# Class 12 NCERT Solutions- Mathematics Part I – Chapter 4 Determinants – Exercise 4.3

Last Updated : 03 Apr, 2024

### Question 1.

(i) [Tex]\begin{vmatrix}3&-4\\0&3\end{vmatrix}  [/Tex]

(ii) [Tex]\begin{vmatrix}a&c\\b&d\end{vmatrix}      [/Tex]

Solution:

(i) [Tex]\begin{vmatrix}3&-4\\0&3\end{vmatrix}  [/Tex]

Finding minors of the elements of the determinant:

Let us assume Mij is Minors of elements aij

M11 = Minor of elements a11 = 3

M12 = Minor of elements a12 = 0

M21 = Minor of elements a21 = âˆ’4

M22 = Minor of elements a22 = 2

Finding cofactor of aij

Let us assume cofactor of aij is Aij Mij

A11 = (âˆ’1)1+1 M11 = (âˆ’1)2 (3) = 3

A12 = (âˆ’1)1+2 M12 = (âˆ’1)3 (0) = 0

A21 = (âˆ’1)2+1 M21 = (âˆ’1)3 (âˆ’4) = 4

A22 = (âˆ’1)2+2 M22 = (âˆ’1)4 (2) = 2

(ii) [Tex]\begin{vmatrix}a&c\\b&d\end{vmatrix}  [/Tex]

Finding minors of the elements of the determinant:

Let us assume Mij is Minors of elements aij

M11 = Minor of element a11 = d

M12 = Minor of elements a12 = b

M21 = Minor of elements a21 = c

M22 = Minor of elements a22 = a

Finding cofactor of aij

Let us assume cofactor of aij is Aij, which is (âˆ’1)i+j Mij

A11 = (âˆ’1)1+1 M11 = (âˆ’1)2 (d) = d

A12 = (âˆ’1)1+2 M12 = (âˆ’1)3 (b) = âˆ’b

A21 = (âˆ’1)2+1 M21 = (âˆ’1)3 (c) = âˆ’c

A22 = (âˆ’1)2+2 M22 = (âˆ’1)4 (a) = a

### Question 2.

(i)[Tex]\begin{vmatrix}1&0&0\\0&1&0\\0&0&1\end{vmatrix}[/Tex]

(ii)[Tex]\begin{vmatrix}1&0&4\\3&5&-5\\0&2&2\end{vmatrix}[/Tex]

Solution:

(i) [Tex]\begin{vmatrix}1&0&0\\0&1&0\\0&0&1\end{vmatrix}[/Tex]

Let us find the Minors and cofactors of the elements:

Assume, Mij is minor of element aij and Aij is cofactor of aij

M11 = Minor of elements a11 =[Tex]\begin{vmatrix} 1 & 0\\0 & 1\end{vmatrix} [/Tex]= 1 âˆ’ 0 = 1 and A11 = 1

M12 = Minor of elements a12 =[Tex]\begin{vmatrix} 0 & 0\\0 & 1\end{vmatrix} [/Tex]= 0 âˆ’ 0 = 0 and A12 = 0

M13 = Minor of elements a13 =[Tex]\begin{vmatrix} 0 & 1\\0 & 0\end{vmatrix} [/Tex]= 0 âˆ’ 0 = 0 and A13 = 0

M21 = Minor of elements a21 =[Tex]\begin{vmatrix} 0 & 0\\0 & 1\end{vmatrix} [/Tex]= 0 âˆ’ 0 = 0 and A21 = 0

M22 = Minor of elements a22 =[Tex]\begin{vmatrix} 1 & 0\\0 & 1\end{vmatrix} [/Tex]= 1 âˆ’ 0 = 1 and A22 = 1

M23 = Minor of elements a23 =[Tex]\begin{vmatrix} 1 & 0\\0 & 0\end{vmatrix} [/Tex]= 0 âˆ’ 0 = 0 and A23 = 0

M31 = Minor of elements a31 =[Tex]\begin{vmatrix} 0 & 0\\1 & 0\end{vmatrix} [/Tex]= 0 âˆ’ 0 = 0 and A31 = 0

M32 = Minor of elements a32 =[Tex]\begin{vmatrix} 1 & 0\\0 & 0\end{vmatrix} [/Tex]= 0 âˆ’ 0 = 0 and A32 = 0

M33 = Minor of elements a33 =[Tex]\begin{vmatrix} 1 & 0\\0 & 1\end{vmatrix} [/Tex]= 1 âˆ’ 0 = 1 and A33 = 1

(ii) [Tex]\begin{vmatrix}1&0&4\\3&5&-5\\0&2&2\end{vmatrix}[/Tex]

Let us find the Minors and cofactors of the elements:

Assume, Mij is minor of element aij and Aij is cofactor of aij

M11 = Minor of elements a11 =[Tex]\begin{vmatrix} 5 & -1\\1 & 2\end{vmatrix} [/Tex]= 10 âˆ’ (âˆ’1) = 11 and A11 = 11

