GSEB Class 12 Maths Solutions Chapter 4 Determinants Exercise 4.4

Question 1. Write minors and cofactors of elements of the following determinants:
(i) \( \left|\begin{array}{cc} 2 & -4 \\ 0 & 3 \end{array}\right| \)
(ii) \( \left|\begin{array}{cc} a & c \\ b & d \end{array}\right| \)
Answer:
(i) For the determinant \( \Delta = \left|\begin{array}{cc} 2 & -4 \\ 0 & 3 \end{array}\right| \):
The minor of \( a_{11} \) is \( M_{11} = 3 \).
The minor of \( a_{12} \) is \( M_{12} = 0 \).
The minor of \( a_{21} \) is \( M_{21} = -4 \).
The minor of \( a_{22} \) is \( M_{22} = 2 \).

The cofactor of \( a_{11} \) is \( A_{11} = (-1)^{1+1} M_{11} = (-1)^2 \times 3 = 3 \).
The cofactor of \( a_{12} \) is \( A_{12} = (-1)^{1+2} M_{12} = (-1)^3 \times 0 = 0 \).
The cofactor of \( a_{21} \) is \( A_{21} = (-1)^{2+1} M_{21} = -(-4) = 4 \).
The cofactor of \( a_{22} \) is \( A_{22} = (-1)^{2+2} M_{22} = 2 \).

(ii) For the determinant \( \Delta = \left|\begin{array}{cc} a & c \\ b & d \end{array}\right| \):
The minor of \( a_{11} \) is \( M_{11} = d \).
The minor of \( a_{12} \) is \( M_{12} = b \).
The minor of \( a_{21} \) is \( M_{21} = c \).
The minor of \( a_{22} \) is \( M_{22} = a \).

The cofactor of \( a_{11} \) is \( A_{11} = (-1)^{1+1} M_{11} = d \).
The cofactor of \( a_{12} \) is \( A_{12} = (-1)^{1+2} M_{12} = -b \).
The cofactor of \( a_{21} \) is \( A_{21} = (-1)^{2+1} M_{21} = -c \).
The cofactor of \( a_{22} \) is \( A_{22} = (-1)^{2+2} M_{22} = a \).
In simple words: For a given determinant, the minor of an element is the determinant of the sub-matrix formed by removing that element's row and column. The cofactor is found by multiplying the minor by \( (-1)^{i+j} \), where \( i \) is the row number and \( j \) is the column number of the element.

Exam Tip: Remember that cofactors combine the minor with a sign based on its position, which alternates in a chessboard pattern (+ - + -). Use the formula \( A_{ij} = (-1)^{i+j} M_{ij} \) accurately.

Question 2. Find the minors and cofactors for the elements of the following determinants:
(i) \( \left|\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right| \)
(ii) \( \left|\begin{array}{ccc} 1 & 0 & 4 \\ 3 & 5 & -1 \\ \theta & 1 & 2 \end{array}\right| \)
Answer:
(i) For the determinant \( \Delta = \left|\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right| \):
**Minors:**
\( M_{11} = \left|\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right| = (1 \times 1) - (0 \times 0) = 1 - 0 = 1 \)
\( M_{12} = \left|\begin{array}{cc} 0 & 0 \\ 0 & 1 \end{array}\right| = (0 \times 1) - (0 \times 0) = 0 - 0 = 0 \)
\( M_{13} = \left|\begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array}\right| = (0 \times 0) - (1 \times 0) = 0 - 0 = 0 \)
\( M_{21} = \left|\begin{array}{cc} 0 & 0 \\ 0 & 1 \end{array}\right| = (0 \times 1) - (0 \times 0) = 0 - 0 = 0 \)
\( M_{22} = \left|\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right| = (1 \times 1) - (0 \times 0) = 1 - 0 = 1 \)
\( M_{23} = \left|\begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array}\right| = (1 \times 0) - (0 \times 0) = 0 - 0 = 0 \)
\( M_{31} = \left|\begin{array}{cc} 0 & 0 \\ 1 & 0 \end{array}\right| = (0 \times 0) - (0 \times 1) = 0 - 0 = 0 \)
\( M_{32} = \left|\begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array}\right| = (1 \times 0) - (0 \times 0) = 0 - 0 = 0 \)
\( M_{33} = \left|\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right| = (1 \times 1) - (0 \times 0) = 1 - 0 = 1 \)

