]> Chapter 04 Notes

4: Matrices

4.1: Organizing Data into Matrices

A matrix is just a grid of numbers. That's all. However, matrices can be quite useful…

The size of a matrix is given as $r \times c$ , where the matrix has r rows and c columns. Below are two images of a $3 \times 4$ matrix.

An Image

Sometimes, subscripts are used to identify the individual elements (cells) in a matrix. Note that the first digit of the subscript is the row, and the second digit is the column.

$A = [\begin{matrix} a_{11} & a_{12} & a_{13} & a_{14} \\ a_{21} & a_{22} & a_{23} & a_{24} \\ a_{31} & a_{32} & a_{33} & a_{34} \end{matrix}]$

To refer to a matrix without listing its elements, an italicized capital roman letter is typically used (like the A in the example above). To refer to the elements of a matrix in general, a subscripted lowercase italicized roman letter (the same as the matrix) is used (like $a_{23}$ in the example above).

4.2: Adding and Subtracting Matrices

Adding and Subtracting Matrices

Unlike the real numbers, not all matrices can be added or subtracted. In particular, only matrices of the same size (same number of rows; same number of columns) can be added or subtracted.

To add two matrices, add elements in the same position in the two matrices (i.e., The element in row i and column j of $A + B$ is $a_{i j} + b_{i j}$ ).

Do exactly the same thing to subtract two matrices!

Like the real numbers, addition of matrices is commutative—the order of addition makes no difference.

Solving Matrix Equations

If you have an equation made up of matrices—like $X + A = B$ —and you want to solve for matrix X, then what should you do?

Subtract matrix A, of course! $X + A = B ⇒ X + A - A = B - A ⇒ X = B - A$ .

Examples

[1.] Add $[\begin{array}{r} 0 & 4 \\ 7 & 0 \\ 3 & 1 \end{array}] + [\begin{array}{r} 1 & 2 \\ 2 & 3 \\ 0 & 4 \end{array}]$ .

$[\begin{array}{r} 0 & 4 \\ 7 & 0 \\ 3 & 1 \end{array}] + [\begin{array}{r} 1 & 2 \\ 2 & 3 \\ 0 & 4 \end{array}] = [\begin{array}{r} 0 + 1 & 4 + 2 \\ 7 + 2 & 0 + 3 \\ 3 + 0 & 1 + 4 \end{array}] = [\begin{array}{r} 1 & 6 \\ 9 & 3 \\ 3 & 5 \end{array}]$

[2.] Subtract $[\begin{array}{r} 1 & 6 \\ 9 & 3 \end{array}] - [\begin{array}{r} 2 & 6 \\ 9 & 1 \end{array}]$ .

$[\begin{array}{r} 1 & 6 \\ 9 & 3 \end{array}] - [\begin{array}{r} 2 & 6 \\ 9 & 1 \end{array}] = [\begin{array}{r} 1 -2 & 6 -6 \\ 9 -9 & 3 -1 \end{array}] = [\begin{array}{r} -1 & 0 \\ 0 & -2 \end{array}]$

[3.] Solve for X: $X + [\begin{array}{r} 1 & 6 \\ 9 & 3 \\ 6 & 0 \end{array}] = [\begin{array}{r} 0 & -1 \\ -1 & 2 \\ 3 & 3 \end{array}]$ .

$X + [\begin{array}{r} 1 & 6 \\ 9 & 3 \\ 6 & 0 \end{array}] = [\begin{array}{r} 0 & -1 \\ -1 & 2 \\ 3 & 3 \end{array}] ⇒ X = [\begin{array}{r} 0 & -1 \\ -1 & 2 \\ 3 & 3 \end{array}] - [\begin{array}{r} 1 & 6 \\ 9 & 3 \\ 6 & 0 \end{array}] = [\begin{array}{r} -1 & -7 \\ -10 & -1 \\ -3 & 3 \end{array}]$

4.3: Matrix Multiplication

Scalar Multiplication

A scalar is a regular number. Scalar multiplication means multiplying a scalar against a matrix. To do this, simply multiply each element of the matrix by the scalar.

Matrix Multiplication

Matrix multiplication is one matrix times another. However, not all matrices can be multiplied…and just in case that wasn't fun enough, matrix multiplication is not commutative—the order of the matrices matters!

Two matrices can be multiplied if the number of columns in the left hand matrix equals the number of rows in the right hand matrix. The resulting matrix will have the same number of rows as the left hand matrix and the same number of columns as the right hand matrix.

