PSEB 12th Class Maths Solutions Chapter 4 Determinants Ex 4.4

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 4 Determinants Ex 4.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.4

Dfrectlon (1 – 2): Write minors and cofactors of the elements of following determinants.

Question 1.
(i) \(\left|\begin{array}{cc}
2 & -4 \\
0 & 3
\end{array}\right|\)

(ii) \(\left|\begin{array}{ll}
\boldsymbol{a} & \boldsymbol{c} \\
\boldsymbol{b} & \boldsymbol{d}
\end{array}\right|\)
Solution.
(i) The given determinants is \(\left|\begin{array}{cc}
2 & -4 \\
0 & 3
\end{array}\right|\)
Minor of elements a_ij is M_ij.
∴ M₁₁ = Minor of element a₁₁ = 3
M₁₂ = Minor of element a₁₂= 0
M₂₁ = Minor of element a₂₁ = – 4
M₂₂ = Minor of element a₂₂ = 2

Cofactor of a_ij is = (- 1)^{i + j} M_ij.

∴ A₁₁ = (- 1)^{1 + 1} M₁₁ = (- 1)² (2) = 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) The given determinant is \(\left|\begin{array}{ll}
a & c \\
b & d
\end{array}\right|\)
Minor of element a_ij is M_ij.
∴ M₁₁ = Minor of element a₁₁ = d
M₁₂ = Minor of element a₁₂= b
M₂₁ = Minor of element a₂₁ = c
M₂₂ = Minor of element a₂₂ = a

Cofactor of a_ij is = (- 1)^{i + j} M_ij.

∴ 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) \(\left|\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right|\)

(ii) \(\left|\begin{array}{ccc}
1 & 0 & 4 \\
3 & 5 & -1 \\
0 & 1 & 2
\end{array}\right|\)
Solution.
100
The given determinant is \(\left|\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right|\)
By the definition of minors and cofactors, we have

M₁₁ = Minor of A₁₁ = \(\left|\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right|\) = 1

M₁₂ = Minor of A₁₂ = \(\left|\begin{array}{ll}
0 & 0 \\
0 & 1
\end{array}\right|\) = 0.

M₁₃ = Minor of A₁₃ = \(\left|\begin{array}{ll}
0 & 1 \\
0 & 0
\end{array}\right|\) = 0

M₂₁ = Minor of A₂₁ = \(\left|\begin{array}{ll}
0 & 0 \\
0 & 1
\end{array}\right|\) = 0

M₂₂ = Minor of A₂₂ = \(\left|\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right|\) = 1

M₂₃ = Minor of A₂₃ = \(\left|\begin{array}{ll}
1 & 0 \\
0 & 0
\end{array}\right|\) = 0

M₃₁ = Minor of A₃₁ = \(\left|\begin{array}{ll}
0 & 0 \\
1 & 0
\end{array}\right|\) = 0

M₃₂ = Minor of A₃₂ = \(\left|\begin{array}{ll}
1 & 0 \\
0 & 0
\end{array}\right|\) = 0

M₃₃ = Minor of A₃₃ = \(\left|\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right|\) = 1.

A₁₁ = Cofactor of a₁₁ = (- 1)^{1 + 1} M₁₁ = 1
A₁₂ = Cofactor of a₁₂ = (- 1)^{1 + 2} M₁₂ = 0
A₁₃ = Cofactor of a₁₃ = (-1)^{1 + 3} M₁₃ = 0
A₂₁ = Cofactor of a₂₁ = (- 1)^{2 + 1} M₂₁ = 0
A₂₂ = Cofactor of a₂₂ = (- 1)^{2 + 2} M₂₂ = 1
A₂₃ = Cofactor of a₂₃ = (- 1)^{2 + 3} M₂₃ = 0
A₃₁ = Cofactor of a₃₁ = (- 1)^{3 + 1} M₃₁ = 0
A₃₂ = Cofactor of a₃₂ = (-1)^{3 + 2} M₃₂ = 0
A₃₃ = Cofactor of a₃₃ = (-1)^{3 + 3} M₃₃ = 1

