NCERT Solutions for Class 12 Maths Chapter 4 Determinants Exercise 4.4

Understanding Minors and Cofactors

To compute determinants of higher orders efficiently, we break them down into smaller elements.

1. Minor ($M_{ij}$): The minor of an element $a_{ij}$ of a determinant is the determinant obtained by deleting its $i$-th row and $j$-th column in which element $a_{ij}$ lies. For a determinant of order $n$ ($n > 1$), the minor of an element is a determinant of order $n - 1$.

1. Cofactor ($A_{ij}$): The cofactor of an element $a_{ij}$ is simply its minor multiplied by a positional sign factor. It is defined as:

$A_{ij} = (-1)^{i+j} M_{ij}$

This $(-1)^{i+j}$ term creates a systematic checkerboard sign pattern over the determinant matrix. For a $3 \times 3$ determinant, the signs alternate as follows:
$\begin{bmatrix} + & - & + \\ - & + & - \\ + & - & + \end{bmatrix}$

Key Identity: If elements of a row (or column) are multiplied with cofactors of any other row (or column), then their sum is zero. For instance, $a_{11}A_{21} + a_{12}A_{22} + a_{13}A_{23} = 0$. However, if they are multiplied with their corresponding cofactors, we obtain the value of the determinant: $\Delta = a_{11}A_{11} + a_{12}A_{12} + a_{13}A_{13}$.

Core Mathematical Terms

Minor of an Element

The determinant of the submatrix left after deleting the row and column containing that specific element.

Cofactor of an Element

The minor of the element scaled by $(-1)^{i+j}$, where $i$ is the row index and $j$ is the column index.

Determinant Expansion via Cofactors

The sum of the products of elements of any row or column with their corresponding cofactors.

Step-by-Step Method to Calculate Cofactors

1. Step 1: Identify Row and Column Indices — Locate the targeted element $a_{ij}$ in the matrix and note down its row number $i$ and column number $j$.
2. Step 2: Eliminate Row $i$ and Column $j$ — Mentally or physically cross out the $i$-th row and $j$-th column of the matrix to isolate the remaining submatrix.
3. Step 3: Calculate the Minor ($M_{ij}$) — Find the determinant of the remaining submatrix. For a $3 \times 3$ matrix, this will require computing a simple $2 \times 2$ determinant.
4. Step 4: Apply the Alternating Sign Factor — Multiply the minor $M_{ij}$ by $(-1)^{i+j}$ to get the cofactor $A_{ij}$. Remember, if $i+j$ is even, $A_{ij} = M_{ij}$. If $i+j$ is odd, $A_{ij} = -M_{ij}$.

Fully Worked Out Examples

• Example 1: 2x2 Determinant Find the minors and cofactors of all elements of the determinant: $\Delta = \begin{vmatrix} 2 & -4 \\ 0 & 3 \end{vmatrix}$ For element $a_{11} = 2$: Minor $M_{11} = |3| = 3$ Cofactor $A_{11} = (-1)^{1+1} M_{11} = 1 \times 3 = 3$ For element $a_{12} = -4$: Minor $M_{12} = |0| = 0$ Cofactor $A_{12} = (-1)^{1+2} M_{12} = -1 \times 0 = 0$ For element $a_{21} = 0$: Minor $M_{21} = |-4| = -4$ Cofactor $A_{21} = (-1)^{2+1} M_{21} = -1 \times (-4) = 4$ For element $a_{22} = 3$: Minor $M_{22} = |2| = 2$ Cofactor $A_{22} = (-1)^{2+2} M_{22} = 1 \times 2 = 2$
• Example 2: 3x3 Determinant - Verification Using the cofactors of elements of the second row, evaluate the determinant: $\Delta = \begin{vmatrix} 5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3 \end{vmatrix}$ Find cofactors of elements of row 2 ($a_{21}=2, a_{22}=0, a_{23}=1$): $A_{21} = (-1)^{2+1} \begin{vmatrix} 3 & 8 \\ 2 & 3 \end{vmatrix} = -1 \times (3(3) - 8(2)) = -1 \times (9 - 16) = 7$ $A_{22} = (-1)^{2+2} \begin{vmatrix} 5 & 8 \\ 1 & 3 \end{vmatrix} = 1 \times (5(3) - 8(1)) = 15 - 8 = 7$ $A_{23} = (-1)^{2+3} \begin{vmatrix} 5 & 3 \\ 1 & 2 \end{vmatrix} = -1 \times (5(2) - 3(1)) = -1 \times (10 - 3) = -7$ * Evaluate $\Delta$: $\Delta = a_{21}A_{21} + a_{22}A_{22} + a_{23}A_{23}$ $\Delta = 2(7) + 0(7) + 1(-7) = 14 + 0 - 7 = 7$ Final Answer: The value of the determinant is $7$.