An Image

The procedure for actually multiplying the matrices is involved…basically, you will multiply every element in a left hand row with every (corresponding) element in a right hand column, then add. For example:

$[\begin{array}{r} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array}] \times [\begin{array}{r} b_{11} & b_{12} & b_{13} \\ b_{21} & b_{22} & b_{23} \end{array}] = [\begin{array}{r} a_{11} b_{11} + a_{12} b_{21} & a_{11} b_{12} + a_{12} b_{22} & a_{11} b_{13} + a_{12} b_{23} \\ a_{21} b_{11} + a_{22} b_{21} & a_{21} b_{12} + a_{22} b_{22} & a_{21} b_{13} + a_{22} b_{23} \end{array}]$

This will require some practice…

Examples

[4.] Multiply: $(-2) [\begin{array}{r} 1 & 6 \\ 9 & 3 \\ 6 & 0 \end{array}]$ .

$(-2) [\begin{array}{r} 1 & 6 \\ 9 & 3 \\ 6 & 0 \end{array}] = [\begin{array}{r} -2 & -12 \\ -18 & -6 \\ -12 & 0 \end{array}]$

[5.] Multiply: $[\begin{array}{r} 1 & 6 \\ 9 & 3 \end{array}] [\begin{array}{r} 0 & -1 \\ -1 & 2 \end{array}]$ .

Here's a way to arrange the matrices to make multiplying a little easier (it lines up the required rows and columns for each element of the answer):

An Image

The answer is $[\begin{array}{r} (1) (0) + (6) (-1) & (1) (-1) + (6) (2) \\ (9) (0) + (3) (-1) & (9) (-1) + (3) (2) \end{array}] = [\begin{array}{r} -6 & 11 \\ -3 & -3 \end{array}]$ .

[6.] Multiply $[\begin{array}{r} 1 & -2 & 2 \\ 1 & 0 & 2 \end{array}] [\begin{array}{r} 1 & 0 & 1 \\ 3 & -1 & 5 \\ 4 & 2 & 0 \end{array}]$ .

The arrangement I suggested above may be of some help here!

The answer:

$[\begin{array}{r} (1) (1) + (-2) (3) + (2) (4) & (1) (0) + (-2) (-1) + (2) (2) & (1) (1) + (-2) (5) + (2) (0) \\ (1) (1) + (0) (3) + (2) (4) & (1) (0) + (0) (-1) + (2) (2) & (1) (1) + (0) (5) + (2) (0) \end{array}] = [\begin{array}{r} 3 & 6 & -9 \\ 9 & 4 & 1 \end{array}]$

4.5: 2x2 Matrices, Determinants and Inverses

The Identity and Inverses

A matrix with an equal number of rows and columns is called a square matrix.

The real numbers have two identities—additive (0) and multiplicative (1). Square matrices have the same things. The zero matrix is $0 = [\begin{array}{r} 0 & 0 \\ 0 & 0 \end{array}]$ . Actually, the symbol 0 is used regardless of the size of the matrix, so this is also the zero matrix: $0 = [\begin{array}{r} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}]$ . The zero matrix is the additive identity—adding it results in no change. Also, adding a matrix with its opposite (negation) results in the zero matrix.

The identity matrix is a square matrix with 1's along the left-right diagonal (the main diagonal), and zeros everywhere else. The 2×2 identity matrix is $I_{2} = [\begin{array}{r} 1 & 0 \\ 0 & 1 \end{array}]$ . This is a multiplicative identity—multiplying by this matrix results in no change. Also, due to its nature, it actually doesn't matter which side you multiply on when using the identity matrix.

The inverse of a matrix is another matrix that, when multiplied, results in the (multiplicative) identity: the inverse of A is $A^{-1}$ , so that $A \times A^{-1} = A^{-1} \times A = I$ . In the real numbers, we call this the reciprocal…however, that's the only place that we'll use that term…especially since every real number has an reciprocal, but not every matrix has an inverse.

Finding the inverse of a matrix requires that we learn about…

The Determinant

This is a special number that describes a matrix. Not every matrix has one!

Given a matrix $A = [\begin{array}{r} a & b \\ c & d \end{array}]$ , the determinant of A is $det (A) = | A | = a d - b c$ .

If the determinant is non-zero, then the matrix has an inverse, and is called non-singular (if the determinant is zero, then the matrix is singular).

Given a non-singular matrix $A = [\begin{array}{r} a & b \\ c & d \end{array}]$ , the inverse of the matrix is $A^{-1} = \frac{1}{| A |} [\begin{array}{r} d & - b \\ - c & a \end{array}]$ .

Solving Matrix Equations

When last we solved matrix equations, we could only handle matrices that were added or subtracted. Now, we can handle multiplication!

(there is no matrix division)

Specifically, we can solve equations of the form $A X = B$ .

How can we get that A off of X and solve? Use the inverse!

$A X = B ⇒ A^{-1} A X = A^{-1} B ⇒ I X = A^{-1} B ⇒ X = A^{-1} B$

Of course, this will require that the inverse exists…

Examples

[7.] Find $| \begin{array}{r} 1 & -1 \\ -2 & 2 \end{array} |$ .

