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Honors: 8. Matrices
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United States of AmericaFL
Grade 912

Honors: 8.05 Applications of inverses and determinants

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Concept summary

A system of equations can be solved by constructing a matrix equation and solving this equation using the inverse matrices.

Matrix Equation

An equation in which a variable stands for a matrix. Solving such equations require matrix operations.

Example:

AX=B

Coefficient Matrix

A matrix containing the coefficients of the variables in a system of equations. Denoted by A in AX=B.

Example:

For the system \begin{cases} 2x-y &=3 \\ 7x-6y &=1 \end{cases} \\ A=\begin{bmatrix} 2 & -1 \\ 7 & -6 \end{bmatrix}

Variable Matrix

A column matrix containing the variables in a system of equations. Denoted by X in \\ AX=B.

Example:

For the system \begin{cases} 2x-y &=3 \\ 7x-6y &=1 \end{cases} \\ X=\begin{bmatrix} x \\ y \end{bmatrix}

Constant Matrix

A column matrix containing the constants in a system of equations. Denoted by B in \\ AX=B.

Example:

For the system \begin{cases} 2x-y &=3 \\ 7x-6y &=1 \end{cases} \\ B=\begin{bmatrix} 3 \\ 1 \end{bmatrix}

Solution Matrix

The matrix resulting from the multiplication A^{-1}B, required to solve the matrix equation \\ AX=B.

The system of equations: \begin{cases} ax+by &=m \\ cx+dy &=n \end{cases} can be written as the matrix equation:\begin{bmatrix} a & b \\ c & d \end{bmatrix}\begin{bmatrix} x\\ y \end{bmatrix}=\begin{bmatrix} m \\ n \end{bmatrix}

In general:

\displaystyle AX=B
\bm{A}
the coefficient matrix
\bm{X}
the variable matrix
\bm{B}
the constant matrix

To solve the matrix equation AX=B, we can use the solution matrix:

\displaystyle X=A^{-1}B
\bm{A^{-1}}
the inverse of the coefficient matrix
\bm{X}
the variable matrix
\bm{B}
the constant matrix
\bm{A^{-1}B}
the solution matrix

Since the inverse of the coefficient is required to find the solution, it must exist, so to have a solution we need: \det(A) \neq 0

We have two interesting applications of determinants.

The equations of a line that passes through two points \left(x_1,y_1\right) and \left(x_2, y_2\right) can be written as a determinant:

\displaystyle \begin{vmatrix} x & y &1\\ x_1 & y_1 &1\\x_2 & y_2 &1\end{vmatrix}=0
\bm{x_i}
The x-coordinate of a point on the line.
\bm{y_i}
The y-coordinate of a point on the line.

Given the vertices of a triangle \left(x_1,y_1\right),\left(x_2, y_2\right), and \left(x_3, y_3\right), we can find the area of the triangle using the formula involving the determinant, where the sign is chosen so that that area is positive.

\displaystyle \text{Area}= \pm \frac{1}{2}\begin{vmatrix} x_1 & y_1 &1\\ x_2 & y_2 &1\\x_3 & y_3 &1\end{vmatrix}
\bm{x_i}
The x-coordinate of a vertex
\bm{y_i}
The y-coordinate of a vertex

Worked examples

Example 1

Consider the following system of equations:

\begin{cases}-7x-y&=7\\9x+3y&=3\end{cases}

a

Write the coefficient, variable, and constant matrix as a matrix equation AX=B.

Approach

We need to find A, the coefficient matrix, X, the variable matrix and B the constant matrix from the system of equations, and then we can form the matrix equation.

Solution

\displaystyle A\displaystyle =\displaystyle \begin{bmatrix} -7 & -1 \\ 9 & 3 \end{bmatrix}From the coefficients of x and y in the system of equations
\displaystyle X\displaystyle =\displaystyle \begin{bmatrix} x \\ y \end{bmatrix}The variables in the system of equations
\displaystyle B\displaystyle =\displaystyle \begin{bmatrix} 7 \\ 3\end{bmatrix}The solutions in the system of equations
\displaystyle AX\displaystyle =\displaystyle BMatrix equation
\displaystyle \begin{bmatrix} -7 & -1 \\ 9 & 3 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix}\displaystyle =\displaystyle \begin{bmatrix} 7 \\ 3\end{bmatrix}The final matrix equation
b

Find A^{-1}B.

