These NCERT Solutions for Class 12 Maths Chapter 4 सारणिक Ex 4.1 Questions and Answers are prepared by our highly skilled subject experts.
NCERT Solutions for Class 12 Maths Chapter 4 सारणिक Ex 4.1
प्रश्न 1.
\(\left|\begin{array}{rr}
2 & 4 \\
-5 & -1
\end{array}\right|\)
हल :
\(\left|\begin{array}{rr}
2 & 4 \\
-5 & -1
\end{array}\right|\) = 2 × (-1) – (- 5) × 4
= – 2 + 20 = 18
प्रश्न 2.
(i) \(\left|\begin{array}{rr}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta
\end{array}\right|\)
(ii) \(\left|\begin{array}{rr}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta
\end{array}\right|\)
हल:
\(\left|\begin{array}{rr}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta
\end{array}\right|\)
= cos2θ – (- sin ) sin θ
=coss2 θ + sin2 θ = 1
(ii) \(\left|\begin{array}{rr}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta
\end{array}\right|\)
= (x2 – x + 1) (x + 1) − (x + 1) (x − 1)
= (x2 – x2 + x + x2 -x+1)(x2-1)
= x2 + 1 − x2 + 1
=x3 – x2 + 2.
प्रश्न 3.
यदि A = \(=\frac{10}{1000}\) तो दिखाइए |24|= 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
|3 A|=27|A|
हल:
A = \(\left[\begin{array}{lll}
1 & 0 & 1 \\
0 & 1 & 2 \\
0 & 0 & 4
\end{array}\right]\)
तो |A| = \(\left[\begin{array}{lll}
1 & 0 & 1 \\
0 & 1 & 2 \\
0 & 0 & 4
\end{array}\right]\)
तृतीय पंक्ति के अनुदिश |A| का प्रसरण करने पर
|A| = 0 \(\left|\begin{array}{ll}
0 & 1 \\
1 & 2
\end{array}\right|-0\left|\begin{array}{ll}
1 & 1 \\
0 & 2
\end{array}\right|+4\left|\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right|\)
0 – 0 + 4 × (1 × 1-0 × 0)
= 4 × 1 = 4
3A = 3 \(\left[\begin{array}{lll}
1 & 0 & 1 \\
0 & 1 & 2 \\
0 & 0 & 4
\end{array}\right]\)
= \(\left[\begin{array}{rrr}
3 \times 1 & 3 \times 0 & 3 \times 1 \\
3 \times 0 & 3 \times 1 & 3 \times 2 \\
3 \times 0 & 3 \times 0 & 3 \times 4
\end{array}\right]=\left[\begin{array}{rrr}
3 & 0 & 3 \\
0 & 3 & 6 \\
0 & 0 & 12
\end{array}\right]\)
∴ |3A|= 0 \(\left|\begin{array}{rrr}
3 & 0 & 3 \\
0 & 3 & 6 \\
0 & 0 & 12
\end{array}\right|\)
|3 A| का तृतीय पंक्ति के अनुदिश प्रसरण करने पर,\
|3 A| = 0 \(\left|\begin{array}{ll}
0 & 3 \\
3 & 6
\end{array}\right|-0\left|\begin{array}{ll}
3 & 3 \\
0 & 6
\end{array}\right|+12\left|\begin{array}{ll}
3 & 0 \\
0 & 3
\end{array}\right|\)
= 0 – 0 + 12 (3 × 3-0 × 0)
= 12 x 9 108
∴ |3A| = 108 तथा | A |= 4
∴27 |A| = 27 × 4 = 108
अत: |3A| = 108 = 27| A |
प्रश्न 5.
निम्नलिखित सारणिकों के मान ज्ञात कीजिए :
हल:
नोट : किसी विषम सममित आव्यूह जिसकी कोटि विषम संख्या हो उसका सारणिक का मान हमेशा शून्य होता है। परंतु यदि कोटि एक सम संख्या (2,4,6,……………) हो तो सारणिक का मान हमेशा एक पर्ण वर्ग होता है।
प्रश्न 6.
यदि A =\(\left[\begin{array}{rrr}
1 & 1 & -2 \\
2 & 1 & -3 \\
5 & 4 & -9
\end{array}\right]\) हो, तो | A | ज्ञात कीजिए ।
हल:
A = \(\left[\begin{array}{rrr}
1 & 1 & -2 \\
2 & 1 & -3 \\
5 & 4 & -9
\end{array}\right]\)
∴ | A | = \(\left[\begin{array}{rrr}
1 & 1 & -2 \\
2 & 1 & -3 \\
5 & 4 & -9
\end{array}\right]\)
| A | का प्रथम पंक्ति के अनुदिश प्रसरण करने पर,
| A | = 1 \(\left|\begin{array}{ll}
1 & -3 \\
4 & -9
\end{array}\right|-1\left|\begin{array}{ll}
2 & -3 \\
5 & -9
\end{array}\right|+(-2)\left|\begin{array}{ll}
2 & 1 \\
5 & 4
\end{array}\right|\)
⇒ | A | = 1 (1 × (9) – 4 × (-3))
– 1(2 × (-9)-5 × (-3)) -2(2 x 4-5 x 1)
⇒ | A| = (- 9 + 12) 1(- 18 + 15) -2 (8 – 5)
= 1 × 3-1 × (-3) – 2 × 3
=3 + 3 – 6
= 6 – 6=0
∴ |A|= 0
प्रश्न 7.
x के मान ज्ञात कीजिए यदि:
(i) \(\left|\begin{array}{ll}
2 & 4 \\
5 & 1
\end{array}\right|=\left|\begin{array}{rr}
2 x & 4 \\
6 & x
\end{array}\right|\)
(ii) \(\left|\begin{array}{ll}
2 & 3 \\
4 & 5
\end{array}\right|=\left|\begin{array}{rr}
x & 3 \\
2 x & 5
\end{array}\right|\)
हल:
(i) \(\left|\begin{array}{cc}
2 & 4 \\
5 & 1
\end{array}\right|=\left|\begin{array}{cc}
2 x & 4 \\
6 & x
\end{array}\right|\)
⇒ 2× 1-5 x 4 = 2x × x – 6 × 4
⇒ 2-20 = 2x2-24
⇒ – 18 = 2x2 – 24
⇒ 2x2 = 24 – 18
⇒ 2x2 = 6
⇒ x2 = 3
∴ = ± \(\sqrt{3}\)
(ii) \(\left|\begin{array}{ll}
2 & 3 \\
4 & 5
\end{array}\right|=\left|\begin{array}{cc}
x & 3 \\
2 x & 5
\end{array}\right|\)
⇒ 2 × 5 – 4 × 3 = x × 5 – 2x × 3
⇒ 10 – 12 = 5x – x
⇒ -2 = – 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 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 18 × 26 × 6 – 18 × 2
⇒ x2-36 = 36 – 36
x2 – 36 = 0
x2 = 36
x =\(\sqrt{36}\)
x= ±6
अतः विकल्प (B) सही है।