M12 = Minor of elements a12 =[Tex]\begin{vmatrix} 3 & -1\\0 & 2\end{vmatrix} [/Tex]= 6 âˆ’ 0 = 6 and A12 = âˆ’6

M13 = Minor of elements a13 =[Tex]\begin{vmatrix} 3 & 5\\0 & 1\end{vmatrix} [/Tex]= 3 âˆ’ 0 = 3 and A13 = 3

M21 = Minor of elements a21 =[Tex]\begin{vmatrix} 0 & 4\\1 & 2\end{vmatrix} [/Tex]= 0 âˆ’ 4 = âˆ’4 and A21 = 4

M22 = Minor of elements a22 =[Tex]\begin{vmatrix} 1 & 4\\0 & 2\end{vmatrix} [/Tex]= 2 âˆ’ 0 = 2 and A22 = 2

M23 = Minor of elements a23 =[Tex]\begin{vmatrix} 1 & 0\\0 & 1\end{vmatrix} [/Tex]= 1 âˆ’ 0 = 1 and A23 = âˆ’1

M31 = Minor of elements a31 =[Tex]\begin{vmatrix} 0 & 4\\5 & -1\end{vmatrix} [/Tex]= 0 âˆ’ 20 = âˆ’20 and A31 = âˆ’20

M32 = Minor of elements a32 =[Tex]\begin{vmatrix} 1 & 4\\3 & -1\end{vmatrix} [/Tex]= âˆ’1 âˆ’ 12 = âˆ’13 and A32 = 13

M33 = Minor of elements a33 =[Tex]\begin{vmatrix} 1 & 0\\3 & 5\end{vmatrix} [/Tex]= 5 âˆ’ 0 = 5 and A33 = 5

### Question 3. Using Cofactors of elements of second row, evaluate â–³?

[Tex]\bigtriangleup=\begin{vmatrix}5&3&8\\2&0&1\\1&2&3\end{vmatrix}[/Tex]

Solution:

Finding the Cofactors of elements of second row:

A21 = Cofactor of elements a21 = (âˆ’1)2+1  [Tex]\begin{vmatrix} 3 & 8\\2 & 3\end{vmatrix} [/Tex]= (âˆ’1)3 (9 âˆ’ 16) = 7

A22 = Cofactor of elements a22 = (âˆ’1)2+2  [Tex]\begin{vmatrix} 5 & 8\\1 & 3\end{vmatrix} [/Tex]= (âˆ’1)4 (15 âˆ’ 8) = 7

A23 = Cofactor of elements a23 = (âˆ’1)2+3  [Tex]\begin{vmatrix} 5 & 3\\1 & 2\end{vmatrix} [/Tex]= (âˆ’1)5 (10 âˆ’ 3) = 7

Now, â–³ = a21 A21 + a22 A22 + a23 A23 = 14 + 0 âˆ’ 7 = 7

### Question 4. Using Cofactors of elements of third column, evaluate â–³?

[Tex]\bigtriangleup = \begin{vmatrix}1&x&yz\\1&y&zx\\1&z&xy\end{vmatrix}[/Tex]

Solution:

Finding the Cofactors of elements of third column:

A13 = Cofactor of elements a13 = (âˆ’1)1+3  [Tex]\begin{vmatrix} 1 & y\\1 & z\end{vmatrix} [/Tex]= (âˆ’1)4 (z âˆ’ y) = âˆ’y

A23 = Cofactor of elements a23 = (âˆ’1)2+3  [Tex]\begin{vmatrix} 1 & x\\1 & z\end{vmatrix} [/Tex]= (âˆ’1)5 (z âˆ’ x) = x âˆ’ z

A33 = Cofactor of elements a33 = (âˆ’1)3+3  [Tex]\begin{vmatrix} 1 & x\\1 & y\end{vmatrix} [/Tex]= (âˆ’1)6 (y âˆ’ x) = y âˆ’ x

Now, â–³ = a13 A13 + a23 A23 + a33 A33

= yz (z âˆ’ y) + zx (x âˆ’ z) + xy (y âˆ’ x)

= (yz2 âˆ’ y2z) + (xy2 âˆ’ xz2) + (xz2 âˆ’ x2y)

= (y âˆ’ z)[âˆ’yz + x(y + z) âˆ’ x2]

= (y âˆ’ z)[âˆ’yz + x (z âˆ’ x) + x (z âˆ’ x)]

= (x âˆ’ y)(y âˆ’ x)(z âˆ’ x)

### Question 5. If [Tex] â–³=\begin{vmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{vmatrix} [/Tex]and Aij is cofactor of aij then value of â–³ is given by:

(A) a11A31 + a12A32 + a13A33

(B) a11A11 + a12A21 + a13A31

(C) a21A11 + a22A12 + a23A13

(D) a11A11 + a21A21 + a31A31

Solution:

Option (D) is correct.

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