GSEB Class 12 Maths Solutions Chapter 4 નિશ્ચાયક Exercise 4.2

GSEB Solutions Class 12 Maths Chapter 4 નિશ્ચાયક Ex 4.2

પ્રશ્ન 1 થી 5 માં નિશ્ચાયકના ગુણધર્મનો ઉપયોગ કરી અને વિસ્તરણ કર્યા સિવાય સાબિત કરો :

Question 1.\[ \left|\begin{array}{lll} x & a & x+a \\ y & b & y+b \\ z & c & z+c \end{array}\right| = 0 \]
Answer:ડા.બા. \( = \left|\begin{array}{lll} x & a & x+a \\ y & b & y+b \\ z & c & z+c \end{array}\right| \)
\( = \left|\begin{array}{lll} x & a & x \\ y & b & y \\ z & c & z \end{array}\right| + \left|\begin{array}{lll} x & a & a \\ y & b & b \\ z & c & c \end{array}\right| \)
In the first determinant, column \(C_1\) and column \(C_3\) are the same. In the second determinant, column \(C_2\) and column \(C_3\) are the same.
Therefore, the value of each determinant is zero.
\( = 0 + 0 \)
\( = 0 \)
In simple words: We split the determinant into two parts. In the first part, the first and third columns are identical, making its value zero. In the second part, the second and third columns are identical, also making its value zero. The sum of these two zeros is zero.

🎯 Exam Tip: Remember that if any two rows or columns of a determinant are identical, its value is zero. This property is crucial for simplifying determinant calculations without expansion.

 

Question 2.\[ \left|\begin{array}{lll} a-b & b-c & c-a \\ b-c & c-a & a-b \\ c-a & a-b & b-c \end{array}\right| = 0 \]
Answer:ડા.બા. \( = \left|\begin{array}{lll} a-b & b-c & c-a \\ b-c & c-a & a-b \\ c-a & a-b & b-c \end{array}\right| \)
Apply the operation \(C_1 \rightarrow C_1 + C_2 + C_3\):
\( = \left|\begin{array}{lll} (a-b)+(b-c)+(c-a) & b-c & c-a \\ (b-c)+(c-a)+(a-b) & c-a & a-b \\ (c-a)+(a-b)+(b-c) & a-b & b-c \end{array}\right| \)
\( = \left|\begin{array}{lll} 0 & b-c & c-a \\ 0 & c-a & a-b \\ 0 & a-b & b-c \end{array}\right| \)
Since all elements in column \(C_1\) are zero, the value of the determinant is zero.
\( = 0 \)
In simple words: We added all three columns together and put the result in the first column. This made all the numbers in the first column turn into zero. When a column is full of zeros, the whole determinant becomes zero.

🎯 Exam Tip: If all elements of any row or column are zero, the value of the determinant is zero. This property can significantly simplify problems involving determinant calculations.

 

Question 3.\[ \left|\begin{array}{lll} 2 & 7 & 65 \\ 3 & 8 & 75 \\ 5 & 9 & 86 \end{array}\right| = 0 \]
Answer:ડા.બા. \( = \left|\begin{array}{lll} 2 & 7 & 65 \\ 3 & 8 & 75 \\ 5 & 9 & 86 \end{array}\right| \)
Apply the operation \(C_3 \rightarrow C_3 - 9C_2\):
\( = \left|\begin{array}{lll} 2 & 7 & 65 - 9(7) \\ 3 & 8 & 75 - 9(8) \\ 5 & 9 & 86 - 9(9) \end{array}\right| \)
\( = \left|\begin{array}{lll} 2 & 7 & 65 - 63 \\ 3 & 8 & 75 - 72 \\ 5 & 9 & 86 - 81 \end{array}\right| \)
\( = \left|\begin{array}{lll} 2 & 7 & 2 \\ 3 & 8 & 3 \\ 5 & 9 & 5 \end{array}\right| \)
Since column \(C_1\) and column \(C_3\) are identical, the value of the determinant is zero.
\( = 0 \)
In simple words: We subtracted nine times the second column from the third column. This made the first and third columns exactly the same. When two columns are identical, the determinant's value is zero.

🎯 Exam Tip: Performing row or column operations like \(C_i \rightarrow C_i + kC_j\) or \(R_i \rightarrow R_i + kR_j\) does not change the value of the determinant. This property is very useful for creating identical rows/columns or rows/columns of zeros to simplify calculations.

 

Question 4.\[ \left|\begin{array}{ccc} 1 & bc & a(b+c) \\ 1 & ca & b(c+a) \\ 1 & ab & c(a+b) \end{array}\right| = 0 \]
Answer:ડા.બા. \( = \left|\begin{array}{ccc} 1 & bc & a(b+c) \\ 1 & ca & b(c+a) \\ 1 & ab & c(a+b) \end{array}\right| \)
First, expand the elements in the third column:
\( = \left|\begin{array}{ccc} 1 & bc & ab+ac \\ 1 & ca & bc+ab \\ 1 & ab & ac+bc \end{array}\right| \)
Apply the operation \(C_3 \rightarrow C_3 + C_2\):
\( = \left|\begin{array}{ccc} 1 & bc & ab+ac+bc \\ 1 & ca & bc+ab+ca \\ 1 & ab & ac+bc+ab \end{array}\right| \)
Now, take the common factor \((ab+bc+ca)\) from column \(C_3\):
\( = (ab+bc+ca) \left|\begin{array}{ccc} 1 & bc & 1 \\ 1 & ca & 1 \\ 1 & ab & 1 \end{array}\right| \)
Since column \(C_1\) and column \(C_3\) are identical, the value of the determinant is zero.
\( = (ab+bc+ca) (0) \)
\( = 0 \)
In simple words: We first multiplied out the terms in the third column. Then, we added the second column to the third column. This made the first and third columns identical, which means the determinant's value is zero.

🎯 Exam Tip: Look for opportunities to create identical rows or columns by performing elementary row/column operations or by taking out common factors. This often leads to a determinant value of zero.

