GSEB Class 12 Maths Solutions Chapter 4 નિશ્ચાયક Exercise 4.4

નીરો આપેલા નિશ્ચાયકના પ્રત્યેક ઘટકના ઉપનિશ્ચાયક અને સહઅવયવ લખો :

Question 1.
(i) \( \left|\begin{array}{cc} 2 & -4 \\ 0 & 3 \end{array}\right| \)
Answer: Let the given determinant be \( A \). We can represent it as \( A = \left|\begin{array}{ll} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array}\right| = \left|\begin{array}{cc} 2 & -4 \\ 0 & 3 \end{array}\right| \).
Minors:
The minor of element 2 (\(a_{11}\)) is \(M_{11} = 3\).
The minor of element -4 (\(a_{12}\)) is \(M_{12} = 0\).
The minor of element 0 (\(a_{21}\)) is \(M_{21} = -4\).
The minor of element 3 (\(a_{22}\)) is \(M_{22} = 2\).
Co-factors:
The co-factor of \(a_{11}\) is \(A_{11} = (-1)^{1+1} (3) = 3\).
The co-factor of \(a_{12}\) is \(A_{12} = (-1)^{1+2} (0) = 0\).
The co-factor of \(a_{21}\) is \(A_{21} = (-1)^{2+1} (-4) = 4\).
The co-factor of \(a_{22}\) is \(A_{22} = (-1)^{2+2} (2) = 2\).
In simple words: To find the minor of an element, calculate the determinant of the smaller matrix left after removing the element's row and column. The co-factor is found by multiplying the minor by \((-1)\) raised to the power of the sum of the element's row and column indices.

🎯 Exam Tip: Accurately identifying the element's row and column indices is crucial for calculating both the minor and especially the co-factor, due to the alternating sign factor.

(ii) \( \left|\begin{array}{ll} a & c \\ b & d \end{array}\right| \)
Answer: Let the given determinant be \( A \). We can write it as \( A = \left|\begin{array}{ll} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array}\right| = \left|\begin{array}{cc} a & c \\ b & d \end{array}\right| \).
Minors:
The minor of element \(a\) (\(a_{11}\)) is \(M_{11} = d\).
The minor of element \(c\) (\(a_{12}\)) is \(M_{12} = b\).
The minor of element \(b\) (\(a_{21}\)) is \(M_{21} = c\).
The minor of element \(d\) (\(a_{22}\)) is \(M_{22} = a\).
Co-factors:
The co-factor of \(a_{11}\) is \(A_{11} = (-1)^{1+1} (d) = d\).
The co-factor of \(a_{12}\) is \(A_{12} = (-1)^{1+2} (b) = -b\).
The co-factor of \(a_{21}\) is \(A_{21} = (-1)^{2+1} (c) = -c\).
The co-factor of \(a_{22}\) is \(A_{22} = (-1)^{2+2} (a) = a\).
In simple words: The minor for each element is the value of the remaining element when its row and column are removed. The co-factor is then found by applying the correct sign based on the element's position.

🎯 Exam Tip: When dealing with symbolic determinants, remember that the minors will also be expressions involving other elements of the determinant. Be careful with signs for co-factors.

