Question
Download Solution PDFConsider the determinant
Δ =
If a13 = yz, a23 = zx, a33 = xy and the minors of a13, a23, a33 are respectively (z − y), (z − x), (y − x) then what is the value of Δ ?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFExplanation:
Given:
Δ =
a13 = yz, a23 = zx, a33 = xy and M13 = (z − y), M23 = (z − x), M33 = (y − x)
Expanding the determinant along Column 3,
⇒ Δ = a13 (a21a32 - a22a31) - a23 (a11a32 - a12a31) + a33 (a11a2 - a12a1)
⇒ Δ = a13M13 - a23M23 + a33M33
where Mij denotes the minor of aij element.
⇒ Δ = yz(z - y) - zx(z - x) + xy(y - x)
⇒ Δ = yz(z - y) - z2x + zx2 + xy2 - x2y
⇒ Δ = yz(z - y) + (xy2 - z2x) + (zx2 - x2y)
⇒ Δ = yz(z - y) + x(y2 - z2) + x2(z - y)
⇒ Δ = (z - y)[yz - x(y + z) + x2]
⇒ Δ = (z - y)[yz - xy - xz + x2]
⇒ Δ = (z - y)[y(z - x) - x(z - x)]
⇒ Δ = (z - y)[(z - x)(y - x)]
Last updated on Jul 8, 2025
->UPSC NDA Application Correction Window is open from 7th July to 9th July 2025.
->UPSC had extended the UPSC NDA 2 Registration Date till 20th June 2025.
-> A total of 406 vacancies have been announced for NDA 2 Exam 2025.
->The NDA exam date 2025 has been announced. The written examination will be held on 14th September 2025.
-> The selection process for the NDA exam includes a Written Exam and SSB Interview.
-> Candidates who get successful selection under UPSC NDA will get a salary range between Rs. 15,600 to Rs. 39,100.
-> Candidates must go through the NDA previous year question paper. Attempting the NDA mock test is also essential.