Paragraph for below question
नीचे दिए गए प्रश्न के लिए अनुच्छेद

If A is a square matrix, then determinant of A is represented by |A|. The value of |A| is equal to zero, if any two rows or columns are identical.

Let, $A = ⎡ ⎢ ⎣ \begin{matrix} x & 3 & 7 4 & x - 1 & 7 4 & 1 & x + 5 \end{matrix} ⎤ ⎥ ⎦,$ then

यदि A एक वर्ग आव्यूह है, तब A के सारणिक को |A| द्वारा दर्शाया जाता है। |A| का मान शून्य है, यदि कोई भी दो पंक्तियाँ या दो स्तम्भ समरूप हैं।

माना, $A = ⎡ ⎢ ⎣ \begin{matrix} x & 3 & 7 4 & x - 1 & 7 4 & 1 & x + 5 \end{matrix} ⎤ ⎥ ⎦,$ तब

Q. If |A| = 0, then one of the possible value of x is

प्रश्न - यदि |A| = 0, तब x का कोई एक संभावित मान है

Zero

शून्य

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

–2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B 4
Solution

Suggest Corrections

Similar questions

A = ⎡ ⎢ ⎣ \begin{matrix} x & 3 & 7 4 & x - 1 & 7 4 & 1 & x + 5 \end{matrix} ⎤ ⎥ ⎦,

A = ⎡ ⎢ ⎣ \begin{matrix} x & 3 & 7 4 & x - 1 & 7 4 & 1 & x + 5 \end{matrix} ⎤ ⎥ ⎦,

adj (k A) = k^{n - 1} (adj A)

| k A | = k^{n} \cdot | A |

adj (k A) = k^{n - 1} (adj A)

| k A | = k^{n} \cdot | A |

A = ⎡ ⎢ ⎣ \begin{matrix} 1 & 3 & 0 2 & 3 & 4 3 & 3 & 4 \end{matrix} ⎤ ⎥ ⎦,

A = ⎡ ⎢ ⎣ \begin{matrix} 1 & 3 & 0 2 & 3 & 4 3 & 3 & 4 \end{matrix} ⎤ ⎥ ⎦,

f (x) = ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{tan x}{sin x} & , & x < 0 a & , & x = 0 \frac{b {log}_{e} (1 + 2 x)}{x} & , & x > 0 \end{matrix}

f (x) = ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{tan x}{sin x} & , & x < 0 a & , & x = 0 \frac{b {log}_{e} (1 + 2 x)}{x} & , & x > 0 \end{matrix}

adj (k A) = k^{n - 1} (adj A)

| k A | = k^{n} \cdot | A |

adj (k A) = k^{n - 1} (adj A)

| k A | = k^{n} \cdot | A |

Join BYJU'S Learning Program