Question 2.
(i) \( \left|\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right| \)
Answer: Let the given determinant be \( A \). We can write it as \( A = \left|\begin{array}{lll} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array}\right| = \left|\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right| \).
Minors:
The minor of element 1 (\(a_{11}\)) is \(M_{11} = \left|\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right| = 1\).
The minor of element 0 (\(a_{12}\)) is \(M_{12} = \left|\begin{array}{cc} 0 & 0 \\ 0 & 1 \end{array}\right| = 0\).
The minor of element 0 (\(a_{13}\)) is \(M_{13} = \left|\begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array}\right| = 0\).
The minor of element 0 (\(a_{21}\)) is \(M_{21} = \left|\begin{array}{cc} 0 & 0 \\ 0 & 1 \end{array}\right| = 0\).
The minor of element 1 (\(a_{22}\)) is \(M_{22} = \left|\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right| = 1\).
The minor of element 0 (\(a_{23}\)) is \(M_{23} = \left|\begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array}\right| = 0\).
The minor of element 0 (\(a_{31}\)) is \(M_{31} = \left|\begin{array}{cc} 0 & 0 \\ 1 & 0 \end{array}\right| = 0\).
The minor of element 0 (\(a_{32}\)) is \(M_{32} = \left|\begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array}\right| = 0\).
The minor of element 1 (\(a_{33}\)) is \(M_{33} = \left|\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right| = 1\).
Co-factors:
The co-factor of \(a_{11}\) is \(A_{11} = (-1)^{1+1} (1) = 1\).
The co-factor of \(a_{12}\) is \(A_{12} = (-1)^{1+2} (0) = 0\).
The co-factor of \(a_{13}\) is \(A_{13} = (-1)^{1+3} (0) = 0\).
The co-factor of \(a_{21}\) is \(A_{21} = (-1)^{2+1} (0) = 0\).
The co-factor of \(a_{22}\) is \(A_{22} = (-1)^{2+2} (1) = 1\).
The co-factor of \(a_{23}\) is \(A_{23} = (-1)^{2+3} (0) = 0\).
The co-factor of \(a_{31}\) is \(A_{31} = (-1)^{3+1} (0) = 0\).
The co-factor of \(a_{32}\) is \(A_{32} = (-1)^{3+2} (0) = 0\).
The co-factor of \(a_{33}\) is \(A_{33} = (-1)^{3+3} (1) = 1\).
In simple words: For a 3x3 determinant, the minor of an element is the determinant of the 2x2 matrix remaining after removing the row and column of that element. The co-factor is then the minor with its correct sign.

🎯 Exam Tip: When working with larger determinants, organize your calculations by rows or columns to avoid errors. For a diagonal matrix like this, many minors and co-factors will be zero or one.

(ii) \( \left|\begin{array}{ccc} 1 & 0 & 4 \\ 3 & 5 & -1 \\ 0 & 1 & 2 \end{array}\right| \)
Answer: Let the given determinant be \( A \). We can write it as \( A = \left|\begin{array}{lll} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array}\right| = \left|\begin{array}{ccc} 1 & 0 & 4 \\ 3 & 5 & -1 \\ 0 & 1 & 2 \end{array}\right| \).
Minors:
The minor of \(a_{11}\) (element 1) is \(M_{11} = \left|\begin{array}{cc} 5 & -1 \\ 1 & 2 \end{array}\right| = (5 \times 2) - (-1 \times 1) = 10 + 1 = 11\).
The minor of \(a_{12}\) (element 0) is \(M_{12} = \left|\begin{array}{cc} 3 & -1 \\ 0 & 2 \end{array}\right| = (3 \times 2) - (-1 \times 0) = 6 - 0 = 6\).
The minor of \(a_{13}\) (element 4) is \(M_{13} = \left|\begin{array}{cc} 3 & 5 \\ 0 & 1 \end{array}\right| = (3 \times 1) - (5 \times 0) = 3 - 0 = 3\).
The minor of \(a_{21}\) (element 3) is \(M_{21} = \left|\begin{array}{cc} 0 & 4 \\ 1 & 2 \end{array}\right| = (0 \times 2) - (4 \times 1) = 0 - 4 = -4\).
The minor of \(a_{22}\) (element 5) is \(M_{22} = \left|\begin{array}{cc} 1 & 4 \\ 0 & 2 \end{array}\right| = (1 \times 2) - (4 \times 0) = 2 - 0 = 2\).
The minor of \(a_{23}\) (element -1) is \(M_{23} = \left|\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right| = (1 \times 1) - (0 \times 0) = 1 - 0 = 1\).
The minor of \(a_{31}\) (element 0) is \(M_{31} = \left|\begin{array}{cc} 0 & 4 \\ 5 & -1 \end{array}\right| = (0 \times -1) - (4 \times 5) = 0 - 20 = -20\).
The minor of \(a_{32}\) (element 1) is \(M_{32} = \left|\begin{array}{cc} 1 & 4 \\ 3 & -1 \end{array}\right| = (1 \times -1) - (4 \times 3) = -1 - 12 = -13\).
The minor of \(a_{33}\) (element 2) is \(M_{33} = \left|\begin{array}{cc} 1 & 0 \\ 3 & 5 \end{array}\right| = (1 \times 5) - (0 \times 3) = 5 - 0 = 5\).
Co-factors:
The co-factor of \(a_{11}\) is \(A_{11} = (-1)^{1+1} (11) = 11\).
The co-factor of \(a_{12}\) is \(A_{12} = (-1)^{1+2} (6) = -6\).
The co-factor of \(a_{13}\) is \(A_{13} = (-1)^{1+3} (3) = 3\).
The co-factor of \(a_{21}\) is \(A_{21} = (-1)^{2+1} (-4) = 4\).
The co-factor of \(a_{22}\) is \(A_{22} = (-1)^{2+2} (2) = 2\).
The co-factor of \(a_{23}\) is \(A_{23} = (-1)^{2+3} (1) = -1\).
The co-factor of \(a_{31}\) is \(A_{31} = (-1)^{3+1} (-20) = -20\).
The co-factor of \(a_{32}\) is \(A_{32} = (-1)^{3+2} (-13) = 13\).
The co-factor of \(a_{33}\) is \(A_{33} = (-1)^{3+3} (5) = 5\).
In simple words: Each minor is the determinant of the smaller matrix formed by removing one row and one column. Each co-factor is then found by applying the alternating sign rule, based on whether the sum of its row and column numbers is even or odd.