 

Question 5.\[ \left|\begin{array}{lll} b+c & q+r & y+z \\ c+a & r+p & z+x \\ a+b & p+q & x+y \end{array}\right| = 2 \left|\begin{array}{lll} a & p & x \\ b & q & y \\ c & r & z \end{array}\right| \]
Answer:ડા.બા. \( = \left|\begin{array}{lll} b+c & q+r & y+z \\ c+a & r+p & z+x \\ a+b & p+q & x+y \end{array}\right| \)
Apply the operation \(R_1 \rightarrow R_1 + R_2 + R_3\):
\( = \left|\begin{array}{lll} (b+c)+(c+a)+(a+b) & (q+r)+(r+p)+(p+q) & (y+z)+(z+x)+(x+y) \\ c+a & r+p & z+x \\ a+b & p+q & x+y \end{array}\right| \)
\( = \left|\begin{array}{lll} 2(a+b+c) & 2(p+q+r) & 2(x+y+z) \\ c+a & r+p & z+x \\ a+b & p+q & x+y \end{array}\right| \)
Take out common factor 2 from row \(R_1\):
\( = 2 \left|\begin{array}{lll} a+b+c & p+q+r & x+y+z \\ c+a & r+p & z+x \\ a+b & p+q & x+y \end{array}\right| \)
Apply the operation \(R_1 \rightarrow R_1 - R_2\):
\( = 2 \left|\begin{array}{lll} (a+b+c)-(c+a) & (p+q+r)-(r+p) & (x+y+z)-(z+x) \\ c+a & r+p & z+x \\ a+b & p+q & x+y \end{array}\right| \)
\( = 2 \left|\begin{array}{lll} b & q & y \\ c+a & r+p & z+x \\ a+b & p+q & x+y \end{array}\right| \)
Apply the operation \(R_3 \rightarrow R_3 - R_1\):
\( = 2 \left|\begin{array}{lll} b & q & y \\ c+a & r+p & z+x \\ (a+b)-b & (p+q)-q & (x+y)-y \end{array}\right| \)
\( = 2 \left|\begin{array}{lll} b & q & y \\ c+a & r+p & z+x \\ a & p & x \end{array}\right| \)
Apply the operation \(R_2 \rightarrow R_2 - R_3\):
\( = 2 \left|\begin{array}{lll} b & q & y \\ (c+a)-a & (r+p)-p & (z+x)-x \\ a & p & x \end{array}\right| \)
\( = 2 \left|\begin{array}{lll} b & q & y \\ c & r & z \\ a & p & x \end{array}\right| \)
Swap \(R_1\) and \(R_3\) (this introduces a negative sign):
\( = -2 \left|\begin{array}{lll} a & p & x \\ c & r & z \\ b & q & y \end{array}\right| \)
Swap \(R_2\) and \(R_3\) (this introduces another negative sign, making it positive again):
\( = (-1)(-2) \left|\begin{array}{lll} a & p & x \\ b & q & y \\ c & r & z \end{array}\right| \)
\( = 2 \left|\begin{array}{lll} a & p & x \\ b & q & y \\ c & r & z \end{array}\right| \)
\( = \) જ.બા.
In simple words: We used several row operations to change the left side determinant. First, we added all rows to the first row, then took out a common factor. Next, we subtracted rows to simplify the elements. Finally, we swapped rows twice to get the desired form on the right side. Each row swap changes the sign of the determinant, so two swaps bring it back to positive.

🎯 Exam Tip: Remember that swapping two rows or two columns in a determinant changes its sign. A sequence of an even number of swaps will retain the original sign, while an odd number of swaps will reverse it. Carefully track the signs when performing multiple swaps.

 

પ્રશ્ન 6 થી 14 માં નિશ્ચાયકના ગુણધર્મનો ઉપયોગ કરીને સાબિત કરો :

Question 6.\[ \left|\begin{array}{ccc} 0 & a & -b \\ -a & 0 & -c \\ b & c & 0 \end{array}\right| = 0 \]
Answer:Let \( D = \left|\begin{array}{ccc} 0 & a & -b \\ -a & 0 & -c \\ b & c & 0 \end{array}\right| \)
Take out -1 common from each row \(R_1, R_2, R_3\):
\( D = (-1)^3 \left|\begin{array}{ccc} 0 & -a & b \\ a & 0 & c \\ -b & -c & 0 \end{array}\right| \)
\( D = -1 \left|\begin{array}{ccc} 0 & -a & b \\ a & 0 & c \\ -b & -c & 0 \end{array}\right| \)
Now, swap the rows and columns (transpose the matrix). The value of the determinant remains the same upon transposition.
\( D = -1 \left|\begin{array}{ccc} 0 & a & -b \\ -a & 0 & -c \\ b & c & 0 \end{array}\right| \)
Notice that the determinant inside is the original determinant D.
So, \( D = -1 \cdot D \)
\( D = -D \)
\( 2D = 0 \)
\( D = 0 \)
Thus, \( \left|\begin{array}{ccc} 0 & a & -b \\ -a & 0 & -c \\ b & c & 0 \end{array}\right| = 0 \)
In simple words: We started with the determinant D. By taking out -1 from each row, we found that D is equal to -1 times the transpose of the original determinant. Since a determinant's value doesn't change when transposed, we got D = -D. This means 2D = 0, so D must be 0.

🎯 Exam Tip: This problem demonstrates a property of skew-symmetric matrices. For an odd-ordered skew-symmetric matrix, its determinant is always zero. Recognizing this can save time in certain cases.

 

Question 7.\[ \left|\begin{array}{ccc} -a^2 & ab & ac \\ ba & -b^2 & bc \\ ca & cb & -c^2 \end{array}\right| = 4a^2b^2c^2 \]
Answer:ડા.બા. \( = \left|\begin{array}{ccc} -a^2 & ab & ac \\ ba & -b^2 & bc \\ ca & cb & -c^2 \end{array}\right| \)
Take common factor 'a' from \(R_1\), 'b' from \(R_2\), and 'c' from \(R_3\):
\( = abc \left|\begin{array}{ccc} -a & b & c \\ a & -b & c \\ a & b & -c \end{array}\right| \)
Take common factor 'a' from \(C_1\), 'b' from \(C_2\), and 'c' from \(C_3\):
\( = abc \cdot abc \left|\begin{array}{ccc} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{array}\right| \)
\( = a^2b^2c^2 \left|\begin{array}{ccc} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{array}\right| \)
Apply the operation \(R_1 \rightarrow R_1 + R_2\):
\( = a^2b^2c^2 \left|\begin{array}{ccc} -1+1 & 1-1 & 1+1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{array}\right| \)
\( = a^2b^2c^2 \left|\begin{array}{ccc} 0 & 0 & 2 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{array}\right| \)
Expand the determinant along \(R_1\):
\( = a^2b^2c^2 [0 - 0 + 2((1)(-1) - (1)(1))] \)
\( = a^2b^2c^2 [2(-1-1)] \)
\( = a^2b^2c^2 [2(-2)] \)
\( = a^2b^2c^2 [-4] \)
\( = -4a^2b^2c^2 \)
*Self-correction: The problem states to prove \(4a^2b^2c^2\), but my calculation yields \(-4a^2b^2c^2\). Let me recheck the determinant expansion or operations.* Let's re-evaluate from \( = a^2b^2c^2 \left|\begin{array}{ccc} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{array}\right| \) Apply \(R_1 \rightarrow R_1 + R_2\): \( = a^2b^2c^2 \left|\begin{array}{ccc} 0 & 0 & 2 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{array}\right| \) Expand along \(R_1\): \( = a^2b^2c^2 \cdot (2 \cdot (1 \cdot 1 - (-1) \cdot 1)) \) \( = a^2b^2c^2 \cdot (2 \cdot (1 + 1)) \) \( = a^2b^2c^2 \cdot (2 \cdot 2) \) \( = 4a^2b^2c^2 \) This matches the RHS. My previous calculation had a sign error in the expansion.
In simple words: First, we took out common factors 'a', 'b', and 'c' from each row, and then again from each column, resulting in \(a^2b^2c^2\) multiplied by a simpler determinant of -1s and 1s. Then, we added the first two rows to make the first row have two zeros, simplifying its expansion. Finally, expanding this determinant gave us the result of \(4a^2b^2c^2\).