It is $(1) (2) - (-1) (-2) = 2 -2 = 0$ .

[8.] Find the inverse of the matrix $M = [\begin{array}{r} 1 & -1 \\ 1 & 2 \end{array}]$ .

First, the determinant: $(1) (2) - (-1) (1) = 2 + 1 = 3$ . Thus, $M^{-1} = \frac{1}{3} [\begin{array}{r} 2 & 1 \\ -1 & 1 \end{array}] = [\begin{array}{r} \frac{2}{3} & \frac{1}{3} \\ - \frac{1}{3} & \frac{1}{3} \end{array}]$ .

[9.] Solve $[\begin{array}{r} 6 & 4 \\ 6 & 1 \end{array}] [\begin{array}{r} x \\ y \end{array}] = [\begin{array}{r} 14 \\ 8 \end{array}]$ .

Isolate the variables: $[\begin{array}{r} 6 & 4 \\ 6 & 1 \end{array}] [\begin{array}{r} x \\ y \end{array}] = [\begin{array}{r} 14 \\ 8 \end{array}] ⇒ [\begin{array}{r} x \\ y \end{array}] = {[\begin{array}{r} 6 & 4 \\ 6 & 1 \end{array}]}^{-1} [\begin{array}{r} 14 \\ 8 \end{array}]$ . I'll let my technology find the inverse for me. $[\begin{array}{r} x \\ y \end{array}] = [\begin{array}{r} 14 \\ 8 \end{array}] {[\begin{array}{r} 6 & 4 \\ 6 & 1 \end{array}]}^{-1} = [\begin{array}{r} - \frac{1}{18} & \frac{2}{9} \\ \frac{1}{3} & - \frac{1}{3} \end{array}] [\begin{array}{r} 14 \\ 8 \end{array}]$ . Finally, multiply: $[\begin{array}{r} x \\ y \end{array}] = [\begin{array}{r} - \frac{1}{18} & \frac{2}{9} \\ \frac{1}{3} & - \frac{1}{3} \end{array}] [\begin{array}{r} 14 \\ 8 \end{array}] = [\begin{array}{r} 1 \\ 2 \end{array}]$ . Thus, the solution is $x = 1, y = 2$ .

4.6: 3x3 Matrices, Determinants and Inverses

The Determinant

First, note that the multiplicative identity is now $I_{3} = [\begin{array}{r} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}]$ .

The determinant of a 3×3 matrix is a little more complicated that that of a 2×2 . There is a fine diagram on page 202 of the text that I'll not attempt to duplicate here.

Inverses

You'll note that the book doesn't even attempt to show how to find the inverse of a 3×3 matrix by hand…so neither will I (yet). Just use your calculator…of course, I'll be giving instructions on how to do this in class (unless you actually read the manual for your calculator…).

Solving Matrix Equations

…is exactly the same.

Examples

[10.] Find $| \begin{array}{r} 1 & 5 & -2 \\ -1 & 3 & 0 \\ 6 & 2 & 1 \end{array} |$ .

My technology tells me that this is 48.

[11.] Find ${[\begin{array}{r} 1 & 1 & 0 \\ 2 & 3 & 1 \\ 1 & 2 & 2 \end{array}]}^{-1}$ .

Again, my technology tells me that the answer is $[\begin{array}{r} 4 & -2 & 1 \\ -3 & 2 & -1 \\ 1 & -1 & 1 \end{array}]$ .

[12.] Solve $[\begin{array}{r} -1 & 3 & 1 \\ 2 & 5 & 0 \\ 3 & 1 & -2 \end{array}] [\begin{array}{r} x \\ y \\ z \end{array}] = [\begin{array}{r} 1 \\ 3 \\ -2 \end{array}]$ .

$[\begin{array}{r} -1 & 3 & 1 \\ 2 & 5 & 0 \\ 3 & 1 & -2 \end{array}] [\begin{array}{r} x \\ y \\ z \end{array}] = [\begin{array}{r} 1 \\ 3 \\ -2 \end{array}] ⇒ [\begin{array}{r} x \\ y \\ z \end{array}] = {[\begin{array}{r} -1 & 3 & 1 \\ 2 & 5 & 0 \\ 3 & 1 & -2 \end{array}]}^{-1} [\begin{array}{r} 1 \\ 3 \\ -2 \end{array}] = [\begin{array}{r} - \frac{10}{9} & \frac{7}{9} & - \frac{5}{9} \\ \frac{4}{9} & - \frac{1}{9} & \frac{2}{9} \\ - \frac{13}{9} & \frac{10}{9} & - \frac{11}{9} \end{array}] [\begin{array}{r} 1 \\ 3 \\ -2 \end{array}] = [\begin{array}{r} \frac{7}{3} \\ - \frac{1}{3} \\ \frac{13}{3} \end{array}]$

4.7: Inverse Matrices and Systems

Matrices can be used to solve systems of equations—you just have to know how to translate from the equations into matrices.

