NCERT Class 12 · Ch 4 Exercise 4.1 · Q4
NCERT Class 12 Maths · Chapter 4 Determinants · Exercise 4.1 Question 4
For A=\(\left[\begin{matrix}1 & 0 & 1\\0 & 1 & 2\\0 & 0 & 4\end{matrix}\right]\), show det(3A)=27 det A.
💡 Concept Hints
- Identify whether this needs determinant expansion, an inverse, or the absolute determinant formula for area.
- For a determinant use signed cofactors. For a system first check invertibility; for an area take half the absolute determinant.
✅ Step-by-Step Solution
Final Answer: \(108\) = 27 × \(4\)
- Expand the determinant along row 1, using cofactors Cᵢⱼ=(−1)ⁱ⁺ʲMᵢⱼ.
- The cofactors in this row are \(\left[\begin{matrix}4 & 0 & 0\end{matrix}\right]\).
- The expansion terms are \(4\), \(0\), \(0\).
- Adding and simplifying gives det A = \(4\).
- Multiplying every entry by 3 multiplies each of the 3 rows by this factor.
- By multilinearity, det(cA)=cⁿdet A for an n×n matrix. Direct evaluation of the scaled matrix gives \(108\).
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