🎯 Exam Tip: Always look for common factors in rows or columns first, as this often simplifies the determinant significantly. Be careful with determinant expansion, especially with signs. Double-check your arithmetic when expanding a 3x3 determinant.

 

Question 8.(i) \( \left|\begin{array}{ccc} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{array}\right| = (a-b)(b-c)(c-a) \)
Answer:ડા.બા. \( = \left|\begin{array}{ccc} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{array}\right| \)
Apply the operation \(R_1 \rightarrow R_1 - R_2\):
\( = \left|\begin{array}{ccc} 1-1 & a-b & a^2-b^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{array}\right| \)
\( = \left|\begin{array}{ccc} 0 & a-b & (a-b)(a+b) \\ 1 & b & b^2 \\ 1 & c & c^2 \end{array}\right| \)
Apply the operation \(R_2 \rightarrow R_2 - R_3\):
\( = \left|\begin{array}{ccc} 0 & a-b & (a-b)(a+b) \\ 1-1 & b-c & b^2-c^2 \\ 1 & c & c^2 \end{array}\right| \)
\( = \left|\begin{array}{ccc} 0 & a-b & (a-b)(a+b) \\ 0 & b-c & (b-c)(b+c) \\ 1 & c & c^2 \end{array}\right| \)
Take out common factor \((a-b)\) from \(R_1\) and \((b-c)\) from \(R_2\):
\( = (a-b)(b-c) \left|\begin{array}{ccc} 0 & 1 & a+b \\ 0 & 1 & b+c \\ 1 & c & c^2 \end{array}\right| \)
Expand the determinant along \(C_1\):
\( = (a-b)(b-c) [1((1)(b+c) - (1)(a+b))] \)
\( = (a-b)(b-c) [b+c-a-b] \)
\( = (a-b)(b-c) [c-a] \)
\( = (a-b)(b-c)(c-a) \)
\( = \) જ.બા.
In simple words: We used row operations to create zeros and common factors. First, we subtracted the second row from the first, then the third row from the second. This allowed us to take out \((a-b)\) and \((b-c)\) as common factors. Finally, we expanded the remaining determinant along the first column to get \((c-a)\), proving the identity.

🎯 Exam Tip: When dealing with determinants involving terms like \(a^2, b^2, c^2\), try to create factors like \((a-b)\), \((b-c)\), \((c-a)\) by applying row/column operations of the form \(R_i \rightarrow R_i - R_j\).

 

Question 8.(ii) \( \left|\begin{array}{ccc} 1 & 1 & 1 \\ a & b & c \\ a^3 & b^3 & c^3 \end{array}\right| = (a-b)(b-c)(c-a)(a+b+c) \)
Answer:ડા.બા. \( = \left|\begin{array}{ccc} 1 & 1 & 1 \\ a & b & c \\ a^3 & b^3 & c^3 \end{array}\right| \)
Apply the operation \(C_1 \rightarrow C_1 - C_2\):
\( = \left|\begin{array}{ccc} 1-1 & 1 & 1 \\ a-b & b & c \\ a^3-b^3 & b^3 & c^3 \end{array}\right| \)
\( = \left|\begin{array}{ccc} 0 & 1 & 1 \\ a-b & b & c \\ (a-b)(a^2+ab+b^2) & b^3 & c^3 \end{array}\right| \)
Apply the operation \(C_2 \rightarrow C_2 - C_3\):
\( = \left|\begin{array}{ccc} 0 & 1-1 & 1 \\ a-b & b-c & c \\ (a-b)(a^2+ab+b^2) & b^3-c^3 & c^3 \end{array}\right| \)
\( = \left|\begin{array}{ccc} 0 & 0 & 1 \\ a-b & b-c & c \\ (a-b)(a^2+ab+b^2) & (b-c)(b^2+bc+c^2) & c^3 \end{array}\right| \)
Take out common factor \((a-b)\) from \(C_1\) and \((b-c)\) from \(C_2\):
\( = (a-b)(b-c) \left|\begin{array}{ccc} 0 & 0 & 1 \\ 1 & 1 & c \\ a^2+ab+b^2 & b^2+bc+c^2 & c^3 \end{array}\right| \)
Expand the determinant along \(R_1\):
\( = (a-b)(b-c) [1((1)(b^2+bc+c^2) - (1)(a^2+ab+b^2))] \)
\( = (a-b)(b-c) [b^2+bc+c^2 - a^2-ab-b^2] \)
\( = (a-b)(b-c) [c^2-a^2 + bc-ab] \)
\( = (a-b)(b-c) [(c-a)(c+a) + b(c-a)] \)
\( = (a-b)(b-c) (c-a) [(c+a)+b] \)
\( = (a-b)(b-c)(c-a)(a+b+c) \)
\( = \) જ.બા.
In simple words: We used column operations to get two zeros in the first row. First, we subtracted the second column from the first, then the third from the second. This allowed us to factor out \((a-b)\) and \((b-c)\). After expanding the remaining determinant, we simplified the expression to find the final factor \((c-a)(a+b+c)\).

🎯 Exam Tip: When proving identities involving symmetric expressions like \(a^3, b^3, c^3\), using column operations to generate factors like \((a-b)\), \((b-c)\), \((c-a)\) is a standard and effective technique. Remember the factorization of \(a^3-b^3 = (a-b)(a^2+ab+b^2)\).

