Gujarat Board GSEB Textbook Solutions Class 12 Maths Chapter 4 Determinants Ex 4.4 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 12 Maths Chapter 4 Determinants Ex 4.4

Write minors and cofactors of elements of the following determinants:

Question 1.

(i) \(\left|\begin{array}{cc} 2 & -4 \\ 0 & 3 \end{array}\right|\)

(ii) \(\left|\begin{array}{cc} a & c \\ b & d \end{array}\right|\)

Solution:

(i) âˆ† = \(\left|\begin{array}{cc} 2 & -4 \\ 0 & 3 \end{array}\right|\)

Minor of a_{11} is M_{11} = 3,

Minor of a_{12} is M_{12} = 0,

Minor of a_{21} is M_{21} = – 4,

Minor of a_{22} is M_{22} = 2.

Cofactor of a_{11} = A_{11} = (-1)^{1+1} M_{11} = (-1)Â² x 3 = 3,

Cofactor of a_{12} = A_{21} = (-1)^{2+1}M_{12} = 0,

Cofactor of a_{21} = A_{21} = (-1)^{2+1} M_{21} = – (- 4) = 4,

Cofactor of a_{22} = A_{22} = (-1)^{2+2} M_{22} = 2.

(ii)

Question 2.

(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|\)

Solution:

Question 3.

Using cofactors of elements of second row, evaluate

âˆ† = \(\left|\begin{array}{rrr}

5 & 3 & 8 \\

2 & 0 & 1 \\

1 & 2 & -3

\end{array}\right|\).

Solution:

Question 4.

Using cofactors of elements of third column, evaluate

âˆ† = \(\left|\begin{array}{rrr} 1 & x & yz \\ 1 & y & zx \\ 1 & z & xy \end{array}\right|\).

Solution:

Elements of third column are yz, zx and xy.

A_{13} = (-1)^{1+3}\(\left|\begin{array}{ll} 1 & y \\ 1 & z \end{array}\right|\) = z – y

A_{23} = (-1)^{2+3}\(\left|\begin{array}{ll} 1 & x \\ 1 & z \end{array}\right|\) = – (z – x) = – z + x

A_{33} = (-1)^{3+3}\(\left|\begin{array}{ll} 1 & x \\ 1 & y \end{array}\right|\) = (y – x)

âˆ´ âˆ† = a_{13}A_{13} + a_{23}A_{23} + a_{33}A_{33}

= yz(z – y) + zx(- z + x) + xy(y – x)

= yzÂ² – yÂ²z – xzÂ² + xÂ²z + xyÂ² – xÂ²y

= (- yÂ²z + yzÂ²) + (xyÂ² – xzÂ²) + (- xÂ²y + xÂ²z)

= – yz(y – z) + x(yÂ² – zÂ²) – xÂ²(y – z)

= (y-z)[- yz + (y + z) – xÂ²]

= (y-z)[-yz + xy+ xz-xÂ²]

= (y – z)[xy – yz – xÂ² + xy]

= (y – z)[z(x – y) – x(x – y)]

= (y – z)(x – y)(z – x)

= (x – y)(y – z)(z – x).

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 value of âˆ† 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}

Solution:

A = Sum of the products of elements of a row or column with the corresponding cofactors.

Here, Part (D) is the required answer, because elements of 1 column are multiplied by their corresponding cofactors.

So, a_{11}A_{11} + a_{21}A_{21} + a_{31}A_{31}.