Approach

First we need to find A^{-1} by finding the determinant of A, \text{det}(A).

Solution

First calculate the determinant of A:

\displaystyle \text{det}(A)\displaystyle =\displaystyle \begin{vmatrix} -7 & -1 \\ 9 & 3 \end{vmatrix}
\displaystyle \text{}\displaystyle =\displaystyle -21+9
\displaystyle \text{}\displaystyle =\displaystyle -12

Next calculate the inverse of A

\displaystyle A^{-1}\displaystyle =\displaystyle \dfrac{1}{\text{det}(A)} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}
\displaystyle \text{}\displaystyle =\displaystyle -\dfrac{1}{12} \begin{bmatrix} 3 & 1 \\ -9 & -7 \end{bmatrix}
\displaystyle \text{}\displaystyle =\displaystyle \begin{bmatrix} -\dfrac{1}{4} & -\dfrac{1}{12} \\\\ \dfrac{3}{4} & \dfrac{7}{12} \end{bmatrix}

Finally calculate the solution matrix: X=A^{-1}B

\displaystyle A^{-1}B\displaystyle =\displaystyle \begin{bmatrix} -\dfrac{1}{4} & -\dfrac{1}{12} \\\\ \dfrac{3}{4} & \dfrac{7}{12} \end{bmatrix} \cdot B
\displaystyle \text{}\displaystyle =\displaystyle \begin{bmatrix} -\dfrac{1}{4} & -\dfrac{1}{12} \\\\ \dfrac{3}{4} & \dfrac{7}{12} \end{bmatrix} \cdot \begin{bmatrix} 7 \\ 3\end{bmatrix}
\displaystyle \text{}\displaystyle =\displaystyle \begin{bmatrix} -2 \\ 7 \end{bmatrix}

Reflection

Alternatively, you could multiply the matrices before multiplying by \dfrac{1}{\text{det}(A)} to avoid working with fractions:

\displaystyle A^{-1}B\displaystyle =\displaystyle -\dfrac{1}{12} \begin{bmatrix} 3 & 1 \\ -9 & -7 \end{bmatrix} \cdot B
\displaystyle \text{}\displaystyle =\displaystyle -\dfrac{1}{12} \begin{bmatrix} 3 & 1 \\ -9 & -7 \end{bmatrix} \cdot \begin{bmatrix} 7 \\ 3\end{bmatrix}
\displaystyle \text{}\displaystyle =\displaystyle -\dfrac{1}{12} \begin{bmatrix} 24 \\ -84 \end{bmatrix}
\displaystyle \text{}\displaystyle =\displaystyle \begin{bmatrix} -2 \\ 7 \end{bmatrix}

Example 2

Dave wants to start a new business. He borrowed a total of \$10\,000 from three different banks which offer different interest rates. One lending at 5\% interest, another at 8\% interest, and the third at 9\% interest. The total annual interest he had to pay on the three loans last year was \$770. The amount invested at 9\% was twice the amount invested at 5\%.

a

Write a system of equations that can be used to find how much Dave borrowed from each bank.

Approach

We want to find how much dave borrowed from three banks so there are 3 unknowns.

  • Let x represent the amount borrowed from the first bank.

  • Let y represent the amount borrowed from the second bank.

  • Let z represent the amount borrowed from the third bank.

This means we need to write 3 equations in terms of x, y and z. The 3 pieces of information we can use to form the equations are:

  1. The total amount borrowed was \$10\,000.

  2. The total interest he had do pay was \$770.

  3. The amount invested at 9\% was twice the amount invested at 5\%.

Solution

\displaystyle x+y+z\displaystyle =\displaystyle 10\,000From the total amount borrowed
\displaystyle 0.05x+0.08y+0.09z\displaystyle =\displaystyle 770From the interest
\displaystyle z\displaystyle =\displaystyle 2yFrom the third point
\displaystyle 2y-z\displaystyle =\displaystyle 0Rearranged

So the system of equations for this problem is given by: \begin{cases} x +y+x = 10\,000 \\ 0.05x+0.08y+0.09z = 770 \\ 2y-z = 0\end{cases}

b

Write the matrix to represent the system of equations in the form AX=B.