Board Exam Pitfalls & Tips

1. The Sign Slip-up: The most common mistake in Exercise 4.4 is forgetting to multiply the minor by $(-1)^{i+j}$ when finding cofactors. Double-check odd indices like $A_{12}, A_{21}, A_{23}, A_{32}$ where the sign MUST change.
2. Zero Elements: If an element in a row/column is $0$, you do not need to expand its cofactor when calculating the final determinant value because $0 \times A_{ij} = 0$. However, if the question explicitly asks you to 'write the cofactors of all elements', you MUST calculate and state the cofactor of that zero element.
3. Property Verification: Questions that ask you to show $a_{11}A_{21} + a_{12}A_{22} + a_{13}A_{23} = 0$ are frequent 2-mark targets. Remember: multiplying elements of one row with the cofactors of another row always yields exactly zero.

Practice Questions with Solutions

• Q: Find the minors and cofactors of the elements of the determinant: $\begin{vmatrix} 1 & 0 & 4 \\ 3 & 5 & -1 \\ 0 & 1 & 2 \end{vmatrix}$ A: Step 1: Let us denote the matrix elements as $a_{ij}$, minors as $M_{ij}$, and cofactors as $A_{ij}$. Step 2: Calculate Minors for all elements: - $M_{11} = \begin{vmatrix} 5 & -1 \\ 1 & 2 \end{vmatrix} = 10 - (-1) = 11$ - $M_{12} = \begin{vmatrix} 3 & -1 \\ 0 & 2 \end{vmatrix} = 6 - 0 = 6$ - $M_{13} = \begin{vmatrix} 3 & 5 \\ 0 & 1 \end{vmatrix} = 3 - 0 = 3$ - $M_{21} = \begin{vmatrix} 0 & 4 \\ 1 & 2 \end{vmatrix} = 0 - 4 = -4$ - $M_{22} = \begin{vmatrix} 1 & 4 \\ 0 & 2 \end{vmatrix} = 2 - 0 = 2$ - $M_{23} = \begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix} = 1 - 0 = 1$ - $M_{31} = \begin{vmatrix} 0 & 4 \\ 5 & -1 \end{vmatrix} = 0 - 20 = -20$ - $M_{32} = \begin{vmatrix} 1 & 4 \\ 3 & -1 \end{vmatrix} = -1 - 12 = -13$ - $M_{33} = \begin{vmatrix} 1 & 0 \\ 3 & 5 \end{vmatrix} = 5 - 0 = 5$ Step 3: Convert to Cofactors by applying $A_{ij} = (-1)^{i+j} M_{ij}$: - $A_{11} = (-1)^{2}(11) = 11$ - $A_{12} = (-1)^{3}(6) = -6$ - $A_{13} = (-1)^{4}(3) = 3$ - $A_{21} = (-1)^{3}(-4) = 4$ - $A_{22} = (-1)^{4}(2) = 2$ - $A_{23} = (-1)^{5}(1) = -1$ - $A_{31} = (-1)^{4}(-20) = -20$ - $A_{32} = (-1)^{5}(-13) = 13$ - $A_{33} = (-1)^{6}(5) = 5$ Final Answer: Minors are: $M_{11}=11, M_{12}=6, M_{13}=3, M_{21}=-4, M_{22}=2, M_{23}=1, M_{31}=-20, M_{32}=-13, M_{33}=5$. Cofactors are: $A_{11}=11, A_{12}=-6, A_{13}=3, A_{21}=4, A_{22}=2, A_{23}=-1, A_{31}=-20, A_{32}=13, A_{33}=5$.