🎯 Exam Tip: Double-check the 2x2 determinant calculations for each minor. A small arithmetic error in a minor will lead to an incorrect co-factor. Pay attention to the sign rule for co-factors.

Question 3.બીજી હારના ઘટકોના સહઅવયવના ઉપયોગથી \( \Delta = \left|\begin{array}{ccc} 5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3 \end{array}\right| \) નું મૂલ્ય શોધો.
Answer: Let the given determinant be \( \Delta = \left|\begin{array}{ccc} 5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3 \end{array}\right| \). We can represent it as \( \left|\begin{array}{lll} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array}\right| \).
We need to find the value of the determinant using the co-factors of the elements of the second row.
The elements of the second row are \(a_{21}\), \(a_{22}\), and \(a_{23}\).
The co-factor of \(a_{21}\) is \(A_{21} = (-1)^{2+1} \left|\begin{array}{cc} 3 & 8 \\ 2 & 3 \end{array}\right| = -( (3 \times 3) - (8 \times 2) ) = -(9 - 16) = -(-7) = 7\).
The co-factor of \(a_{22}\) is \(A_{22} = (-1)^{2+2} \left|\begin{array}{cc} 5 & 8 \\ 1 & 3 \end{array}\right| = ( (5 \times 3) - (8 \times 1) ) = (15 - 8) = 7\).
The co-factor of \(a_{23}\) is \(A_{23} = (-1)^{2+3} \left|\begin{array}{cc} 5 & 3 \\ 1 & 2 \end{array}\right| = -( (5 \times 2) - (3 \times 1) ) = -(10 - 3) = -7\).
Now, the value of the determinant \( \Delta \) is calculated as:
\( \Delta = a_{21} A_{21} + a_{22} A_{22} + a_{23} A_{23} \)
\( = 2(7) + 0(7) + 1(-7) \)
\( = 14 + 0 - 7 = 7 \)
In simple words: To find the determinant's value using co-factors of a specific row or column, multiply each element in that row/column by its corresponding co-factor, and then add these products together. This method simplifies calculating the determinant.

🎯 Exam Tip: When expanding a determinant along a row or column, carefully identify the elements and their corresponding co-factors. A row or column with a zero element can simplify calculations significantly, as its term will be zero.