 

Question 9.\[ \left|\begin{array}{ccc} x & x^2 & yz \\ y & y^2 & zx \\ z & z^2 & xy \end{array}\right| = (x-y)(y-z)(z-x)(xy+yz+zx) \]
Answer:ડા.બા. \( = \left|\begin{array}{ccc} x & x^2 & yz \\ y & y^2 & zx \\ z & z^2 & xy \end{array}\right| \)
Multiply \(R_1\) by x, \(R_2\) by y, and \(R_3\) by z. To compensate, divide the determinant by xyz:
\( = \frac{1}{xyz} \left|\begin{array}{ccc} x^2 & x^3 & xyz \\ y^2 & y^3 & xyz \\ z^2 & z^3 & xyz \end{array}\right| \)
Take out common factor xyz from \(C_3\):
\( = \frac{xyz}{xyz} \left|\begin{array}{ccc} x^2 & x^3 & 1 \\ y^2 & y^3 & 1 \\ z^2 & z^3 & 1 \end{array}\right| \)
\( = \left|\begin{array}{ccc} x^2 & x^3 & 1 \\ y^2 & y^3 & 1 \\ z^2 & z^3 & 1 \end{array}\right| \)
Apply the operation \(R_1 \rightarrow R_1 - R_2\):
\( = \left|\begin{array}{ccc} x^2-y^2 & x^3-y^3 & 0 \\ y^2 & y^3 & 1 \\ z^2 & z^3 & 1 \end{array}\right| \)
Apply the operation \(R_2 \rightarrow R_2 - R_3\):
\( = \left|\begin{array}{ccc} x^2-y^2 & x^3-y^3 & 0 \\ y^2-z^2 & y^3-z^3 & 0 \\ z^2 & z^3 & 1 \end{array}\right| \)
Factorize the terms:
\( = \left|\begin{array}{ccc} (x-y)(x+y) & (x-y)(x^2+xy+y^2) & 0 \\ (y-z)(y+z) & (y-z)(y^2+yz+z^2) & 0 \\ z^2 & z^3 & 1 \end{array}\right| \)
Take out common factor \((x-y)\) from \(R_1\) and \((y-z)\) from \(R_2\):
\( = (x-y)(y-z) \left|\begin{array}{ccc} x+y & x^2+xy+y^2 & 0 \\ y+z & y^2+yz+z^2 & 0 \\ z^2 & z^3 & 1 \end{array}\right| \)
Expand the determinant along \(C_3\):
\( = (x-y)(y-z) [1((x+y)(y^2+yz+z^2) - (y+z)(x^2+xy+y^2))] \)
Now simplify the term inside the bracket:
\( (x+y)(y^2+yz+z^2) - (y+z)(x^2+xy+y^2) \)
\( = (xy^2+xyz+xz^2+y^3+y^2z+yz^2) - (yx^2+xy^2+y^3+zx^2+xyz+yz^2) \)
\( = xy^2+xyz+xz^2+y^3+y^2z+yz^2 - yx^2-xy^2-y^3-zx^2-xyz-yz^2 \)
\( = xz^2+y^2z - yx^2-zx^2 \)
\( = z(xz+y^2) - x(yx+zx) \)
\( = z^2x+y^2z - xy^2-x^2z \)
\( = z^2x - x^2z + y^2z - xy^2 \)
\( = xz(z-x) + y^2(z-x) \)
\( = (z-x)(xz+y^2) \) - *This isn't leading to the desired factor \(xy+yz+zx\)* Let's re-examine the expansion and the simplification of the bracketed term. From: \( = (x-y)(y-z) \left|\begin{array}{ccc} x+y & x^2+xy+y^2 & 0 \\ y+z & y^2+yz+z^2 & 0 \\ z^2 & z^3 & 1 \end{array}\right| \) Expanding along \(C_3\), we get: \( = (x-y)(y-z) [1 \cdot ((x+y)(y^2+yz+z^2) - (y+z)(x^2+xy+y^2))] \) Let's simplify the bracket term: \(B = (x+y)(y^2+yz+z^2) - (y+z)(x^2+xy+y^2)\) \(B = (xy^2 + xyz + xz^2 + y^3 + y^2z + yz^2) - (yx^2 + xy^2 + y^3 + zx^2 + xyz + yz^2)\) Cancel common terms: \(xy^2\), \(xyz\), \(y^3\), \(yz^2\) \(B = xz^2 + y^2z - yx^2 - zx^2 \) \(B = xz(z-x) + y(y z - x y) \) \(B = xz(z-x) + y^2(z) - xy^2 \) This path seems complicated. Let's try to manipulate the determinant before final expansion differently. Let's retry from \( = (x-y)(y-z) \left|\begin{array}{ccc} x+y & x^2+xy+y^2 & 0 \\ y+z & y^2+yz+z^2 & 0 \\ z^2 & z^3 & 1 \end{array}\right| \) Apply \(R_2 \rightarrow R_2 - R_1\): \( = (x-y)(y-z) \left|\begin{array}{ccc} x+y & x^2+xy+y^2 & 0 \\ (y+z)-(x+y) & (y^2+yz+z^2)-(x^2+xy+y^2) & 0 \\ z^2 & z^3 & 1 \end{array}\right| \) \( = (x-y)(y-z) \left|\begin{array}{ccc} x+y & x^2+xy+y^2 & 0 \\ z-x & yz+z^2-x^2-xy & 0 \\ z^2 & z^3 & 1 \end{array}\right| \) Simplify the second element in \(R_2\): \(yz+z^2-x^2-xy = (z^2-x^2) + yz-xy = (z-x)(z+x) + y(z-x) = (z-x)(z+x+y)\) So the determinant becomes: \( = (x-y)(y-z) \left|\begin{array}{ccc} x+y & x^2+xy+y^2 & 0 \\ z-x & (z-x)(x+y+z) & 0 \\ z^2 & z^3 & 1 \end{array}\right| \) Take out common factor \((z-x)\) from \(R_2\): \( = (x-y)(y-z)(z-x) \left|\begin{array}{ccc} x+y & x^2+xy+y^2 & 0 \\ 1 & x+y+z & 0 \\ z^2 & z^3 & 1 \end{array}\right| \) Expand along \(C_3\): \( = (x-y)(y-z)(z-x) [1 \cdot ((x+y)(x+y+z) - (1)(x^2+xy+y^2))] \) Simplify the bracket term: \( (x+y)(x+y+z) - (x^2+xy+y^2) \) \( = (x+y)^2 + z(x+y) - (x^2+xy+y^2) \) \( = x^2+2xy+y^2 + xz+yz - x^2-xy-y^2 \) \( = xy + xz + yz \) So the bracket term simplifies to \(xy+yz+zx\).
Therefore,
\( = (x-y)(y-z)(z-x)(xy+yz+zx) \)
\( = \) જ.બા.
In simple words: We changed the determinant to make two columns the same, allowing us to factor out \(xyz\). Then, we performed row operations to create zeros in the third column and factor out \((x-y)\), \((y-z)\), and \((z-x)\). Finally, we expanded the determinant and simplified the remaining expression to get \((xy+yz+zx)\), which completes the proof.

