Rbse Solutions Class 12 Maths: Exercise 4.3

Here are the solutions for Exercise 4.3, focusing on finding Minors and Cofactors and using them to evaluate determinants.

The Minor ($M_{ij}$) of an element $a_{ij}$ is the determinant of the submatrix obtained by deleting the $i$-th row and $j$-th column.

The Cofactor ($A_{ij}$) of an element $a_{ij}$ is given by $A_{ij} = (-1)^{i+j} M_{ij}$.

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1. Write Minors and Cofactors of the elements of the following determinants:

(i) $\begin{vmatrix} 2 & -4 \\ 0 & 3 \end{vmatrix}$

The elements are $a_{11}=2, a_{12}=-4, a_{21}=0, a_{22}=3$.

Element	Minor (Mij​)	Cofactor (Aij​=(−1)i+jMij​)
$a_{11}=2$	$M_{11} = 3$	$A_{11} = (-1)^{1+1}(3) = 3$
$a_{12}=-4$	$M_{12} = 0$	$A_{12} = (-1)^{1+2}(0) = 0$
$a_{21}=0$	$M_{21} = -4$	$A_{21} = (-1)^{2+1}(-4) = 4$
$a_{22}=3$	$M_{22} = 2$	$A_{22} = (-1)^{2+2}(2) = 2$

(ii) $\begin{vmatrix} a & c \\ b & d \end{vmatrix}$

The elements are $a_{11}=a, a_{12}=c, a_{21}=b, a_{22}=d$.

Element	Minor (Mij​)	Cofactor (Aij​=(−1)i+jMij​)
$a_{11}=a$	$M_{11} = d$	$A_{11} = (-1)^{1+1}(d) = d$
$a_{12}=c$	$M_{12} = b$	$A_{12} = (-1)^{1+2}(b) = -b$
$a_{21}=b$	$M_{21} = c$	$A_{21} = (-1)^{2+1}(c) = -c$
$a_{22}=d$	$M_{22} = a$	$A_{22} = (-1)^{2+2}(a) = a$

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2. Write Minors and Cofactors of the elements of the following determinants:

(i) $\begin{vmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix}$

Element	Minor (Mij​)	Cofactor (Aij​)
$a_{11}=1$	$M_{11} = \begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix} = 1$	$A_{11} = 1$
$a_{12}=0$	$M_{12} = \begin{vmatrix} 0 & 0 \\ 0 & 1 \end{vmatrix} = 0$	$A_{12} = 0$
$a_{13}=0$	$M_{13} = \begin{vmatrix} 0 & 1 \\ 0 & 0 \end{vmatrix} = 0$	$A_{13} = 0$
$a_{21}=0$	$M_{21} = \begin{vmatrix} 0 & 0 \\ 0 & 1 \end{vmatrix} = 0$	$A_{21} = 0$
$\mathbf{a_{22}=1}$	$\mathbf{M_{22} = \begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix} = 1}$	$\mathbf{A_{22} = 1}$
$a_{23}=0$	$M_{23} = \begin{vmatrix} 1 & 0 \\ 0 & 0 \end{vmatrix} = 0$	$A_{23} = 0$
$a_{31}=0$	$M_{31} = \begin{vmatrix} 0 & 0 \\ 1 & 0 \end{vmatrix} = 0$	$A_{31} = 0$
$a_{32}=0$	$M_{32} = \begin{vmatrix} 1 & 0 \\ 0 & 0 \end{vmatrix} = 0$	$A_{32} = 0$
$a_{33}=1$	$M_{33} = \begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix} = 1$	$A_{33} = 1$

(ii) $\begin{vmatrix} 1 & 0 & 4 \\ 3 & 5 & -1 \\ 0 & 1 & 2 \end{vmatrix}$

Element	Minor (Mij​)	Cofactor (Aij​)
$a_{11}=1$	$M_{11} = \begin{vmatrix} 5 & -1 \\ 1 & 2 \end{vmatrix} = 10 – (-1) = 11$	$A_{11} = 11$
$a_{12}=0$	$M_{12} = \begin{vmatrix} 3 & -1 \\ 0 & 2 \end{vmatrix} = 6 – 0 = 6$	$A_{12} = -6$
$a_{13}=4$	$M_{13} = \begin{vmatrix} 3 & 5 \\ 0 & 1 \end{vmatrix} = 3 – 0 = 3$	$A_{13} = 3$
$a_{21}=3$	$M_{21} = \begin{vmatrix} 0 & 4 \\ 1 & 2 \end{vmatrix} = 0 – 4 = -4$	$A_{21} = -(-4) = 4$
$a_{22}=5$	$M_{22} = \begin{vmatrix} 1 & 4 \\ 0 & 2 \end{vmatrix} = 2 – 0 = 2$	$A_{22} = 2$
$a_{23}=-1$	$M_{23} = \begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix} = 1 – 0 = 1$	$A_{23} = -(1) = -1$
$a_{31}=0$	$M_{31} = \begin{vmatrix} 0 & 4 \\ 5 & -1 \end{vmatrix} = 0 – 20 = -20$	$A_{31} = -20$
$a_{32}=1$	$M_{32} = \begin{vmatrix} 1 & 4 \\ 3 & -1 \end{vmatrix} = -1 – 12 = -13$	$A_{32} = -(-13) = 13$
$a_{33}=2$	$M_{33} = \begin{vmatrix} 1 & 0 \\ 3 & 5 \end{vmatrix} = 5 – 0 = 5$	$A_{33} = 5$

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3. Using Cofactors of elements of second row, evaluate $\Delta = \begin{vmatrix} 5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3 \end{vmatrix}$.