**Cofactors:**
\( A_{11} = (-1)^{1+1} M_{11} = 1 \times 1 = 1 \)
\( A_{12} = (-1)^{1+2} M_{12} = -1 \times 0 = 0 \)
\( A_{13} = (-1)^{1+3} M_{13} = 1 \times 0 = 0 \)
\( A_{21} = (-1)^{2+1} M_{21} = -1 \times 0 = 0 \)
\( A_{22} = (-1)^{2+2} M_{22} = 1 \times 1 = 1 \)
\( A_{23} = (-1)^{2+3} M_{23} = -1 \times 0 = 0 \)
\( A_{31} = (-1)^{3+1} M_{31} = 1 \times 0 = 0 \)
\( A_{32} = (-1)^{3+2} M_{32} = -1 \times 0 = 0 \)
\( A_{33} = (-1)^{3+3} M_{33} = 1 \times 1 = 1 \)

(ii) For the determinant \( \Delta = \left|\begin{array}{ccc} 1 & 0 & 4 \\ 3 & 5 & -1 \\ \theta & 1 & 2 \end{array}\right| \):
**Minors:**
\( M_{11} = \left|\begin{array}{cc} 5 & -1 \\ 1 & 2 \end{array}\right| = (5 \times 2) - (-1 \times 1) = 10 - (-1) = 10 + 1 = 11 \)
\( M_{12} = \left|\begin{array}{cc} 3 & -1 \\ 0 & 2 \end{array}\right| = (3 \times 2) - (-1 \times 0) = 6 - 0 = 6 \)
\( M_{13} = \left|\begin{array}{cc} 3 & 5 \\ 0 & 1 \end{array}\right| = (3 \times 1) - (5 \times 0) = 3 - 0 = 3 \)
\( M_{21} = \left|\begin{array}{cc} 0 & 4 \\ 1 & 2 \end{array}\right| = (0 \times 2) - (4 \times 1) = 0 - 4 = -4 \)
\( M_{22} = \left|\begin{array}{cc} 1 & 4 \\ 0 & 2 \end{array}\right| = (1 \times 2) - (4 \times 0) = 2 - 0 = 2 \)
\( M_{23} = \left|\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right| = (1 \times 1) - (0 \times 0) = 1 - 0 = 1 \)
\( M_{31} = \left|\begin{array}{cc} 0 & 4 \\ 5 & -1 \end{array}\right| = (0 \times -1) - (4 \times 5) = 0 - 20 = -20 \)
\( M_{32} = \left|\begin{array}{cc} 1 & 4 \\ 3 & -1 \end{array}\right| = (1 \times -1) - (4 \times 3) = -1 - 12 = -13 \)
\( M_{33} = \left|\begin{array}{cc} 1 & 0 \\ 3 & 5 \end{array}\right| = (1 \times 5) - (0 \times 3) = 5 - 0 = 5 \)

**Cofactors:**
\( A_{11} = (-1)^{1+1} M_{11} = 1 \times 11 = 11 \)
\( A_{12} = (-1)^{1+2} M_{12} = -1 \times 6 = -6 \)
\( A_{13} = (-1)^{1+3} M_{13} = 1 \times 3 = 3 \)
\( A_{21} = (-1)^{2+1} M_{21} = -1 \times (-4) = 4 \)
\( A_{22} = (-1)^{2+2} M_{22} = 1 \times 2 = 2 \)
\( A_{23} = (-1)^{2+3} M_{23} = -1 \times 1 = -1 \)
\( A_{31} = (-1)^{3+1} M_{31} = 1 \times (-20) = -20 \)
\( A_{32} = (-1)^{3+2} M_{32} = -1 \times (-13) = 13 \)
\( A_{33} = (-1)^{3+3} M_{33} = 1 \times 5 = 5 \)
In simple words: When calculating minors, remember to cover the row and column of the element and find the determinant of the remaining sub-matrix. For cofactors, apply the \( (-1)^{i+j} \) sign multiplier correctly to each minor.

Exam Tip: For 3x3 determinants, ensure you properly identify the 2x2 sub-matrix for each minor. Pay close attention to the sign changes when computing cofactors.

Question 3. Using cofactors of elements of second row, evaluate \( \Delta = \left|\begin{array}{rrr} 5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & -3 \end{array}\right| \).
Answer:
The given determinant is \( \Delta = \left|\begin{array}{rrr} 5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & -3 \end{array}\right| \).
We need to evaluate the determinant by using the cofactors of the elements in the second row. The elements in the second row are \( a_{21}=2 \), \( a_{22}=0 \), and \( a_{23}=1 \).

