In this article, we share MP Board Class 12th Maths Book Solutions Chapter 4 सारणिक Ex 4.1 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 4.1

**प्रश्न 1 से 2 तक में सारणिकों का मान ज्ञात कीजिए**

प्रश्न 1.

\(\left|\begin{array}{cc}{2} & {4} \\ {-5} & {-1}\end{array}\right|\)

हल:

\(\left|\begin{array}{cc}{2} & {4} \\ {-5} & {-1}\end{array}\right|\) = 2 × (-1) – 4 × (-5) = -2 + 20 = 18

प्रश्न 2.

हल:

(i) \(\left|\begin{array}{cc}{\cos \theta} & {-\sin \theta} \\ {\sin \theta} & {\cos \theta}\end{array}\right|\)

=cose θ cosθ – (sin θ) × (-sin θ)

= cos^{2}θ + sin^{2}θ

= 1

(ii) \(\left|\begin{array}{cc}{x^{2}-x+1} & {x-1} \\ {x+1} & {x+1}\end{array}\right|\)

= (x^{2} – x + 1)(x + 1) – (x + 1)(x – 1)

= (x + 1) – (x^{2} – 1) = x^{3} + 1 – x^{2} + 1

= x^{3} – x^{2} + 2

प्रश्न 3.

यदि A = \(\left[\begin{array}{ll}{1} & {2} \\ {4} & {2}\end{array}\right]\), तो दिखाइए |2A| = 4|A|

हल:

प्रश्न 4. यदि A = \(\left|\begin{array}{lll}{1} & {0} & {1} \\ {0} & {1} & {2} \\ {0} & {0} & {4}\end{array}\right|\) हो, तो दिखाइए |3A| = 27|A|

हलः

प्रश्न 5.

निम्नलिखित सारणिकों का मान ज्ञात कीजिए-

हल:

= 0[0-(-3) × 3][-1 × 0 – (-3) × (-2)] + 2[3 × (-1) – (-2) × 0]

= 0 × 9 – 1[0 – 6] + 2[-3 – 0]

= 0 – 1 × (-6) + 2 × (-3)

= 6 – 6 = 0

= 2[2 × 0 – (-1) × (-5)] + 1[0 × 0 – (-1) × 3] – 2[0 × (-5) – 2 × 3]

= 2[0 – 5] + 1[0 + 3] – 2 × [0 – 6]

= 2 × (-5) + 1 × 3 – 2 × (-6)

= -10 + 3 + 12 = -10 + 15 = 5

प्रश्न 6.

यदि A = \(\left|\begin{array}{ccc}{1} & {1} & {-2} \\ {2} & {1} & {-3} \\ {5} & {4} & {-9}\end{array}\right|\), हो तो |A| ज्ञात कीजिए।

हल:

प्रश्न 7.

x के मान ज्ञात कीजिए यदि

हल:

⇒ 2 – 20 = 2x^{2} – 24

⇒ -18 = 2x^{2} – 24

⇒ 2x^{2} – 24 +18 = 0

⇒ 2x^{2} -6 =0

⇒ x^{2} = 3

⇒ x = ±\( \sqrt{{3}} \)

⇒ 10 – 12 = 5x – 6x

-2x^{3} = – x

x = 2

प्रश्न 8.

यदि \(\left|\begin{array}{cc}{x} & {2} \\ {18} & {x}\end{array}\right|=\left|\begin{array}{cc}{6} & {2} \\ {18} & {6}\end{array}\right|\) हो तो x बराबर है-

(A) 6

(B) ±6

(C) -6

(D) 0

हल:

\(\left|\begin{array}{cc}{x} & {2} \\ {18} & {x}\end{array}\right|=\left|\begin{array}{cc}{6} & {2} \\ {18} & {6}\end{array}\right|\)

⇒ x × x -2 × 18 = 6 × 6 – 2 × 18

⇒ x^{2} – 36 = 36 – 36

⇒ x^{2} – 36 = 0

⇒ x = 36

∴ x = ±6

अतः विकल्प (B) सही है।