• Q: For the determinant: $\Delta = \begin{vmatrix} 2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7 \end{vmatrix}$ Verify that $a_{11}A_{31} + a_{12}A_{32} + a_{13}A_{33} = 0$. A: Step 1: Write down the first-row elements: $a_{11} = 2$, $a_{12} = -3$, $a_{13} = 5$. Step 2: Find the cofactors of the third-row elements ($A_{31}, A_{32}, A_{33}$): - $A_{31} = (-1)^{3+1} \begin{vmatrix} -3 & 5 \\ 0 & 4 \end{vmatrix} = 1 \times (-12 - 0) = -12$ - $A_{32} = (-1)^{3+2} \begin{vmatrix} 2 & 5 \\ 6 & 4 \end{vmatrix} = -1 \times (8 - 30) = -1 \times (-22) = 22$ - $A_{33} = (-1)^{3+3} \begin{vmatrix} 2 & -3 \\ 6 & 0 \end{vmatrix} = 1 \times (0 - (-18)) = 18$ Step 3: Calculate the linear combination: $a_{11}A_{31} + a_{12}A_{32} + a_{13}A_{33}$ $= 2(-12) + (-3)(22) + 5(18)$ $= -24 - 66 + 90$ $= -90 + 90 = 0$ Final Answer: Since the calculated value is exactly $0$, the relation is verified.
• Q: If $\Delta = \begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{vmatrix}$ and $A_{ij}$ is the cofactor of $a_{ij}$, then the value of $\Delta$ is given by which linear combination? A: Step 1: Recall that the determinant $\Delta$ is found by summing the products of the elements of any row or column with their corresponding cofactors. Step 2: Analyze options of combinations. Any formulation where the element index and cofactor index match perfectly represents the determinant value. Step 3: Thus, along Column 1, we have: $\Delta = a_{11}A_{11} + a_{21}A_{21} + a_{31}A_{31}$ Or along Row 1: $\Delta = a_{11}A_{11} + a_{12}A_{12} + a_{13}A_{13}$ Final Answer: $\Delta$ is given by the sum of elements of any row or column multiplied by their corresponding cofactors (e.g., $a_{11}A_{11} + a_{21}A_{21} + a_{31}A_{31}$).
• Q: Using cofactors of elements of third column, evaluate the determinant: $\Delta = \begin{vmatrix} 1 & x & yz \\ 1 & y & zx \\ 1 & z & xy \end{vmatrix}$ A: Step 1: Identify third column elements: $a_{13} = yz$, $a_{23} = zx$, $a_{33} = xy$. Step 2: Find the cofactors of these elements: - $A_{13} = (-1)^{1+3} \begin{vmatrix} 1 & y \\ 1 & z \end{vmatrix} = (z - y)$ - $A_{23} = (-1)^{2+3} \begin{vmatrix} 1 & x \\ 1 & z \end{vmatrix} = -(z - x) = (x - z)$ - $A_{33} = (-1)^{3+3} \begin{vmatrix} 1 & x \\ 1 & y \end{vmatrix} = (y - x)$ Step 3: Expand the determinant: $\Delta = a_{13}A_{13} + a_{23}A_{23} + a_{33}A_{33}$ $\Delta = yz(z - y) + zx(x - z) + xy(y - x)$ $\Delta = yz^2 - y^2z + zx^2 - z^2x + xy^2 - x^2y$ Step 4: Factorize the expression: Grouping terms: $= yz(z - y) + x(z^2 - y^2) + x^2(y - z)$ $= (y - z) [ -yz + x(y + z) - x^2 ]$ $= (y - z) [ -yz + xy + xz - x^2 ]$ $= (y - z) [ x(y - x) - z(y - x) ]$ $= (y - z)(y - x)(x - z)$ $= (x - y)(y - z)(z - x)$ Final Answer: The simplified value of the determinant is $(x - y)(y - z)(z - x)$.

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