cc} 2 & -1 \\ -5 & 0 \end{array}\right| = (2)(0) - (-1)(-5) = 0 - 5 = -5 \). For -1, its minor determinant is \( \left|\begin{array}{cc} 0 & -1 \\ 3 & 0 \end{array}\right| = (0)(0) - (-1)(3) = 0 - (-3) = 3 \). For -2, its minor determinant is \( \left|\begin{array}{cc} 0 & 2 \\ 3 & -5 \end{array}\right| = (0)(-5) - (2)(3) = 0 - 6 = -6 \). So, the determinant is \( 2(-5) - (-1)(3) + (-2)(-6) = -10 + 3 + 12 = 5 \).
\[ \left|\begin{array}{ccc} 2 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0 \end{array}\right| \] \[ = 2\left|\begin{array}{cc} 2 & -1 \\ -5 & 0 \end{array}\right| - (-1)\left|\begin{array}{cc} 0 & -1 \\ 3 & 0 \end{array}\right| + (-2)\left|\begin{array}{cc} 0 & 2 \\ 3 & -5 \end{array}\right| \] \[ = 2(0 - 5) + 1(0 - (-3)) - 2(0 - 6) \] \[ = 2(-5) + 1(3) - 2(-6) \] \[ = -10 + 3 + 12 \] \[ = 5 \]In simple words: Expand the determinant using the first row. Multiply each element by its small determinant, remembering the correct signs. Add the results to find the total value.

🎯 Exam Tip: Double-check all multiplications and subtractions, especially with negative numbers, to avoid calculation errors.

 

Question 6. જો A = \( \left[\begin{array}{III} 1 & 1 & -2 \\ 2 & 1 & -3 \\ 5 & 4 & -9 \end{array}\right] \) હોય, તો \(|A|\) શોધો.

Answer: We expand the determinant along the first row. For the element 1 (first row, first column), its minor determinant is \( \left|\begin{array}{cc} 1 & -3 \\ 4 & -9 \end{array}\right| = (1)(-9) - (-3)(4) = -9 - (-12) = -9 + 12 = 3 \). For the element 1 (first row, second column), its minor determinant is \( \left|\begin{array}{cc} 2 & -3 \\ 5 & -9 \end{array}\right| = (2)(-9) - (-3)(5) = -18 - (-15) = -18 + 15 = -3 \). For the element -2 (first row, third column), its minor determinant is \( \left|\begin{array}{cc} 2 & 1 \\ 5 & 4 \end{array}\right| = (2)(4) - (1)(5) = 8 - 5 = 3 \). So, the determinant is \( 1(3) - 1(-3) + (-2)(3) = 3 + 3 - 6 = 0 \).
\[ |A| = \left|\begin{array}{III} 1 & 1 & -2 \\ 2 & 1 & -3 \\ 5 & 4 & -9 \end{array}\right| \] \[ = 1\left|\begin{array}{ll} 1 & -3 \\ 4 & -9 \end{array}\right| - 1\left|\begin{array}{ll} 2 & -3 \\ 5 & -9 \end{array}\right| - 2\left|\begin{array}{ll} 2 & 1 \\ 5 & 4 \end{array}\right| \] \[ = 1 ((-9) - (-12)) - 1 ((-18) - (-15)) - 2 (8 - 5) \] \[ = 1(-9 + 12) - 1(-18 + 15) - 2(3) \] \[ = 1(3) - 1(-3) - 2(3) \] \[ = 3 + 3 - 6 \] \[ = 0 \]In simple words: Find the determinant of the given 3x3 matrix by expanding along the first row. Multiply each element by its smaller 2x2 determinant, paying attention to positive and negative signs. All the terms cancel out, giving a final value of zero.

🎯 Exam Tip: A determinant of zero indicates that the matrix is singular, meaning it does not have an inverse. This can be a useful property to recognize.

 

પ્રશ્ન 7. xનું મૂલ્ય શોધો :

Question 7.

