In this article, we share MP Board Class 12th Maths Book Solutions Chapter 4 सारणिक Ex 4.4 Pdf, These solutions are solved by subject experts from the latest MP Board books.

MP Board Class 12th Maths Book Solutions Chapter 4 सारणिक Ex 4.4

निम्नलिखित सारणिकों के अवयवों के उपसारणिक एवं सहखण्ड लिखिए।
प्रश्न 1.
MP Board Class 12th Maths Book Solutions Chapter 4 सारणिक Ex 4.4 img 1
हल:
\(\left|\begin{array}{cc}{2} & {-4} \\ {0} & {3}\end{array}\right|\)
a11 का उपसारणिक M11 = 3
a12 का उपसारणिक M12 = 0
a21 का उपसारणिक M21 = -4
a22 का उपसाराणिक M22 = 2
a11 का सहखण्ड = A11 = (-1)1+1
M11 = (-1)2 × 3 =3
a12 का सहखण्ड = A12 = (-1)1+2
M12 = (-1)3 × 0 = 0
a21 का सहखण्ड = A13 = (-1)2 + 1
M21 = (-1)3 × (-4) = 4
a22 का सहखण्ड = A22 = (-1)2+2
M22 = (-1)4 × 2 = 2

(ii) यहाँ \(\left|\begin{array}{ll}{a} & {c} \\ {b} & {d}\end{array}\right|\) सारणिक \(\left|\begin{array}{ll}{a} & {c} \\ {b} & {d}\end{array}\right|\) के अवयवों के उपसारणिक निम्न हैं-
M11 = d
M12 = b
M21 = c
M22 = a
इसलिए सहखंड निम्न होंगे-
A11 = d
A12 = -b
A21 = -c
तथा A22 = 1

MP Board Solutions

प्रश्न 2.
MP Board Class 12th Maths Book Solutions Chapter 4 सारणिक Ex 4.4 img 2
हल:
MP Board Class 12th Maths Book Solutions Chapter 4 सारणिक Ex 4.4 img 3
MP Board Class 12th Maths Book Solutions Chapter 4 सारणिक Ex 4.4 img 4
A11 = (-1)1+1 M11 = (-1)2 × 1 = 1
A12 = 11+2 M12 = (-1)3 × 0= 0
A13 = (-1)1+3 M13 = (-1)4 × 0 = 0
A21 = (-1)2+1 M21 = (-1)3 × 0 = 0
A22 = (-1)2+2 M22 = (-1)4 × 1 = 1
A23 = (-1)2+3 M23 = (-1)5 × 0 = 0
A31 =(-1)3+1 M31 = (-1)4 × 0 = 0
A32 = (-1)3+2 M32 = (-1)5 × 0= 0
A33 = (-1)3+3 M33 = (-1)6 × 1 = 1

(ii) यहाँ \(\left|\begin{array}{ccc}{1} & {0} & {4} \\ {3} & {5} & {-1} \\ {0} & {1} & {2}\end{array}\right|\) उपसारणिक और सहखण्ड की परिभाषां से a1 का उपसारणिक M1 = \(\left|\begin{array}{cc}{5} & {-1} \\ {1} & {2}\end{array}\right|\) = 10 + 1 = 11
हल:
MP Board Class 12th Maths Book Solutions Chapter 4 सारणिक Ex 4.4 img 5
MP Board Class 12th Maths Book Solutions Chapter 4 सारणिक Ex 4.4 img 6
MP Board Class 12th Maths Book Solutions Chapter 4 सारणिक Ex 4.4 img 7

प्रश्न 3.
दूसरी पंक्ति के अवयवों के सहखण्डों का प्रयोग करके ∆ = \(\left|\begin{array}{lll}{5} & {3} & {8} \\ {2} & {0} & {1} \\ {1} & {2} & {3}\end{array}\right|\) का मान ज्ञात कीजिए।
हल:
MP Board Class 12th Maths Book Solutions Chapter 4 सारणिक Ex 4.4 img 8
दूसरी पंक्ति से सारणिक का विस्तार करने पर,
∆ = a21 A21 + a22 A22 + a23 A23
= 2 × 7 + 0 × 7 + 1 × (-7)
= 14 + 0 – 7 = 7

MP Board Solutions

प्रश्न 4.
तीसरे स्तम्भ के अवयवों के सहखण्डों का प्रयोग करके ∆ = \(\left|\begin{array}{ccc}{1} & {x} & {y z} \\ {1} & {y} & {z x} \\ {1} & {z} & {x y}\end{array}\right|\) का मान ज्ञात कीजिए।
हल:
MP Board Class 12th Maths Book Solutions Chapter 4 सारणिक Ex 4.4 img 9
A23 = (-1)2+3 \(\left|\begin{array}{ll}{1} & {x} \\ {1} & {z}\end{array}\right|\) = (-1)5 [z – y]
= -(z – x)
A33 = (-1)3+3 \(\left|\begin{array}{ll}{1} & {x} \\ {1} & {y}\end{array}\right|\) = (-1)6 [y – x] = (y – x)
∴ ∆ = a13 A13 + a23 A23 + a33 A33
= yz (z – y) + zx (-z + x) + xy(y – x)
= yz2 – y2z – xz2 + x2z + xy2 – x2y
= (-y2z + yz2) + (y2 – xz2) + (-x2y + x2z)
= – yz (y – z) + x(y2 – z2 ) – x2(y – z)
= (y – z)[-yz + x (y + z) – x2]
= (y – z)[z (x – y) – x(x – y)]
= (y – z)(x – y)(z – x)
= (x – y)(y – z)(z – x)

प्रश्न 5.
यदि ∆ = \(\left|\begin{array}{lll}{a_{11}} & {a_{12}} & {a_{13}} \\ {a_{21}} & {a_{22}} & {a_{23}} \\ {a_{31}} & {a_{32}} & {a_{33}}\end{array}\right|\) और aij का सहखण्ड Aij हो तो ∆ का मान निम्नलिखित रूप में व्यक्त किया जाता है
(A) A11 A31 + a12 A32 + a13 A33
(B) a11 A11 + a12 A21 + a13 A31
(C) a21 A11 + a22 A12 + a23 A13
(D) a11 A11 + a21 A21 + a31 A31
हल:
∆ = किसी पंक्ति (या स्तम्भ) के अवयवों तथा उनके संगत सहखण्डों के गुणन का योग
C1 स्तम्भ के अवयव (a11, a21, a31)
इनमें सहखण्ड a11, A21, A31
⇒ ∆ = a11 A11 + a21 A21 + a31 A31
अत: विकल्प (D) सही है।

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