Standardized Test Practice
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To solve a matrix equation, multiply both sides by the inverse of the coefficient matrix.
15/2
The left-hand side of the equation is the product of the coefficient matrix and the variable matrix.
[
cc
4 & 3
2 & 2
]
*
[
c
a
b
]
=
[
c
21
9
]
1/2 [ cc 2 & - 3 - 2 & 4 ] * LHS = 1/2 [ cc 2 & - 3 - 2 & 4 ] * RHS
Multiply matrices
Multiply matrices
Now that the left-hand side is only the variable matrix, we can simplify the right-hand side to find the values of the variables.
Multiply matrices
Multiply
Add terms
Multiply matrix by 1/2
The desired a value is 152.
To find the inverse of a 2* 2 matrix, we use the corresponding formula.
| Matrix | Inverse |
|---|---|
| A= [ cc a & b c & d ] | A^(- 1)=1/ad-bc [ cc d & - b - c & a ] where ad-bc ≠0 |
The expression ad-bc is known as the determinant of a 2* 2 matrix. Because it is in the denominator of a fraction, if the determinant is zero, the matrix cannot have an inverse. Consider our coefficient matrix. [ cc 4 & 3 2 & 2 ] Let's calculate its determinant.
Since the determinant is not zero, the matrix has an inverse. We can now apply the formula for the inverse. Note that we usually refer to the determinant using the notation ad-bc=det(A).
Substitute values
Remove parentheses
For simplicity, instead of multiplying the matrix by the scalar, we will leave the expression as it is.
In our exercise, we had to multiply a 2* 2 matrix by a 2* 1 matrix. This is possible because the number of columns of the first matrix is equal to the number of rows of the second one. Moreover, the product has as many rows as the first matrix and as many columns as the second one. Thus, the product is a 2* 1 matrix. [ cc a & b c & d ] * [ c x y ] = [ c ax+by cx+dy ] 2 * 2 2 * 1 2 * 1 Let's multiply the matrices of the exercise.
Multiply matrices
Multiply
Add terms