First, we compute the cofactors for these elements as provided in the source:
\( A_{21} = (-1)^{2+1} \left|\begin{array}{cc} 3 & 8 \\ 2 & 3 \end{array}\right| = -((3 \times 3) - (8 \times 2)) = -(9 - 16) = -(-7) = 7 \)
\( A_{22} = (-1)^{2+2} \left|\begin{array}{cc} 5 & 8 \\ 1 & 3 \end{array}\right| = ((5 \times 3) - (8 \times 1)) = (15 - 8) = 7 \)
\( A_{23} = (-1)^{2+3} \left|\begin{array}{cc} 5 & 3 \\ 1 & 2 \end{array}\right| = -((5 \times 2) - (3 \times 1)) = -(10 - 3) = -7 \)

Now, we can find the value of the determinant \( \Delta \) by using the formula for expansion along the second row:
\( \Delta = a_{21}A_{21} + a_{22}A_{22} + a_{23}A_{23} \)
\( \Delta = (2 \times 7) + (0 \times 7) + (1 \times -7) \)
\( \Delta = 14 + 0 - 7 \)
\( \Delta = 7 \)
In simple words: To evaluate a determinant, you can use the cofactors of any row or column. Here, we chose the second row. We calculated the cofactor for each number in that row. Then, we multiplied each number by its cofactor and added those results together to get the determinant's final value.

Exam Tip: When evaluating determinants using cofactor expansion, remember that the sign of the cofactor depends on its position. Choose the row or column with the most zeros to simplify calculations.

Question 4. Using cofactors of elements of third column, evaluate \( \Delta = \left|\begin{array}{ccc} 1 & x & yz \\ 1 & y & zx \\ 1 & z & xy \end{array}\right| \).
Answer:
The given determinant is \( \Delta = \left|\begin{array}{ccc} 1 & x & yz \\ 1 & y & zx \\ 1 & z & xy \end{array}\right| \).
We will evaluate the determinant using the cofactors of the elements of the third column. The elements of the third column are \( a_{13}=yz \), \( a_{23}=zx \), and \( a_{33}=xy \).

First, we find the cofactors for these elements:
\( A_{13} = (-1)^{1+3} \left|\begin{array}{cc} 1 & y \\ 1 & z \end{array}\right| = (1 \times z) - (y \times 1) = z - y \)
\( A_{23} = (-1)^{2+3} \left|\begin{array}{cc} 1 & x \\ 1 & z \end{array}\right| = -((1 \times z) - (x \times 1)) = -(z - x) = x - z \)
\( A_{33} = (-1)^{3+3} \left|\begin{array}{cc} 1 & x \\ 1 & y \end{array}\right| = (1 \times y) - (x \times 1) = y - x \)

Now, we calculate the determinant \( \Delta \) using the formula for expansion along the third column:
\( \Delta = a_{13}A_{13} + a_{23}A_{23} + a_{33}A_{33} \)
\( \Delta = yz(z - y) + zx(x - z) + xy(y - x) \)
\( \Delta = yz^2 - y^2z + x^2z - xz^2 + xy^2 - x^2y \)
\( \Delta = (-y^2z + yz^2) + (xy^2 - xz^2) + (x^2z - x^2y) \)
\( \Delta = -yz(y - z) + x(y^2 - z^2) - x^2(y - z) \)
\( \Delta = (y - z)[-yz + x(y + z) - x^2] \)
\( \Delta = (y - z)[-yz + xy + xz - x^2] \)
\( \Delta = (y - z)[xy - x^2 + xz - yz] \)
\( \Delta = (y - z)[x(y - x) + z(x - y)] \)
\( \Delta = (y - z)[x(y - x) - z(y - x)] \)
\( \Delta = (y - z)(y - x)(x - z) \)
\( \Delta = (x - y)(y - z)(z - x) \)
In simple words: We found the cofactors for each element in the third column. Then, we multiplied each element by its cofactor and added all these products together. After simplifying the resulting algebraic expression, we got the final factorized form of the determinant.

Exam Tip: When dealing with symbolic determinants, calculate cofactors carefully. Be extra cautious with algebraic expansion and factorization to ensure accuracy and reach the simplified form.

Question 5. If \( A = \left|\begin{array}{lll} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array}\right| \) and \( A_{ij} \) is the cofactor of \( a_{ij} \), then the value of \( A \) is given by
(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) \( a_{11}A_{11} + a_{21}A_{21} + a_{31}A_{31} \)
In simple words: The value of a determinant can be calculated by multiplying each element of a column (or row) by its matching cofactor, and then adding these products together. Option (d) shows this rule applied to the first column.

Exam Tip: Remember the expansion rule for determinants: the sum of the products of elements of any row or column with their *corresponding* cofactors equals the determinant's value. Using elements of one row/column with cofactors of *another* row/column will always result in zero.