Discussion
Let \(A = \begin{bmatrix} a & -1 & -a \\ 0 & 1 & -1 \\ 1 & 0 & 4 \end{bmatrix}\). If \(|A| = 26\) , then the value of a is equal to
Answer: A
\(a =\) 5.
Expand along the first row:
\[ \begin{aligned} |A| &= a(4 - 0) + 1(0 + 1) - a(0 - 1) \\ &= 4a + 1 + a = 5a + 1 \\ 5a + 1 &= 26 \;\Rightarrow\; a = 5 \end{aligned} \]
Dropping the sign of the third cofactor gives \(3a + 1 = 26\), which has no integer answer among the options, and the other values do not satisfy \(5a + 1 = 26\).
Check: with \(a = 5\), the expansion gives \(5 \times 4 + 1 + 5 = 26\).
No comment is present. Be the first to comment.
Loading…
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development