Question

Prove that if ad - bc is not equal to 0 and C1 and C2 are given constants, then the system of linear equations ax1 + bx2 = C1, cx1 + dx2 = C2 has a unique solution. On the other hand, if ad - bc = 0, are there necessarily infinitely many solutions?

Name: prove that if adbc not equal to 0 and c1 and c2 are given constants then the system of linear equations ax1 bx2 c1 cx1 dx2 c2 has a unique solution on the other hand if ad bc 0 are there nec 63482
Uploaded: 2023-03-28T01:01:42-08:00
Duration: 3 min 5 s
Channel: Supreeta N
Description: prove that if adbc not equal to 0 and c1 and c2 are given constants then the system of linear equations ax1 bx2 c1 cx1 dx2 c2 has a unique solution on the other hand if ad bc 0 are there nec 63482

          Prove that if ad - bc is not equal to 0 and C1 and C2 are given constants, then the system of linear equations ax1 + bx2 = C1, cx1 + dx2 = C2 has a unique solution. On the other hand, if ad - bc = 0, are there necessarily infinitely many solutions?

Added by Christopher L.

Elementary and Intermediate Algebra

Alan S. Tussy, R. David Gustafson 5th Edition

Instant Answer

Solved by Expert Supreeta N

03/28/2023

Step 1

First, we can write the given system of linear equations as a matrix equation: $$ \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \begin{pmatrix} C_1 \\ C_2 \end{pmatrix} $$ Now, let's find the determinant of the Show more…

Show all steps

Thanks for your feedback!

Prove that if ad - bc is not equal to 0 and C1 and C2 are given constants, then the system of linear equations ax1 + bx2 = C1, cx1 + dx2 = C2 has a unique solution. On the other hand, if ad - bc = 0, are there necessarily infinitely many solutions?