Pls solve these 2 questions for me:-

Q: Without expanding, evaluat the determinant:

1. 1 3 9 27

3 9 27 1

9 27 1 3

27 1 3 9

2) 67 19 21

39 13 14

81 24 26

Question

Pls solve these 2 questions for me:-
Q: Without expanding, evaluat the determinant:
1 1 3 9 27
3 9 27 1
9 27 1 3
27 1 3 9
2) 67 19 21
39 13 14
81 24 26

Gursheen kaur. · Accepted Answer

$2) L e t Δ = | \begin{matrix} 67 & 19 & 21 \\ 39 & 13 & 14 \\ 81 & 24 & 26 \end{matrix} | f i r s t b r e a k d o w n t h e g i v e n \det e r \min a n t, t h a t i s, Δ = | \begin{matrix} 10 + 57 & 19 & 21 \\ 0 + 39 & 13 & 14 \\ 9 + 72 & 24 & 26 \end{matrix} |$

By the property of determinant, if each element of a column of a determinant is expressed as the sum of two terms, the determinant can be expressed as the sum of two determinants.

Therefore,

$Δ = | \begin{matrix} 10 & 19 & 21 \\ 0 & 13 & 14 \\ 9 & 24 & 26 \end{matrix} | + | \begin{matrix} 57 & 19 & 21 \\ 39 & 13 & 14 \\ 72 & 24 & 26 \end{matrix} | n o w, t a k e t h e \det e r m i a n t | \begin{matrix} 57 & 19 & 21 \\ 39 & 13 & 14 \\ 72 & 24 & 26 \end{matrix} | = | \begin{matrix} 0 & 19 & 21 \\ 0 & 13 & 14 \\ 0 & 24 & 26 \end{matrix} |, a p p l y i n g C_{1} \to (- 3) C_{2} = 0, a s e a c h e l e m e n t i n a c o l u m n o f a \det e r \min a n t i s 0. t h e r e f o r e, Δ = | \begin{matrix} 10 & 19 & 21 \\ 0 & 13 & 14 \\ 9 & 24 & 26 \end{matrix} | a g a i n r e w r i t e i t a s Δ = | \begin{matrix} 10 & 19 & 19 \\ 0 & 13 & 13 \\ 9 & 24 & 24 \end{matrix} | + | \begin{matrix} 10 & 19 & 2 \\ 0 & 13 & 1 \\ 9 & 24 & 2 \end{matrix} | w h e r e, | \begin{matrix} 10 & 19 & 19 \\ 0 & 13 & 13 \\ 9 & 24 & 24 \end{matrix} | = 0, a s t w o c o l u m n s o f \det e r \min a n t h a v e s a m e e l e m e n t s Δ = 0 + | \begin{matrix} 10 & 19 & 2 \\ 0 & 13 & 1 \\ 9 & 24 & 2 \end{matrix} | = | \begin{matrix} 10 & 19 & 2 \\ 0 & 13 & 1 \\ 9 & 24 & 2 \end{matrix} |$

Now, use minors to calculate the value of the determinant. Consider column one and let its elements be a1, a2, a3 . thus their corresponding minors are A1, A2 and A3.

substituting the values, we get,

a1A1-a2A2+a3A3

thus, the second term = 0 , that is

10A1-0A2+9A3

Also, we find the minors A1 and A3 by eliminating from the determinant the row and column that a1 and a3 are calculated.

$A_{1} = | \begin{matrix} 13 & 1 \\ 24 & 2 \end{matrix} |, A_{3} = | \begin{matrix} 19 & 2 \\ 13 & 1 \end{matrix} |$

$a n d Δ = | \begin{matrix} 67 & 19 & 21 \\ 39 & 13 & 14 \\ 81 & 24 & 26 \end{matrix} | = | \begin{matrix} 10 & 19 & 2 \\ 0 & 13 & 1 \\ 9 & 24 & 2 \end{matrix} | = 10 | \begin{matrix} 13 & 1 \\ 24 & 2 \end{matrix} | + 9 | \begin{matrix} 19 & 2 \\ 13 & 1 \end{matrix} | = 10 (26 - 24) + 9 (19 - 26) = 10 (2) + 9 (- 7) = 20 - 63 Δ = - 43$

Pls solve these 2 questions for me:- Q: Without expanding, evaluat the determinant: 1. 1 3 9 27 3 9 27 1 9 27 1 3 27 1 3 9 2) 67 19 21 39 13 14 81 24 26

Pls solve these 2 questions for me:-

Q: Without expanding, evaluat the determinant:

1. 1 3 9 27

3 9 27 1

9 27 1 3

27 1 3 9

2) 67 19 21

39 13 14

81 24 26