🎯 Exam Tip: When faced with products of terms like \((x-y)(y-z)(z-x)\), it's often effective to perform row or column operations of the form \(R_i \rightarrow R_i - R_j\) or \(C_i \rightarrow C_i - C_j\) to factor out these differences.

 

Question 10.(i) \( \left|\begin{array}{ccc} x+4 & 2x & 2x \\ 2x & x+4 & 2x \\ 2x & 2x & x+4 \end{array}\right| = (5x+4)(4-x)^2 \)
Answer:ડા.બા. \( = \left|\begin{array}{ccc} x+4 & 2x & 2x \\ 2x & x+4 & 2x \\ 2x & 2x & x+4 \end{array}\right| \)
Apply the operation \(C_1 \rightarrow C_1 + C_2 + C_3\):
\( = \left|\begin{array}{ccc} (x+4)+2x+2x & 2x & 2x \\ 2x+(x+4)+2x & x+4 & 2x \\ 2x+2x+(x+4) & 2x & x+4 \end{array}\right| \)
\( = \left|\begin{array}{ccc} 5x+4 & 2x & 2x \\ 5x+4 & x+4 & 2x \\ 5x+4 & 2x & x+4 \end{array}\right| \)
Take out common factor \((5x+4)\) from \(C_1\):
\( = (5x+4) \left|\begin{array}{ccc} 1 & 2x & 2x \\ 1 & x+4 & 2x \\ 1 & 2x & x+4 \end{array}\right| \)
Apply the operation \(R_1 \rightarrow R_1 - R_2\):
\( = (5x+4) \left|\begin{array}{ccc} 1-1 & 2x-(x+4) & 2x-2x \\ 1 & x+4 & 2x \\ 1 & 2x & x+4 \end{array}\right| \)
\( = (5x+4) \left|\begin{array}{ccc} 0 & x-4 & 0 \\ 1 & x+4 & 2x \\ 1 & 2x & x+4 \end{array}\right| \)
Apply the operation \(R_2 \rightarrow R_2 - R_3\):
\( = (5x+4) \left|\begin{array}{ccc} 0 & x-4 & 0 \\ 1-1 & x+4-2x & 2x-(x+4) \\ 1 & 2x & x+4 \end{array}\right| \)
\( = (5x+4) \left|\begin{array}{ccc} 0 & x-4 & 0 \\ 0 & 4-x & x-4 \\ 1 & 2x & x+4 \end{array}\right| \)
Take out common factor \((x-4)\) from \(R_1\) and \((4-x)\) from \(R_2\). Note that \((4-x) = -(x-4)\).
\( = (5x+4) (x-4) \left|\begin{array}{ccc} 0 & 1 & 0 \\ 0 & -1 & x-4 \\ 1 & 2x & x+4 \end{array}\right| \)
Wait, let's restart from here. This is simpler.
\( = (5x+4) \left|\begin{array}{ccc} 0 & x-4 & 0 \\ 1 & x+4 & 2x \\ 1 & 2x & x+4 \end{array}\right| \)
Expand along \(R_1\):
\( = (5x+4) [-(x-4)(1(x+4) - 1(2x))] \)
\( = (5x+4) [-(x-4)(x+4-2x)] \)
\( = (5x+4) [-(x-4)(4-x)] \)
Since \(-(4-x) = x-4\), we have \(-(x-4)(4-x) = (x-4)(x-4) = (x-4)^2\).
Or, \(-(x-4)(4-x) = (-(x-4))(4-x) = (4-x)(4-x) = (4-x)^2\).
\( = (5x+4)(4-x)^2 \)
\( = \) જ.બા.
In simple words: We started by adding all columns to the first column, which allowed us to factor out \((5x+4)\). Then, we created zeros in the first row by subtracting the second row from the first. Expanding the determinant along the first row, we simplified the terms to arrive at \((4-x)^2\), thus proving the identity.

🎯 Exam Tip: When all rows (or columns) have the same sum, adding all columns (or rows) to one column (or row) creates a common factor, which can then be taken out to simplify the determinant significantly.

 

Question 10.(ii) \( \left|\begin{array}{ccc} y+k & y & y \\ y & y+k & y \\ y & y & y+k \end{array}\right| = k^2(3y+k) \)
Answer:ડા.બા. \( = \left|\begin{array}{ccc} y+k & y & y \\ y & y+k & y \\ y & y & y+k \end{array}\right| \)
Apply the operation \(C_1 \rightarrow C_1 + C_2 + C_3\):
\( = \left|\begin{array}{ccc} (y+k)+y+y & y & y \\ y+(y+k)+y & y+k & y \\ y+y+(y+k) & y & y+k \end{array}\right| \)
\( = \left|\begin{array}{ccc} 3y+k & y & y \\ 3y+k & y+k & y \\ 3y+k & y & y+k \end{array}\right| \)
Take out common factor \((3y+k)\) from \(C_1\):
\( = (3y+k) \left|\begin{array}{ccc} 1 & y & y \\ 1 & y+k & y \\ 1 & y & y+k \end{array}\right| \)
Apply the operation \(R_1 \rightarrow R_1 - R_2\):
\( = (3y+k) \left|\begin{array}{ccc} 1-1 & y-(y+k) & y-y \\ 1 & y+k & y \\ 1 & y & y+k \end{array}\right| \)
\( = (3y+k) \left|\begin{array}{ccc} 0 & -k & 0 \\ 1 & y+k & y \\ 1 & y & y+k \end{array}\right| \)
Expand the determinant along \(R_1\):
\( = (3y+k) [0 - (-k)(1(y+k) - 1(y)) + 0] \)
\( = (3y+k) [k(y+k-y)] \)
\( = (3y+k) [k(k)] \)
\( = k^2(3y+k) \)
\( = \) જ.બા.
In simple words: We first added all columns into the first column to get a common factor \((3y+k)\), which we then took out. Next, we subtracted the second row from the first to create zeros in the first row. Expanding this simplified determinant along the first row, we found the remaining factor to be \(k^2\), which proved the identity.

🎯 Exam Tip: Similar to the previous problem, if the sum of elements in each row or column is the same, adding them up into one row/column is a powerful first step to extract a common factor and simplify the determinant.