(ii) The given determinant is \(\left|\begin{array}{ccc}
1 & 0 & 4 \\
3 & 5 & -1 \\
0 & 1 & 2
\end{array}\right|\)
By definition of minors and cofactors, we have

M₁₁ = Minor of a₁₁
= \(\left|\begin{array}{cc}
5 & -1 \\
1 & 2
\end{array}\right|\) = 10 + 1 = 11

M₁₂ = Minor of a₁₂
= \(\left|\begin{array}{cc}
3 & -1 \\
0 & 2
\end{array}\right|\) = 6 – 0 = 6

M₁₃ = Minor of a₁₃
= \(\left|\begin{array}{ll}
3 & 5 \\
0 & 1
\end{array}\right|\) = 3 – 0 = 3

M₂₁ = Minor of a₂₁
= \(\left|\begin{array}{ll}
0 & 4 \\
1 & 2
\end{array}\right|\) = 0 – 4 = – 4

M₂₂ = Minor of a₂₂
= \(\left|\begin{array}{ll}
1 & 4 \\
0 & 2
\end{array}\right|\) = 2 – 0 = 2

M₂₃ = Minor of a₂₃
= \(\left|\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right|\) = 1 – 0 = 1

M₃₁ = Minor of a₃₁
= \(\left|\begin{array}{cc}
0 & 4 \\
5 & -1
\end{array}\right|\) = 0 – 20 = – 20

M₃₂ = Minor of a₃₂
= \(\left|\begin{array}{cc}
1 & 4 \\
3 & -1
\end{array}\right|\) = – 1 – 12 = – 13

M₃₃ = Minor of a₃₃
= \(\left|\begin{array}{ll}
1 & 0 \\
3 & 5
\end{array}\right|\) = 5 – 0 = 5.

Cofactor of a_ij = (- 1)^{i + j} M_ij

A₁₁ = Cofactor of a₁₁ = (- 1)^{1 + 1} M₁₁ = 11
A₁₂ = Cofactor of a₁₂ = (- 1)^{1 + 2} M₁₂ = – 6
A₁₃ = Cofactor of a₁₃ = (- 1)^{1 + 3} M₁₃ = 3
A₂₁ = Cofactor of a₂₁ = (- 1)^{2 + 1} M₂₁ = 4
A₂₂ = Cofactor of a₂₂ = (- 1)^{2 + 2} M₂₂ = 2
A₂₃ = Cofactor of a₂₃ = (- 1)^{2 + 3} M₂₃ = – 1
A₃₁ = Cofactor of a₃₁ = (- 1)^{3 + 1} M₃₁ = – 20
A₃₂ = Cofactor of a₃₂ = (- 1)^{3 + 2} M₃₂ = 13
A₃₃ = Cofactor of a₃₃ = (- 1)^{3 + 3} M₃₃ = 5

Question 3.
Using cofactors of elements of second row, evaluate ∆ = \(\left|\begin{array}{lll}
5 & 3 & 8 \\
2 & 0 & 1 \\
1 & 2 & 3
\end{array}\right|\).
Solution.
The given determinant is \(\left|\begin{array}{lll}
5 & 3 & 8 \\
2 & 0 & 1 \\
1 & 2 & 3
\end{array}\right|\)
We have,
M₂₁ = \(\left|\begin{array}{ll}
3 & 8 \\
2 & 3
\end{array}\right|\) = 9 – 16 = – 7
∴ A₂₁ = Cofactor of a₂₁ = (- 1)^{2 + 1} = 7

M₂₂ = \(\left|\begin{array}{ll}
5 & 8 \\
1 & 3
\end{array}\right|\) = 15 – 8 = 7
∴ A₂₂ = Cofactor of a₂₂ = (- 1)^{2 + 1} = 7

M₂₃ = \(\left|\begin{array}{ll}
5 & 3 \\
1 & 2
\end{array}\right|\)
∴ A₂₃ = Cofactor of a₂₃ = (- 1)^{2 + 3} M₂₃ = – 7
We know that is equal to the sum of the product of the elements of the second row with their corresponding cofactors.
∴ ∆ = a₂₁A₂₁ + a₂₂A₂₂ + a₂₃A₂₃
= 2 (7) + 0 (7) + 1 (- 7) = 14 – 7 = 7.

