If Blake wanted to use the Substitution method to solve the following system of equations, what could be a good first
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<em>Note: you did not add the answer options, so I am solving this question generally to solve your query, which anyways would clear your concept.</em>
Answer:
Please check the explanation.
Step-by-step explanation:
Given the equation
All we need is to find any random value of 'x' and then solve for 'y'.
For example, putting x=2
Thus, (2, 17) is one of the many ordered pairs solutions of the equation.
Similarly, putting x=3
Thus, (3, 21) is one of the many ordered pairs solutions of the equation.
First, we have to make sure that the number of columns in the first matrix is equal to the number of rows in the second matrix.
![\left[\begin{array}{cc}1&-3&2&0\\\end{array}\right] * \left[\begin{array}{ccc}2&3&4\\1&2&3\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%26-3%262%260%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%2A%20%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%263%264%5C%5C1%262%263%5Cend%7Barray%7D%5Cright%5D%20)
Since this is true, we can continue to solve the problem.
To multiply two matrices, multiply each row element in the first matrix by each column element in the second matrix. For example:
1*2 = 2
-3*1=-3
Then we add them to get our new matrix element.
-3+2=-1
Then we move to the next column of the second matrix.
1*3=3
-3*2=-6
-6+3=-3
Then the final column of the second matrix.
1*4=4
-3*3=-9
-9+4=-5
Our matrix so far:
![\left[\begin{array}{ccc}-1&-3&-5\\x&x&x\end{array}\right]](https://tex.z-dn.net/?f=%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-1%26-3%26-5%5C%5Cx%26x%26x%5Cend%7Barray%7D%5Cright%5D%20)
We do the same for the bottom row of the first matrix.
<em>First Column</em>
2*2=4
0*1=0
4+0=4
<em>Second Column
</em>2*3=6
0*2=0
6+0=6
<em>Third Column</em>
2*4=8
0*3=0
8+0=8
Our final matrix is:
![\left[\begin{array}{ccc}-1&-3&-5\\4&6&8\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-1%26-3%26-5%5C%5C4%266%268%5Cend%7Barray%7D%5Cright%5D)
:)
Since this is true, we can continue to solve the problem.
To multiply two matrices, multiply each row element in the first matrix by each column element in the second matrix. For example:
1*2 = 2
-3*1=-3
Then we add them to get our new matrix element.
-3+2=-1
Then we move to the next column of the second matrix.
1*3=3
-3*2=-6
-6+3=-3
Then the final column of the second matrix.
1*4=4
-3*3=-9
-9+4=-5
Our matrix so far:
We do the same for the bottom row of the first matrix.
<em>First Column</em>
2*2=4
0*1=0
4+0=4
<em>Second Column
</em>2*3=6
0*2=0
6+0=6
<em>Third Column</em>
2*4=8
0*3=0
8+0=8
Our final matrix is:
:)