We want to find the values of a, b, c, and d such that the given matrix product is equal to a 2x2 identity matrix. We will solve a system of equations to find:

b = 1
a = -1
c = -2
d = -1

<h3>Presenting the equation:</h3>

Basically, we want to solve:

$\left[\begin{array}{cc}-1&2\\a&1\end{array}\right]*\left[\begin{array}{cc}b&c\\1&d\end{array}\right] = \left[\begin{array}{cc}1&0\\0&1\end{array}\right]$

The matrix product will be:

$\left[\begin{array}{cc}-b + 2&-c + 2d\\a*b + 1&a*c + d\end{array}\right]$

Then we must have:

-b + 2 = 1

This means that:

b = 2 - 1 = 1

b = 1

We also need to have:

a*b + 1 = 0

we know the value of b, so we just have:

a*1 + b = 0

a = -1

Now the two remaining equations are:

-c + 2d = 0

a*c + d = 1

Replacing the value of a we get:

-c + 2d = 0

-c + d = 1

Isolating c in the first equation we get:

c = 2d

Replacing that in the other equation we get:

-(2d) + d = 1

-d = 1

d = -1

Then:

c = 2d = 2*(-1) = -2

c = -2

So the values are:

b = 1
a = -1
c = -2
d = -1

If you want to learn more about systems of equations, you can read:

brainly.com/question/13729904

4 0

2 years ago

Does -2(z+3)-z=-z-4(z+2) have one,zero or infinite solutions

Andrew [12]

Hello from MrBillDoesMath!

Answer:

One solution (z = -1)

Discussion:

-2(z+3)-z=-z-4(z+2) =>

-2z -6 -z = -z -4z - 8 =>

-3z -6 = -5z -8 => add 6 to both sides

-2z = -5z -2 => add 5z to both sides

3z = -5z +5z -3 =>

3z = -3 =>

z = -1

Thank you,

MrB

6 0

3 years ago