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Class 12 NCERT Solutions – Mathematics Part I – Chapter 4 Determinants – Exercise 4.4
  • Last Updated : 25 Jan, 2021

Write minors and cofactors of the elements of the following determinants: 

Question 1. 

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

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

Solution: 

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

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 = −4

M22 = Minor of elements a22 = 2

Finding cofactor of aij

Let us assume cofactor of aij is Aij Mij

A11 = (−1)1+1 M11 = (−1)2 (3) = 3



A12 = (−1)1+2 M12 = (−1)3 (0) = 0

A21 = (−1)2+1 M21 = (−1)3 (−4) = 4

A22 = (−1)2+2 M22 = (−1)4 (2) = 2

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

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 (−1)i+j Mij

A11 = (−1)1+1 M11 = (−1)2 (d) = d

A12 = (−1)1+2 M12 = (−1)3 (b) = −b

A21 = (−1)2+1 M21 = (−1)3 (c) = −c

A22 = (−1)2+2 M22 = (−1)4 (a) = a

Question 2.

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

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

Solution:

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

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 =\begin{vmatrix} 1 & 0\\0 & 1\end{vmatrix} = 1 − 0 = 1 and A11 = 1

M12 = Minor of elements a12 =\begin{vmatrix} 0 & 0\\0 & 1\end{vmatrix} = 0 − 0 = 0 and A12 = 0

M13 = Minor of elements a13 =\begin{vmatrix} 0 & 1\\0 & 0\end{vmatrix} = 0 − 0 = 0 and A13 = 0

M21 = Minor of elements a21 =\begin{vmatrix} 0 & 0\\0 & 1\end{vmatrix} = 0 − 0 = 0 and A21 = 0

M22 = Minor of elements a22 =\begin{vmatrix} 1 & 0\\0 & 1\end{vmatrix} = 1 − 0 = 1 and A22 = 1

M23 = Minor of elements a23 =\begin{vmatrix} 1 & 0\\0 & 0\end{vmatrix} = 0 − 0 = 0 and A23 = 0

M31 = Minor of elements a31 =\begin{vmatrix} 0 & 0\\1 & 0\end{vmatrix} = 0 − 0 = 0 and A31 = 0

M32 = Minor of elements a32 =\begin{vmatrix} 1 & 0\\0 & 0\end{vmatrix} = 0 − 0 = 0 and A32 = 0

M33 = Minor of elements a33 =\begin{vmatrix} 1 & 0\\0 & 1\end{vmatrix} = 1 − 0 = 1 and A33 = 1

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

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 =\begin{vmatrix} 5 & -1\\1 & 2\end{vmatrix} = 10 − (−1) = 11 and A11 = 11

M12 = Minor of elements a12 =\begin{vmatrix} 3 & -1\\0 & 2\end{vmatrix} = 6 − 0 = 6 and A12 = −6

M13 = Minor of elements a13 =\begin{vmatrix} 3 & 5\\0 & 1\end{vmatrix} = 3 − 0 = 3 and A13 = 3

M21 = Minor of elements a21 =\begin{vmatrix} 0 & 4\\1 & 2\end{vmatrix} = 0 − 4 = −4 and A21 = 4

M22 = Minor of elements a22 =\begin{vmatrix} 1 & 4\\0 & 2\end{vmatrix} = 2 − 0 = 2 and A22 = 2

M23 = Minor of elements a23 =\begin{vmatrix} 1 & 0\\0 & 1\end{vmatrix} = 1 − 0 = 1 and A23 = −1

M31 = Minor of elements a31 =\begin{vmatrix} 0 & 4\\5 & -1\end{vmatrix} = 0 − 20 = −20 and A31 = −20

M32 = Minor of elements a32 =\begin{vmatrix} 1 & 4\\3 & -1\end{vmatrix} = −1 − 12 = −13 and A32 = 13

M33 = Minor of elements a33 =\begin{vmatrix} 1 & 0\\3 & 5\end{vmatrix} = 5 − 0 = 5 and A33 = 5

Question 3. Using Cofactors of elements of second row, evaluate △?

\bigtriangleup=\begin{vmatrix}5&3&8\\2&0&1\\1&2&3\end{vmatrix}

Solution:

Finding the Cofactors of elements of second row:

A21 = Cofactor of elements a21 = (−1)2+1  \begin{vmatrix} 3 & 8\\2 & 3\end{vmatrix} = (−1)3 (9 − 16) = 7

A22 = Cofactor of elements a22 = (−1)2+2  \begin{vmatrix} 5 & 8\\1 & 3\end{vmatrix} = (−1)4 (15 − 8) = 7

A23 = Cofactor of elements a23 = (−1)2+3  \begin{vmatrix} 5 & 3\\1 & 2\end{vmatrix} = (−1)5 (10 − 3) = 7

Now, △ = a21 A21 + a22 A22 + a23 A23 = 14 + 0 − 7 = 7

Question 4. Using Cofactors of elements of third column, evaluate △?

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

Solution:

Finding the Cofactors of elements of third column:

A13 = Cofactor of elements a13 = (−1)1+3  \begin{vmatrix} 1 & y\\1 & z\end{vmatrix} = (−1)4 (z − y) = −y

A23 = Cofactor of elements a23 = (−1)2+3  \begin{vmatrix} 1 & x\\1 & z\end{vmatrix} = (−1)5 (z − x) = x − z

A33 = Cofactor of elements a33 = (−1)3+3  \begin{vmatrix} 1 & x\\1 & y\end{vmatrix} = (−1)6 (y − x) = y − x

Now, △ = a13 A13 + a23 A23 + a33 A33

= yz (z − y) + zx (x − z) + xy (y − x)

= (yz2 − y2z) + (xy2 − xz2) + (xz2 − x2y)

= (y − z)[−yz + x(y + z) − x2]

= (y − z)[−yz + x (z − x) + x (z − x)]

= (x − y)(y − x)(z − x)

Question 5. If  △=\begin{vmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{vmatrix} and Aij is cofactor of aij then value of △ 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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