 

Question 11.(i) \( \left|\begin{array}{ccc} a-b-c & 2a & 2a \\ 2b & b-c-a & 2b \\ 2c & 2c & c-a-b \end{array}\right| = (a+b+c)^3 \)
Answer:ડા.બા. \( = \left|\begin{array}{ccc} a-b-c & 2a & 2a \\ 2b & b-c-a & 2b \\ 2c & 2c & c-a-b \end{array}\right| \)
Apply the operation \(R_1 \rightarrow R_1 + R_2 + R_3\):
\( = \left|\begin{array}{ccc} (a-b-c)+2b+2c & 2a+(b-c-a)+2c & 2a+2b+(c-a-b) \\ 2b & b-c-a & 2b \\ 2c & 2c & c-a-b \end{array}\right| \)
\( = \left|\begin{array}{ccc} a+b+c & a+b+c & a+b+c \\ 2b & b-c-a & 2b \\ 2c & 2c & c-a-b \end{array}\right| \)
Take out common factor \((a+b+c)\) from \(R_1\):
\( = (a+b+c) \left|\begin{array}{ccc} 1 & 1 & 1 \\ 2b & b-c-a & 2b \\ 2c & 2c & c-a-b \end{array}\right| \)
Apply the operation \(C_1 \rightarrow C_1 - C_2\):
\( = (a+b+c) \left|\begin{array}{ccc} 1-1 & 1 & 1 \\ 2b-(b-c-a) & b-c-a & 2b \\ 2c-2c & 2c & c-a-b \end{array}\right| \)
\( = (a+b+c) \left|\begin{array}{ccc} 0 & 1 & 1 \\ b+c+a & b-c-a & 2b \\ 0 & 2c & c-a-b \end{array}\right| \)
Apply the operation \(C_2 \rightarrow C_2 - C_3\):
\( = (a+b+c) \left|\begin{array}{ccc} 0 & 1-1 & 1 \\ b+c+a & (b-c-a)-2b & 2b \\ 0 & 2c-(c-a-b) & c-a-b \end{array}\right| \)
\( = (a+b+c) \left|\begin{array}{ccc} 0 & 0 & 1 \\ b+c+a & -(a+b+c) & 2b \\ 0 & c+a+b & c-a-b \end{array}\right| \)
Expand the determinant along \(R_1\):
\( = (a+b+c) [1((b+c+a)(c+a+b) - (-(a+b+c))(0))] \)
\( = (a+b+c) [(a+b+c)(a+b+c)] \)
\( = (a+b+c) (a+b+c)^2 \)
\( = (a+b+c)^3 \)
\( = \) જ.બા.
In simple words: We added all three rows to the first row, which created a common factor of \((a+b+c)\). After taking this out, we performed column operations to get two zeros in the first row. Then, we expanded the determinant along the first row and simplified the terms to prove the identity \((a+b+c)^3\).

🎯 Exam Tip: When the diagonal elements are complex and off-diagonal elements are simpler, applying \(R_1 \rightarrow R_1 + R_2 + R_3\) (or similar column operations) can often create a common factor, simplifying the determinant significantly.

 

Question 11.(ii) \( \left|\begin{array}{ccc} x+y+2z & x & y \\ z & y+z+2x & y \\ z & x & z+x+2y \end{array}\right| = 2(x+y+z)^3 \)
Answer:ડા.બા. \( = \left|\begin{array}{ccc} x+y+2z & x & y \\ z & y+z+2x & y \\ z & x & z+x+2y \end{array}\right| \)
Apply the operation \(C_1 \rightarrow C_1 + C_2 + C_3\):
\( = \left|\begin{array}{ccc} (x+y+2z)+x+y & x & y \\ z+(y+z+2x)+y & y+z+2x & y \\ z+x+(z+x+2y) & x & z+x+2y \end{array}\right| \)
\( = \left|\begin{array}{ccc} 2x+2y+2z & x & y \\ 2x+2y+2z & y+z+2x & y \\ 2x+2y+2z & x & z+x+2y \end{array}\right| \)
Take out common factor \(2(x+y+z)\) from \(C_1\):
\( = 2(x+y+z) \left|\begin{array}{ccc} 1 & x & y \\ 1 & y+z+2x & y \\ 1 & x & z+x+2y \end{array}\right| \)
Apply the operation \(R_2 \rightarrow R_2 - R_1\):
\( = 2(x+y+z) \left|\begin{array}{ccc} 1 & x & y \\ 1-1 & (y+z+2x)-x & y-y \\ 1 & x & z+x+2y \end{array}\right| \)
\( = 2(x+y+z) \left|\begin{array}{ccc} 1 & x & y \\ 0 & y+z+x & 0 \\ 1 & x & z+x+2y \end{array}\right| \)
Apply the operation \(R_3 \rightarrow R_3 - R_1\):
\( = 2(x+y+z) \left|\begin{array}{ccc} 1 & x & y \\ 0 & x+y+z & 0 \\ 1-1 & x-x & (z+x+2y)-y \end{array}\right| \)
\( = 2(x+y+z) \left|\begin{array}{ccc} 1 & x & y \\ 0 & x+y+z & 0 \\ 0 & 0 & x+y+z \end{array}\right| \)
Expand the determinant along \(C_1\):
\( = 2(x+y+z) [1((x+y+z)(x+y+z) - (0)(0))] \)
\( = 2(x+y+z) [(x+y+z)^2] \)
\( = 2(x+y+z)^3 \)
\( = \) જ.બા.
In simple words: We added all columns to the first column to get a common factor of \(2(x+y+z)\), which we factored out. Then, we used row operations to create zeros in the first column, making the determinant a triangular matrix. The value of a triangular determinant is the product of its diagonal elements. Multiplying these elements gave us \((x+y+z)^2\), which, when combined with the factored term, resulted in \(2(x+y+z)^3\).

🎯 Exam Tip: Converting a determinant into an upper or lower triangular form is a very efficient way to calculate its value, as the determinant is simply the product of the diagonal elements. Row/column operations are key to achieving this form.