Approach

We need to find A, the coefficient matrix, X, the variable matrix and B the constant matrix from the system of equations: \begin{cases} x +y+x = 10\,000 \\ 0.05x+0.08y+0.09z = 770 \\ 2y-z = 0\end{cases} and then we can form the matrix equation.

Solution

\displaystyle A\displaystyle =\displaystyle \begin{bmatrix} 1 & 1 & 1 \\ 0.05 & 0.08 & 0.09\\ 2 & 0 & -1\end{bmatrix}Coefficient matrix
\displaystyle X\displaystyle =\displaystyle \begin{bmatrix} x \\ y \\z \end{bmatrix}Variable matrix
\displaystyle B\displaystyle =\displaystyle \begin{bmatrix} 10\,000 \\ 770\\0\end{bmatrix}constant matrix

Now we can write the matrix equation:

\displaystyle AX\displaystyle =\displaystyle B
\displaystyle \begin{bmatrix} 1 & 1 & 1 \\ 0.05 & 0.08 & 0.09\\ 2 & 0 & -1\end{bmatrix} \begin{bmatrix} x \\ y \\z \end{bmatrix}\displaystyle =\displaystyle \begin{bmatrix} 10\,000 \\ 770\\0\end{bmatrix}
c

Solve the matrix equation to find the amount Dave borrowed from each bank.

Approach

The inverse matrix method can be used to solve for the matrix variable X by multiplying A^{-1} on both sides of the equation to find X=A^{-1}B. But first we must find A^{-1} using technology.

Solution

\displaystyle A^{-1}\displaystyle =\displaystyle \begin{bmatrix} 8 & -100 & -1 \\ -23 & 300 & 4\\ 16 & -200 & -3\end{bmatrix}Found using technology
\displaystyle X\displaystyle =\displaystyle A^{-1}BFormula for X
\displaystyle \begin{bmatrix} x \\ y\\ z \end{bmatrix}\displaystyle =\displaystyle \begin{bmatrix} 8 & -100 & -1 \\ -23 & 300 & 4\\ 16 & -200 & -3\end{bmatrix} \cdot \begin{bmatrix} 10\,000 \\ 770\\0\end{bmatrix}
\displaystyle \text{}\displaystyle =\displaystyle \begin{bmatrix} 3000 \\ 1000\\ 6000 \end{bmatrix}

Therefore, Dave borrowed \$3000 at 5\% interest, \$1000 at 8\% interest, and \$6000 at 9\% interest.

Example 3

Find the equation of the line that passes through (-2, 0) and (5, 1) by using the equation: \begin{vmatrix} x & y &1\\ x_1 & y_1 &1\\x_2 & y_2 &1\end{vmatrix}=0

Approach

From the points (-2, 0) and (5, 1), we can see that x_1 = -2, y_1 = 0, x_2 = 5 and y_2=1. We can substitute these values into the given equation.

Solution

\displaystyle \begin{vmatrix} x & y &1\\ x_1 & y_1 &1\\x_2 & y_2 &1\end{vmatrix}\displaystyle =\displaystyle 0Given
\displaystyle \begin{vmatrix} x & y &1\\ -2 & 0 &1\\5 & 1 &1\end{vmatrix}\displaystyle =\displaystyle 0Substitute x_1 = -2, y_1 = 0, x_2 = 5, y_2=1
\displaystyle x \begin{vmatrix} 0 & 1 \\ 1 & 1 \end{vmatrix}-y\begin{vmatrix} -2 & 1 \\ 5 & 1 \end{vmatrix}+1\begin{vmatrix} -2 & 0 \\ 5 & 1 \end{vmatrix}\displaystyle =\displaystyle 0
\displaystyle -x-y(-7)+1(-2)\displaystyle =\displaystyle 0
\displaystyle -x+7y-2\displaystyle =\displaystyle 0
\displaystyle x-7y\displaystyle =\displaystyle -2Standard form

The equations of the line that passes through the given points is x-7y=-2

Outcomes

MA.912.NSO.4.2

Given a mathematical or real-world context, represent and solve a system of two- or three-variable linear equations using matrices.

MA.912.NSO.4.4

Solve mathematical and real-world problems using the inverse and determinant of matrices.

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