![]() Chapter 9: Matrices and DeterminantsStudy Guide |
Section 9.1
Hints:
1. A rectangular array of numbers arranged in rows and
columns and placed in brackets is called a matrix. The numbers
inside the matrix are called elements. When you give the dimensions
of a matrix, you are giving the number of its rows and columns. The
number of rows is always listed first.
Ex.:
The matrix in the example has 2 rows and 3 columns. You would say it is
a
matrix.
2. An augmented matrix can be used to write a system of equations. You may think of it as a kind of shorthand notation. In an augmented matrix a vertical bar separates the columns of the matrix into two groups. The coefficients of the system are placed to the left of the bar, and the constants are placed to the right of the bar.
Ex.: 
3. A matrix with 1s down the diagonal from upper left to lower right and 0s below the 1s is said to be in row-echelon form. If you can change a system's augmented matrix into row-echelon form, then you will have the solution to the system. Below is an example of a row-echelon matrix.
Ex.:
In order to turn an augmented matrix into a row-echelon form, you should use row operations. Row operations are given on page 771.
4. The process of using row operations to solve linear systems is called Gaussian elimination. The steps for using Gaussian elimination are on page 773. Remember the numbers to the left of the bar are the coefficients of the variables! The final steps after getting the augmented matrix in row-echelon form are to use back substitution to solve for the remaining variables. For example, the row-echelon matrix on the right below can be converted into the system given on the right and then solved by back substitution.
Ex.:

5. The method called Gauss-Jordan elimination just carries Gaussian Eeimination a little farther. When you use Gauss-Jordan elimination, you use row operations to obtain a matrix to the left of the bar with 1s down the diagonal from upper left to lower right and 0s in every position above and below each 0. When you have done this, the solution to the system will be the column to the right of the bar.
Ex.:
The solution to the system given by this matrix is (-1, 3, 2).
The steps for Gauss-Jordan elimination are on page 778.
| Additional Exercises: | |
1.![]() |
Q: Solve the system using Gaussian elimination. |
2.![]() |
Q: Solve the system using Gauss-Jordan elimination.
|
Section 9.2
Hints:
1. Linear systems can have one solution,
no solution, or infinitely many solutions. Gaussian elimination
can be used to determine the number of solutions for systems with three or more
equations.
2. When using Gaussian elimination, if the bottom row to the left of the line becomes all 0s and the bottom row to the right of the line is a number that is NOT zero, then the system has no solution.
Ex.: 
3. When using Gaussian elimination, if the bottom row to the left and right of the line becomes all 0s, then the system has infinitely many solutions.
Ex.:
In a system with three equations and three unknowns and infinitely many solutions, it is possible to find an algebraic expression for x and y in terms of z. In order to do this, you should first drop the bottom row of zeros; then, take the remaining two rows and convert them back into a system. This system can then be solved for x and y in terms of z.
Section 9.3
Hints:
is in row 3 and column two of the matrix.
4. Properties of matrix addition are similar to the properties of with adding real numbers. These properties are given on page 795.
5. Any matrix can be multiplied by a real number. These real numbers are called scalars. When multiplying by a scalar, you multiply every element in the matrix by the scalar.
6. When multiplying matrices, you do not multiply corresponding elements. Instead, you multiply rows by columns. For the product of two matrices to be defined, the number of columns of the first matrix must equal the number of rows of the second matrix. The order of the answer matrix is the number of rows in the first matrix by the number of columns in the second matrix.
Ex. Order--First Matrix Order--Second Matrix Order--Answer Matrix
2 x
3
3 x
4
2 (rows from 1st) x 4
(columns from 2nd)
Notice that the columns in the first matrix match the rows in the second matrix. Since this is the case, these two matrices can be multiplied together. There are 2 rows in the first matrix and 4 columns in the second matrix; therefore, the answer matrix will be 2 x 4.
7. Remember! Matrix multiplication is not commutative. AB is not necessarily the same thing as BA.
| Additional Exercises: | |
1.![]() |
Q: Find 3A - 4B. |
2.![]() |
Q: Find AB. |
3.![]() |
Q: Find A(B - C). |
Section 9.4
Hints:
1. A n x n square matrix whose main
diagonal elements are 1s, while all other elements are 0s, is called the
multiplicative identity matrix of order n. The symbol used to
designate this multiplicative identity matrix is
.
Ex.:
2. Some matrices have a multiplicative inverse. When
you multiply a matrix by its multiplicative inverse, you get an identity
matrix,
.
This is similar to getting a 1 when you multiply a real number by its
reciprocal. The symbol used for the multiplicative inverse of matrix A is
.
A nonsquare matrix cannot have a multiplicative
inverse.
3. The process for finding the multiplicative inverse of a 2 x 2 matrix is
If
, then
.
The matrix is invertible if and only if ad - bc does not equal zero. If you look at the process carefully, you will see that the first factor requires you to multiply along the diagonals of the matrix and subtract. To find the second factor, which is a matrix, reverse a and d and then negate b and c.
4. The procedure for finding the multiplicative inverse of any invertible matrix is on page 813. This procedure requires you to form an augmented matrix and then perform row operations.
5. You may solve a system using a multiplicative inverse.
If
,
where A is the coefficient matrix and
B is the constant matrix, has a
unique solution,
.
To solve the linear system of equations, you multiply
and B to find X.
Additional Exercises:
1.

Q: Find
.
2.

Q: Find
.

3.

Q: Write the linear system as a matrix equation and then solve
the system using the inverse of the coefficient matrix.

Section 9.5
Hints:
is found by computing
.
represents its determinant.
, then

, then
.
Each third order determinant is found by expansion by minors.
6. When using Cramer's Rule, if D = 0 and at least one of the determinants in the numerator is not 0, then the system is inconsistent. The solution is the empty set. If D = 0 and all the determinants in the numerators are 0, then the equations in the system are dependent. The system has infinitely many solutions. This hold true regardless of the size of the system.