 

Question 12.\[ \left|\begin{array}{ccc} 1 & x & x^2 \\ x^2 & 1 & x \\ x & x^2 & 1 \end{array}\right| = (1-x^3)^2 \]
Answer:ડા.બા. \( = \left|\begin{array}{ccc} 1 & x & x^2 \\ x^2 & 1 & x \\ x & x^2 & 1 \end{array}\right| \)
Apply the operation \(C_1 \rightarrow C_1 + C_2 + C_3\):
\( = \left|\begin{array}{ccc} 1+x+x^2 & x & x^2 \\ x^2+1+x & 1 & x \\ x+x^2+1 & x^2 & 1 \end{array}\right| \)
Take out common factor \((1+x+x^2)\) from \(C_1\):
\( = (1+x+x^2) \left|\begin{array}{ccc} 1 & x & x^2 \\ 1 & 1 & x \\ 1 & x^2 & 1 \end{array}\right| \)
Apply the operation \(R_1 \rightarrow R_1 - R_2\):
\( = (1+x+x^2) \left|\begin{array}{ccc} 1-1 & x-1 & x^2-x \\ 1 & 1 & x \\ 1 & x^2 & 1 \end{array}\right| \)
\( = (1+x+x^2) \left|\begin{array}{ccc} 0 & x-1 & x(x-1) \\ 1 & 1 & x \\ 1 & x^2 & 1 \end{array}\right| \)
Take out common factor \((x-1)\) from \(R_1\):
\( = (1+x+x^2)(x-1) \left|\begin{array}{ccc} 0 & 1 & x \\ 1 & 1 & x \\ 1 & x^2 & 1 \end{array}\right| \)
Apply the operation \(R_2 \rightarrow R_2 - R_3\):
\( = (1+x+x^2)(x-1) \left|\begin{array}{ccc} 0 & 1 & x \\ 1-1 & 1-x^2 & x-1 \\ 1 & x^2 & 1 \end{array}\right| \)
\( = (1+x+x^2)(x-1) \left|\begin{array}{ccc} 0 & 1 & x \\ 0 & (1-x)(1+x) & -(1-x) \\ 1 & x^2 & 1 \end{array}\right| \)
Take out common factor \((1-x)\) from \(R_2\):
\( = (1+x+x^2)(x-1)(1-x) \left|\begin{array}{ccc} 0 & 1 & x \\ 0 & 1+x & -1 \\ 1 & x^2 & 1 \end{array}\right| \)
Expand the determinant along \(C_1\):
\( = (1+x+x^2)(x-1)(1-x) [1((1)(-1) - (x)(1+x))] \)
\( = (1+x+x^2)(x-1)(1-x) [-1 - x - x^2] \)
\( = (1+x+x^2)(x-1)(1-x) [-(1+x+x^2)] \)
We know that \((1-x^3) = (1-x)(1+x+x^2)\).
So, \((x-1)(1-x)\) can be written as \(-(1-x)(1-x) = -(1-x)^2\).
Let's re-arrange the factors:
\( = (1+x+x^2)[-(1-x)] \cdot [-(1+x+x^2)] (1-x) \)
\( = (1+x+x^2)(1-x) \cdot (1+x+x^2)(1-x) \)
\( = (1-x^3)(1-x^3) \)
\( = (1-x^3)^2 \)
\( = \) જ.બા.
In simple words: We started by adding all columns to the first column to get a common factor \((1+x+x^2)\). Then, we used row operations to create zeros in the first column and factored out \((x-1)\) and \((1-x)\). Finally, we expanded the determinant and multiplied all the collected factors. Recognizing that \((1-x)(1+x+x^2)\) is \((1-x^3)\), we arrived at the result \((1-x^3)^2\).

🎯 Exam Tip: Remember the algebraic identity \(a^3-b^3 = (a-b)(a^2+ab+b^2)\). This is frequently useful when terms like \(1+x+x^2\) and \(x-1\) appear in determinant problems, allowing for consolidation of factors.

 

Question 13.\[ \left|\begin{array}{ccc} 1+a^2-b^2 & 2ab & -2b \\ 2ab & 1-a^2+b^2 & 2a \\ 2b & -2a & 1-a^2-b^2 \end{array}\right| = (1+a^2+b^2)^3 \]
Answer:ડા.બા. \( = \left|\begin{array}{ccc} 1+a^2-b^2 & 2ab & -2b \\ 2ab & 1-a^2+b^2 & 2a \\ 2b & -2a & 1-a^2-b^2 \end{array}\right| \)
Apply the operation \(R_1 \rightarrow R_1 + bR_3\):
\( = \left|\begin{array}{ccc} (1+a^2-b^2)+b(2b) & 2ab+b(-2a) & -2b+b(1-a^2-b^2) \\ 2ab & 1-a^2+b^2 & 2a \\ 2b & -2a & 1-a^2-b^2 \end{array}\right| \)
\( = \left|\begin{array}{ccc} 1+a^2-b^2+2b^2 & 2ab-2ab & -2b+b-a^2b-b^3 \\ 2ab & 1-a^2+b^2 & 2a \\ 2b & -2a & 1-a^2-b^2 \end{array}\right| \)
\( = \left|\begin{array}{ccc} 1+a^2+b^2 & 0 & -b-a^2b-b^3 \\ 2ab & 1-a^2+b^2 & 2a \\ 2b & -2a & 1-a^2-b^2 \end{array}\right| \)
\( = \left|\begin{array}{ccc} 1+a^2+b^2 & 0 & -b(1+a^2+b^2) \\ 2ab & 1-a^2+b^2 & 2a \\ 2b & -2a & 1-a^2-b^2 \end{array}\right| \)
Apply the operation \(R_2 \rightarrow R_2 - aR_3\):
\( = \left|\begin{array}{ccc} 1+a^2+b^2 & 0 & -b(1+a^2+b^2) \\ 2ab-a(2b) & (1-a^2+b^2)-a(-2a) & 2a-a(1-a^2-b^2) \\ 2b & -2a & 1-a^2-b^2 \end{array}\right| \)
\( = \left|\begin{array}{ccc} 1+a^2+b^2 & 0 & -b(1+a^2+b^2) \\ 0 & 1-a^2+b^2+2a^2 & 2a-a+a^3+ab^2 \\ 2b & -2a & 1-a^2-b^2 \end{array}\right| \)
\( = \left|\begin{array}{ccc} 1+a^2+b^2 & 0 & -b(1+a^2+b^2) \\ 0 & 1+a^2+b^2 & a+a^3+ab^2 \\ 2b & -2a & 1-a^2-b^2 \end{array}\right| \)
\( = \left|\begin{array}{ccc} 1+a^2+b^2 & 0 & -b(1+a^2+b^2) \\ 0 & 1+a^2+b^2 & a(1+a^2+b^2) \\ 2b & -2a & 1-a^2-b^2 \end{array}\right| \)
Take out common factor \((1+a^2+b^2)\) from \(R_1\) and \(R_2\):
\( = (1+a^2+b^2)^2 \left|\begin{array}{ccc} 1 & 0 & -b \\ 0 & 1 & a \\ 2b & -2a & 1-a^2-b^2 \end{array}\right| \)
Expand the determinant:
\( = (1+a^2+b^2)^2 [1((1)(1-a^2-b^2) - (a)(-2a)) - 0 + (-b)((0)(-2a) - (1)(2b))] \)
\( = (1+a^2+b^2)^2 [1-a^2-b^2+2a^2 - b(-2b)] \)
\( = (1+a^2+b^2)^2 [1+a^2-b^2+2b^2] \)
\( = (1+a^2+b^2)^2 [1+a^2+b^2] \)
\( = (1+a^2+b^2)^3 \)
\( = \) જ.બા.
In simple words: We performed row operations \(R_1 \rightarrow R_1 + bR_3\) and \(R_2 \rightarrow R_2 - aR_3\) to create a common factor \((1+a^2+b^2)\) in the first two rows. We factored it out twice, making the determinant simpler. Finally, we expanded the remaining 3x3 determinant, which resulted in another \((1+a^2+b^2)\) term, leading to the proof of \((1+a^2+b^2)^3\).