Question 4.ત્રીજા સ્તંભના ઘટકોના અવયવના ઉપયોગથી \( \Delta = \left|\begin{array}{ccc} 1 & x & y z \\ 1 & y & z x \\ 1 & z & x y \end{array}\right| \) નું મૂલ્ય શોધો.
Answer: Let the given determinant be \( \Delta = \left|\begin{array}{ccc} 1 & x & y z \\ 1 & y & z x \\ 1 & z & x y \end{array}\right| \). We can represent it as \( \left|\begin{array}{lll} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array}\right| \).
We need to find the value of the determinant using the co-factors of the elements of the third column.
The elements of the third column are \(a_{13}\), \(a_{23}\), and \(a_{33}\).
The co-factor of \(a_{13}\) is \(A_{13} = (-1)^{1+3} \left|\begin{array}{cc} 1 & y \\ 1 & z \end{array}\right| = (1 \times z) - (y \times 1) = z - y\).
The co-factor of \(a_{23}\) is \(A_{23} = (-1)^{2+3} \left|\begin{array}{cc} 1 & x \\ 1 & z \end{array}\right| = - ( (1 \times z) - (x \times 1) ) = -(z - x) = x - z\).
The co-factor of \(a_{33}\) is \(A_{33} = (-1)^{3+3} \left|\begin{array}{cc} 1 & x \\ 1 & y \end{array}\right| = ( (1 \times y) - (x \times 1) ) = y - x\).
Now, the value of the determinant \( \Delta \) is calculated as:
\( \Delta = a_{13} A_{13} + a_{23} A_{23} + a_{33} A_{33} \)
\( = yz(z - y) + zx(x - z) + xy(y - x) \)
\( = z^2y - y^2z + x^2z - z^2x + y^2x - x^2y \)
Rearranging terms to factor:
\( = yz(z - y) - zx(z - x) - xy(x - y) \)
We can factor by grouping or by noting the cyclical symmetry. Let's try to get a common factor of \((x-y)\), \((y-z)\), \((z-x)\).
\( = yz(z - y) + zx(x - z) + xy(y - x) \)
\( = yz(z - y) - xy(z - y) - z^2x + z^2y + x^2z - x^2y \)
\( = (z - y)(yz - xy) - zx(z - x) + yx(z - x) \)
\( = y(z - y)(z - x) - x(z - x)(z - y) \)
This isn't directly leading to the provided solution factoring. Let's follow the solution's factorization steps which might involve adding and subtracting `xyz`.
The given solution's factorization is:
\( = yz(x - y) - xy(x - y) - z^2(x - y) + zx(x - y) \)
\( = (x - y) (yz - xy - z^2 + zx) \)
\( = (x - y) [y(z - x) - z(z - x)] \)
\( = (x - y) (y - z) (z - x) \)
In simple words: By expanding the determinant using co-factors of the third column, we get an algebraic expression. Factoring this expression strategically, we can find the determinant's value as a product of differences of x, y, and z.

🎯 Exam Tip: When dealing with algebraic determinants, look for patterns that allow for common factors. Expanding along a column or row with simple elements can make the initial sum easier to factorize later.

Question 5.પ્રશ્ન 5 માં વિધાન સારું બને તે રીતે આપેલ વિકલ્પોમાંથી સોગ્ય વિકલ્પ પસંદ કરો :
જો \( \Delta = \left|\begin{array}{lll} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array}\right| \) ને \(a_{ij}\) નો સહઅવયવ \(A_{ij}\) હોય, તો \( \Delta \) નું મૂલ્ય ............
(A) \(a_{11} A_{31} + a_{12} A_{32} + a_{13} A_{33}\)
(B) \(a_{11} A_{11} + a_{12} A_{21} + a_{13} A_{31}\)
(C) \(a_{21} A_{11} + a_{22} A_{12} + a_{23} A_{13}\)
(D) \(a_{11} A_{11} + a_{21} A_{21} + a_{31} A_{31}\)
Answer: (D) \(a_{11} A_{11} + a_{21} A_{21} + a_{31} A_{31}\)
The value of a determinant is found by summing the products of the elements of any row or column with their corresponding co-factors. For the determinant to be correctly calculated, the subscript of the element (\(a_{ij}\)) must match the subscript of its co-factor (\(A_{ij}\)). Option (D) correctly sums the products of elements from the first column with their corresponding co-factors: \(a_{11} A_{11}\) (first element, first co-factor), \(a_{21} A_{21}\) (second element, second co-factor), and \(a_{31} A_{31}\) (third element, third co-factor).
In simple words: The value of a determinant is found by multiplying each element in a row or column by its own co-factor, then adding all these results together. It is very important that the element and its co-factor come from the exact same position in the matrix.

🎯 Exam Tip: Remember Laplace expansion. The determinant is the sum of products of elements of any row/column with their *corresponding* co-factors. If elements of one row/column are multiplied by co-factors of *another* row/column, the sum will always be zero.