Question 4.
Using cofactors of elements of third column, evaluate
∆ = \(\left|\begin{array}{lll}
1 & x & y z \\
1 & y & z x \\
1 & x & x y
\end{array}\right|\)
Solution.
The given determinant is \(\left|\begin{array}{lll}
1 & x & y z \\
1 & y & z x \\
1 & x & x y
\end{array}\right|\)
We have,
M₁₃ = \(\left|\begin{array}{ll}
1 & y \\
1 & z
\end{array}\right|\) = z – y

M₂₃ = \(\left|\begin{array}{ll}
1 & x \\
1 & z
\end{array}\right|\) = z – x

M₃₃ = \(\left|\begin{array}{ll}
1 & x \\
1 & y
\end{array}\right|\) = y – x

∴ A₁₃ = Cofactor of a₁₃ = (- 1)^{1 + 3} M₁₃ = (z – y)
A₂₃ = Cofactor of a₂₃ = (- 1)^{2 + 3} M₂₃ = – (z – x) = (x – z)
A₃₃ = Cofactor of a₃₃ = (- 1 )^{3 + 3} M₃₃ = (y – x)
We know that ∆ is equal to the sum of the product of the elements of the second row with their corresponding cofactors.

∴ ∆ = a₁₃A₁₃ + a₂₃A₂₃ + a₃₃A₃₃
= yz (z – y) + zx (x – z) + xy (y – x)
= yz² – y²z + x²z – xz² + xy² – x²y
= (x²z – y²z) + (yz² – xz²) + (xy² – x²y)
= z(x² – y²) + z² (y – x) + xy (y – x)
= z(x – y)(x + y) + z² (y – x) + xy (y – x)
= (x – y)[zx + zy – z² – xy]
= (x – y) [z(x – z) + y(z – x)]
= (x – y) (z – x)[- z + y]
= (x – y) (y – z) (z – x)
Hence, A = (x – y)(y – z)(z – x)

Question 5.
If ∆ = \(\left|\begin{array}{lll}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33}
\end{array}\right|\) and A_ij is cofactor of a_ij, then value of ∆ is given by
(A) a₁₁A₃₁ + a₁₂A₃₂ + a₁₃A₃₃
(B) a₁₁A₁₁ + a₁₂A₂₁ + a₁₃A₃₁
(C) a₂₁A₁₁ + a₂₂A₁₂ + a₂₃A₁₃
(D) a₁₁A₁₁ + a₂₁A₂₁ + a₃₁A₃₁
Solution.
∆ is equal to the sum of the products of the elements of a row (or a column) with their corresponding cofactors.
∆ = a₁₁A₁₁ + a₁₂A₁₂ + a₁₃A₁₃ or
a₂₁A₂₁ + a₂₂A₂₂ + a₂₃A₂₃ or
a₃₁A₃₁ + a₃₂A₃₂ + a₃₃A₃₃ or

a₁₁A₁₁ + a₂₁A₂₁ + a₃₁A₃₁ or
a₁₂A₁₂ + a₂₂A₂₂ + a₃₂A₃₂ or
a₁₃A₁₃ + a₂₃A₂₃ + a₃₃A₃₃

Sum of the products of the elements of first column with their corresponding cofactors is ∆ = a₁₁A₁₁ + a₂₁A₂₁ + a₃₁A₃₁
Hence, the correct answer is (D).