The determinant $\Delta$ is evaluated using the elements of the second row ($a_{21}=2, a_{22}=0, a_{23}=1$) and their corresponding cofactors ($A_{21}, A_{22}, A_{23}$):

$$\Delta = a_{21} A_{21} + a_{22} A_{22} + a_{23} A_{23}$$

1. Find Cofactors of $R_2$:
   • $A_{21} = (-1)^{2+1} M_{21} = – \begin{vmatrix} 3 & 8 \\ 2 & 3 \end{vmatrix} = -[3(3) – 8(2)] = -[9 – 16] = -(-7) = 7$
   • $A_{22} = (-1)^{2+2} M_{22} = + \begin{vmatrix} 5 & 8 \\ 1 & 3 \end{vmatrix} = 5(3) – 8(1) = 15 – 8 = 7$
   • $A_{23} = (-1)^{2+3} M_{23} = – \begin{vmatrix} 5 & 3 \\ 1 & 2 \end{vmatrix} = -[5(2) – 3(1)] = -[10 – 3] = -(7) = -7$
2. Evaluate $\Delta$:$$\Delta = a_{21} A_{21} + a_{22} A_{22} + a_{23} A_{23}$$$$\Delta = 2(7) + 0(7) + 1(-7)$$$$\Delta = 14 + 0 – 7$$$$\mathbf{\Delta = 7}$$

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4. Using Cofactors of elements of third column, evaluate $\Delta = \begin{vmatrix} 1 & x & yz \\ 1 & y & zx \\ 1 & z & xy \end{vmatrix}$.

The determinant $\Delta$ is evaluated using the elements of the third column ($a_{13}=yz, a_{23}=zx, a_{33}=xy$) and their corresponding cofactors ($A_{13}, A_{23}, A_{33}$):

$$\Delta = a_{13} A_{13} + a_{23} A_{23} + a_{33} A_{33}$$

1. Find Cofactors of $C_3$:
   • $A_{13} = (-1)^{1+3} M_{13} = + \begin{vmatrix} 1 & y \\ 1 & z \end{vmatrix} = z – y$
   • $A_{23} = (-1)^{2+3} M_{23} = – \begin{vmatrix} 1 & x \\ 1 & z \end{vmatrix} = -(z – x) = x – z$
   • $A_{33} = (-1)^{3+3} M_{33} = + \begin{vmatrix} 1 & x \\ 1 & y \end{vmatrix} = y – x$
2. Evaluate $\Delta$:$$\Delta = yz(z – y) + zx(x – z) + xy(y – x)$$$$\Delta = yz^2 – y^2z + x^2z – xz^2 + xy^2 – x^2y$$Factor the expression by grouping terms with common factors (e.g., $z^2, z$):$$\Delta = z^2(y – x) + z(x^2 – y^2) + xy(y – x)$$Using the identity $x^2 – y^2 = (x – y)(x + y)$:$$\Delta = z^2(y – x) + z(x – y)(x + y) + xy(y – x)$$Factor out $-(x – y) = (y – x)$:$$\Delta = (y – x)z^2 – z(y – x)(x + y) + xy(y – x)$$Factor out the common term $(y – x)$:$$\Delta = (y – x) [z^2 – z(x + y) + xy]$$$$\Delta = (y – x) [z^2 – zx – zy + xy]$$Factor the quadratic expression by grouping:$$\Delta = (y – x) [z(z – x) – y(z – x)]$$$$\Delta = (y – x) (z – x) (z – y)$$Rewriting in standard cyclic form $(x-y)(y-z)(z-x)$:$$\Delta = -(x – y) \cdot -(x – z) \cdot -(y – z) \implies \text{Wait! Let’s simplify:}$$$$\Delta = (y – x) (z – x) (z – y)$$$$\mathbf{\Delta = (x – y)(y – z)(z – x)}$$

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5. 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 value of $\Delta$ is given by

The determinant can be evaluated by summing the product of elements of any row or column with their corresponding cofactors.

$$\Delta = a_{ij} A_{ij} + a_{i(j+1)} A_{i(j+1)} + a_{i(j+2)} A_{i(j+2)} \quad \text{(Expansion along Row } i \text{)}$$

$$\Delta = a_{i j} A_{i j} + a_{(i+1) j} A_{(i+1) j} + a_{(i+2) j} A_{(i+2) j} \quad \text{(Expansion along Column } j \text{)}$$

Let’s check the options:

(A) $a_{11} A_{31} + a_{12} A_{32} + a_{13} A_{33}$: Products of elements of $\mathbf{R_1}$ with cofactors of $\mathbf{R_3}$. This equals $\mathbf{0}$.

(B) $a_{11} A_{11} + a_{12} A_{21} + a_{13} A_{31}$: Products of elements of $\mathbf{R_1}$ with cofactors of $\mathbf{C_1}$. This equals $\mathbf{0}$.

(C) $a_{21} A_{11} + a_{22} A_{12} + a_{23} A_{13}$: Products of elements of $\mathbf{R_2}$ with cofactors of $\mathbf{R_1}$. This equals $\mathbf{0}$.

(D) $a_{11} A_{11} + a_{21} A_{21} + a_{31} A_{31}$: Products of elements of $\mathbf{C_1}$ with cofactors of $\mathbf{C_1}$. This is the expansion along the first column.

$$\text{The correct option is } \mathbf{(D)}$$