Answer:
Option C
Step-by-step explanation:
We are given a coefficient matrix along and not the solution matrix
Since solution matrix is not given we cannot check for infinity solutions.
But we can check whether coefficient matrix is 0 or not
If coefficient matrix is zero, the system is inconsistent and hence no solution.
Option A)
|A|=![\left[\begin{array}{ccc}4&2&6\\2&1&3\\-2&3&-4\end{array}\right] =0](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D4%262%266%5C%5C2%261%263%5C%5C-2%263%26-4%5Cend%7Barray%7D%5Cright%5D%20%3D0)
since II row is a multiple of I row
Hence no solution or infinite
OPtion B
|B|=![\left[\begin{array}{ccc}2&0&-2\\-7&1&5\\4&-2&0\end{array}\right] \\=2(10)-2(10)=0](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%260%26-2%5C%5C-7%261%265%5C%5C4%26-2%260%5Cend%7Barray%7D%5Cright%5D%20%5C%5C%3D2%2810%29-2%2810%29%3D0)
Hence no solution or infinite
Option C
![\left[\begin{array}{ccc}6&0&-2\\-2&0&6\\1&-2&0\end{array}\right] \\=2(36-2)=68](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D6%260%26-2%5C%5C-2%260%266%5C%5C1%26-2%260%5Cend%7Barray%7D%5Cright%5D%20%5C%5C%3D2%2836-2%29%3D68)
Hence there will be a unique solution
Option D
![\left[\begin{array}{ccc}5&10&5\\4&1&4\\-1&-2&-1\end{array}\right] \\=2(10)-2(10)=0](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D5%2610%265%5C%5C4%261%264%5C%5C-1%26-2%26-1%5Cend%7Barray%7D%5Cright%5D%20%5C%5C%3D2%2810%29-2%2810%29%3D0)
=0
(since I row is -5 times III row)
Hence there will be no or infinite solution
Option C is the correct answer
Answer:
the first one
Step-by-step explanation:
Answer:
Um, can you add the picture of the number line?
Answer:
The surface area of a box that is 6 by 4 inches is 24in²
Answer:
3p³ + 2p² – 3p – 11
Step-by-step explanation:
From the question given above, the following data were obtained:
Side 1 (S₁) = –1(p + 5)
Side 2 (S₂) = 2(p² – 3)
Side 3 (S₃) = 3p³ – 2p
Perimeter (P) =?
The perimeter of the triangle can be obtained as follow
P = S₁ + S₂ + S₃
P = –1(p + 5) + 2(p² – 3) + 3p³ – 2p
Clear bracket
P = –p – 5 + 2p² – 6 + 3p³ – 2p
Rearrange
P = 3p³ + 2p² – 2p – p – 6 – 5
P = 3p³ + 2p² – 3p – 11
Therefore, the perimeter of the triangle is 3p³ + 2p² – 3p – 11
