Let's solve the given matrix equality step-by-step:
We are given:
[tex]\[
\left(\begin{array}{cc}
2^x & 3^y \\
x z & y z
\end{array}\right)
=
\left(\begin{array}{ll}
16 & 27 \\
16 & 12
\end{array}\right)
\][/tex]
We need to find the values of [tex]\(x\)[/tex], [tex]\(y\)[/tex], and [tex]\(z\)[/tex] that satisfy this equality.
### Step 1: Solve for [tex]\(x\)[/tex] and [tex]\(y\)[/tex]
First, consider the elements of the matrix in the first row.
#### Solving for [tex]\(x\)[/tex]
The element at the first row, first column gives us the equation:
[tex]\[
2^x = 16
\][/tex]
We know that [tex]\(16\)[/tex] can be written as a power of [tex]\(2\)[/tex]:
[tex]\[
16 = 2^4
\][/tex]
Therefore, we have:
[tex]\[
2^x = 2^4
\][/tex]
By equating the exponents, we find:
[tex]\[
x = 4
\][/tex]
#### Solving for [tex]\(y\)[/tex]
The element at the first row, second column gives us the equation:
[tex]\[
3^y = 27
\][/tex]
We know that [tex]\(27\)[/tex] can be written as a power of [tex]\(3\)[/tex]:
[tex]\[
27 = 3^3
\][/tex]
Therefore, we have:
[tex]\[
3^y = 3^3
\][/tex]
By equating the exponents, we find:
[tex]\[
y = 3
\][/tex]
### Step 2: Solve for [tex]\(z\)[/tex]
Next, we substitute [tex]\(x = 4\)[/tex] and [tex]\(y = 3\)[/tex] into the elements of the second row to find [tex]\(z\)[/tex].
The element at the second row, first column gives us the equation:
[tex]\[
x z = 16
\][/tex]
Substituting [tex]\(x = 4\)[/tex]:
[tex]\[
4z = 16
\][/tex]
Solving for [tex]\(z\)[/tex]:
[tex]\[
z = \frac{16}{4} = 4
\][/tex]
The element at the second row, second column gives us the equation:
[tex]\[
y z = 12
\][/tex]
Substituting [tex]\(y = 3\)[/tex]:
[tex]\[
3z = 12
\][/tex]
Solving for [tex]\(z\)[/tex]:
[tex]\[
z = \frac{12}{3} = 4
\][/tex]
Since both approaches for finding [tex]\(z\)[/tex] yield the same result, we have our final values.
### Conclusion
The solution to the matrix equality is:
[tex]\[
x = 4, \quad y = 3, \quad z = 4
\][/tex]
