Prove that if BA=I then BA=AB.

Mathematics

1 answer:

Zanzabum3 years ago

4 0

Answer with Step-by-step explanation:

Since we have given that

$BA=I$

As we know that

AA⁻¹ = I (A is invertible matrix)

Multiplying A⁻¹ on the both the sides:

$BAA^{-1}=IA^{-1}\\\\B=A^{-1}$

Using the above result, we get that

$BA=I=AA^{-1}\\\\BA=AB$

Therefore, BA = AB

Hence, proved.

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What is the answer for y=2x-5 and 4x-y=7

Dvinal [7]

Substitute the first equation into the second equation. You will get:

4x - (2x - 5) = 7

Distribute the negative sign into the parenthesis:

4x - 2x + 5 = 7

Simplify

2x + 5 = 7

Subtract 5 on both sides

2x = 2

x = 1

Now, substitute x = 1 into the first equation:

y = 2(1) - 5

y = 2 - 5

y = -3

The solution to the system of equations is (1, -3).

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3 years ago

Graph the solution set

antiseptic1488 [7]

Answer:

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The true inclusion is where all the ranges intersect. Be sure to color this the darkest or with a special pattern.