(i) \( \left|\begin{array}{ll} 2 & 4 \\ 5 & 1 \end{array}\right|=\left|\begin{array}{cc} 2 x & 4 \\ 6 & x \end{array}\right| \)
Answer: First, calculate the determinant on the left side: \( (2)(1) - (4)(5) = 2 - 20 = -18 \). Next, calculate the determinant on the right side: \( (2x)(x) - (4)(6) = 2x^2 - 24 \). Set the two determinants equal to each other: \( -18 = 2x^2 - 24 \). Add 24 to both sides: \( -18 + 24 = 2x^2 \implies 6 = 2x^2 \). Divide by 2: \( x^2 = 3 \). Take the square root of both sides: \( x = \pm\sqrt{3} \).
\[ \left|\begin{array}{ll} 2 & 4 \\ 5 & 1 \end{array}\right|=\left|\begin{array}{cc} 2 x & 4 \\ 6 & x \end{array}\right| \] \[ 2(1) - 4(5) = 2x(x) - 4(6) \] \[ 2 - 20 = 2x^2 - 24 \] \[ -18 = 2x^2 - 24 \] \[ 2x^2 = -18 + 24 \] \[ 2x^2 = 6 \] \[ x^2 = 3 \] \[ x = \pm\sqrt{3} \]In simple words: Calculate the value of the determinant on each side of the equation. Set these values equal to each other and solve for x.

🎯 Exam Tip: Remember to consider both positive and negative roots when taking the square root of a variable, as \(x^2 = k\) has solutions \(x = \pm\sqrt{k}\).

 

Question 7.

(ii) \( \left|\begin{array}{ll} 2 & 3 \\ 4 & 5 \end{array}\right|=\left|\begin{array}{cc} x & 3 \\ 2 x & 5 \end{array}\right| \)
Answer: First, calculate the determinant on the left side: \( (2)(5) - (3)(4) = 10 - 12 = -2 \). Next, calculate the determinant on the right side: \( (x)(5) - (3)(2x) = 5x - 6x = -x \). Set the two determinants equal to each other: \( -2 = -x \). Multiply both sides by -1 to find x: \( x = 2 \).
\[ \left|\begin{array}{ll} 2 & 3 \\ 4 & 5 \end{array}\right|=\left|\begin{array}{cc} x & 3 \\ 2 x & 5 \end{array}\right| \] \[ 2(5) - 3(4) = x(5) - 3(2x) \] \[ 10 - 12 = 5x - 6x \] \[ -2 = -x \] \[ x = 2 \]In simple words: Find the value of each determinant. Set them equal to each other and solve the simple equation for x.

🎯 Exam Tip: Always be careful with arithmetic operations, especially subtractions, to avoid simple calculation errors that can lead to incorrect values for x.

 

પ્રશ્ન 8 માં વિધાન સાચું બને તે રીતે આપેલ વિોમાંથી યોગ્ય વિકલ્પ પસંદ કરો :

Question 8. જો \( \left|\begin{array}{cc} x & 2 \\ 18 & x \end{array}\right|=\left|\begin{array}{cc} 6 & 2 \\ 18 & 6 \end{array}\right] \) હોય, તો \(x =\)

(A) 6
(B) ±6
(C) -6
(D) 0
Answer: Calculate the determinant on the left side: \( (x)(x) - (2)(18) = x^2 - 36 \). Calculate the determinant on the right side: \( (6)(6) - (2)(18) = 36 - 36 = 0 \). Set the two determinants equal to each other: \( x^2 - 36 = 0 \). Add 36 to both sides: \( x^2 = 36 \). Take the square root of both sides: \( x = \pm6 \). Therefore, the correct option is (B).
\[ \left|\begin{array}{cc} x & 2 \\ 18 & x \end{array}\right|=\left|\begin{array}{cc} 6 & 2 \\ 18 & 6 \end{array}\right| \] \[ x(x) - 2(18) = 6(6) - 2(18) \] \[ x^2 - 36 = 36 - 36 \] \[ x^2 - 36 = 0 \] \[ x^2 = 36 \] \[ x = \pm6 \]
Answer: (B) ±6In simple words: Find the value of both determinants. Set the two resulting expressions equal to each other and solve the equation to find the possible values of x.

🎯 Exam Tip: When solving for a variable squared, always remember that there will be both a positive and a negative root. Failing to include both can lead to an incorrect answer, especially in multiple-choice questions.