🎯 Exam Tip: When the determinant has quadratic terms like \(1+a^2-b^2\) and \(2ab\), combining rows/columns by multiplying with 'a' or 'b' (e.g., \(R_i \rightarrow R_i + k R_j\)) is a common strategy to generate the target factor \((1+a^2+b^2)\).

 

Question 14.\[ \left|\begin{array}{ccc} a^2+1 & ab & ac \\ ab & b^2+1 & bc \\ ca & cb & c^2+1 \end{array}\right| = 1+a^2+b^2+c^2 \]
Answer:ડા.બા. \( = \left|\begin{array}{ccc} a^2+1 & ab & ac \\ ab & b^2+1 & bc \\ ca & cb & c^2+1 \end{array}\right| \)
Multiply \(R_1\) by a, \(R_2\) by b, and \(R_3\) by c. To compensate, divide the determinant by abc:
\( = \frac{1}{abc} \left|\begin{array}{ccc} a(a^2+1) & a^2b & a^2c \\ ab^2 & b(b^2+1) & b^2c \\ ac^2 & bc^2 & c(c^2+1) \end{array}\right| \)
Take out common factor 'a' from \(C_1\), 'b' from \(C_2\), and 'c' from \(C_3\):
\( = \frac{abc}{abc} \left|\begin{array}{ccc} a^2+1 & a^2 & a^2 \\ b^2 & b^2+1 & b^2 \\ c^2 & c^2 & c^2+1 \end{array}\right| \)
\( = \left|\begin{array}{ccc} a^2+1 & a^2 & a^2 \\ b^2 & b^2+1 & b^2 \\ c^2 & c^2 & c^2+1 \end{array}\right| \)
Apply the operation \(R_1 \rightarrow R_1 + R_2 + R_3\):
\( = \left|\begin{array}{ccc} (a^2+1)+b^2+c^2 & a^2+(b^2+1)+c^2 & a^2+b^2+(c^2+1) \\ b^2 & b^2+1 & b^2 \\ c^2 & c^2 & c^2+1 \end{array}\right| \)
\( = \left|\begin{array}{ccc} 1+a^2+b^2+c^2 & 1+a^2+b^2+c^2 & 1+a^2+b^2+c^2 \\ b^2 & b^2+1 & b^2 \\ c^2 & c^2 & c^2+1 \end{array}\right| \)
Take out common factor \((1+a^2+b^2+c^2)\) from \(R_1\):
\( = (1+a^2+b^2+c^2) \left|\begin{array}{ccc} 1 & 1 & 1 \\ b^2 & b^2+1 & b^2 \\ c^2 & c^2 & c^2+1 \end{array}\right| \)
Apply the operation \(C_2 \rightarrow C_2 - C_1\):
\( = (1+a^2+b^2+c^2) \left|\begin{array}{ccc} 1 & 1-1 & 1 \\ b^2 & (b^2+1)-b^2 & b^2 \\ c^2 & c^2-c^2 & c^2+1 \end{array}\right| \)
\( = (1+a^2+b^2+c^2) \left|\begin{array}{ccc} 1 & 0 & 1 \\ b^2 & 1 & b^2 \\ c^2 & 0 & c^2+1 \end{array}\right| \)
Apply the operation \(C_3 \rightarrow C_3 - C_1\):
\( = (1+a^2+b^2+c^2) \left|\begin{array}{ccc} 1 & 0 & 1-1 \\ b^2 & 1 & b^2-b^2 \\ c^2 & 0 & (c^2+1)-c^2 \end{array}\right| \)
\( = (1+a^2+b^2+c^2) \left|\begin{array}{ccc} 1 & 0 & 0 \\ b^2 & 1 & 0 \\ c^2 & 0 & 1 \end{array}\right| \)
This is a lower triangular matrix. Its determinant is the product of diagonal elements.
\( = (1+a^2+b^2+c^2) (1 \cdot 1 \cdot 1) \)
\( = 1+a^2+b^2+c^2 \)
\( = \) જ.બા.
In simple words: We first multiplied each row by a, b, and c respectively, and then divided the determinant by abc. This allowed us to factor out a, b, c from each column, bringing back the common factor. Then, we added all rows to the first row to get a common factor \((1+a^2+b^2+c^2)\). After taking this out, we performed column operations to create zeros and simplify the determinant into a triangular form. The value of this triangular determinant is 1, proving the identity.

🎯 Exam Tip: A common strategy for these types of problems is to manipulate rows/columns to achieve a common factor like \((1+a^2+b^2+c^2)\). Multiplying and then dividing by the row/column elements can expose such factors.

 

Question 15. A એ 3 × 3 કક્ષાનો ચોરસ શ્રેણિક હોય, તો \(|kA| =\) ..........
(A) \(k|A|\)
(B) \(k^2|A|\)
(C) \(k^3|A|\)
(D) \(3k|A|\)
Answer: (C) \(k^3|A|\)
In simple words: When you multiply a square matrix A of size 3x3 by a scalar k, the determinant of the new matrix \(|kA|\) is equal to k raised to the power of 3, multiplied by the determinant of A.

🎯 Exam Tip: Remember the property that for any square matrix A of order n, \(|kA| = k^n|A|\). This is a common formula tested in determinant questions.

 

Question 16. નીચે આપેલામાંથી કયું વિધાન સત્ય છે ?
(A) નિશ્ચાયક એ ચોરસ શ્રેણિક છે.
(B) નિશ્ચાયક એ શ્રેણિક સાથે સંકળાયેલ એક સંખ્યા છે.
(C) નિશ્ચાયક એ ચોરસ મેબ્રિક સાથે સંકળાયેલ એક સંખ્યા છે.
(D) આમાંથી કોઈ નહિ.
Answer: (C) નિશ્ચાયક એ ચોરસ મેબ્રિક સાથે સંકળાયેલ એક સંખ્યા છે.
In simple words: A determinant is a specific number that is connected only to a square matrix. It is not the matrix itself, but a value derived from it.

🎯 Exam Tip: It is crucial to understand that a determinant is a numerical value associated with a square matrix, not the matrix itself. This fundamental definition is key to solving theoretical questions.