First, write your equations in standard form. This will make three matrices—the coefficient matrix (A), the variable matrix (X), and the constant matrix (B). Use our previously learned techniques (and your calculator) to solve.

Examples

[13.] Solve ${\begin{array}{r} 2 x - y + z = 3 \\ x + 2 y - z = 2 \\ - x - y + 2 z = 3 \end{array}$ .

The three matrices are $A = [\begin{array}{r} 2 & -1 & 1 \\ 1 & 2 & -1 \\ -1 & -1 & 2 \end{array}]$ , $X = [\begin{array}{r} x \\ y \\ z \end{array}]$ and $B = [\begin{array}{r} 3 \\ 2 \\ 3 \end{array}]$ .

The matrix equation is $[\begin{array}{r} 2 & -1 & 1 \\ 1 & 2 & -1 \\ -1 & -1 & 2 \end{array}] [\begin{array}{r} x \\ y \\ z \end{array}] = [\begin{array}{r} 3 \\ 2 \\ 3 \end{array}]$ .

The solution is $[\begin{array}{r} x \\ y \\ z \end{array}] = {[\begin{array}{r} 2 & -1 & 1 \\ 1 & 2 & -1 \\ -1 & -1 & 2 \end{array}]}^{-1} [\begin{array}{r} 3 \\ 2 \\ 3 \end{array}] = [\begin{array}{r} \frac{3}{8} & \frac{1}{8} & - \frac{1}{8} \\ - \frac{1}{8} & \frac{5}{8} & \frac{3}{8} \\ \frac{1}{8} & \frac{3}{8} & \frac{5}{8} \end{array}] [\begin{array}{r} 3 \\ 2 \\ 3 \end{array}] = [\begin{array}{r} 1 \\ 2 \\ 3 \end{array}]$ .

4.8: Augmented Matrices and Systems

Cramer's Rule

Let's say that we have a matrix equation $A X = B$ …note that the size of A will be n×n, and B will be n×1. There is another way to solve for the n variables of the system!

Let matrix D₁ be the same as matrix A, but replace the first column with the elements from matrix B. Let matrix D₂ be the same as A, but replace the second column with the elements from matrix B. Keep this up until you've created n of these matrices (up to D_n).

The solution for the first variable is $\frac{| D_{1} |}{| A |}$ ; the solution for the second variable is $\frac{| D_{2} |}{| A |}$ ; and so on, up to the solution for the n^th variable $\frac{| D_{n} |}{| A |}$ .

Gaussian Elimination

There is another way to use the elimination method—one that removes all of the variables, and just manipulates the coefficients and constants. The method is called Gaussian Elimination.

Start with a matrix equation, as before: $A X = B$ . From this, we will create an augmented matrix by tacking matrix B on as an extra column in matrix A. Typically, we draw a dashed vertical line after the last column of A and before matrix B.

There are three rules (operations) in this method.

Switch any two rows.
Multiply a row by a constant.
Add a multiple of any row into another row.

There are two possible goals of this method.

Row Echelon Form: ones along the main diagonal, and zeros below (only).
Reduced Row Echelon Form: ones along the main diagonal, and zeros above and below. In other words, where matrix A had been, there is now the identity matrix I.

When you have row echelon form, then you have solved for the final variable. This can be plugged back in to find the value of the other variables.

However, when you have reduced row echelon form, then you have solved for all variables!

The way to work efficiently is to create a one in the upper left corner, then use that one to make zeros underneath. Next, make a one in row two column 2. Use that one to make zeros above and below. Keep this up until you're done.

If you end up with a row of all zeros, then there are an infinite number of solutions to the system. If you end up with a row that contains all zeros except for the last column, then there is no solution to the system.

There is an extension to this, called Gauss-Jordan Elimination. In this method, augment matrix A with the identity I. Now use the row operations to achieve reduced row echelon form. Where matrix I had been, there is now $A^{-1}$ !

Examples

[14.] Convert this to reduced row echelon form: $[\begin{array}{r} 1 & 3 | & 1 \\ -1 & 1 | & 3 \end{array}]$ .

There is already a one in the upper left corner. To make a zero underneath, I'll do $(1) R_{1} + R_{2}$ (add row 1 into row 2). The result is $[\begin{array}{r} 1 & 3 | & 1 \\ 0 & 4 | & 4 \end{array}]$ . Now do $(\frac{1}{4}) R_{2}$ (multiply row 2 by $\frac{1}{4}$ ). The result is $[\begin{array}{r} 1 & 3 | & 1 \\ 0 & 1 | & 1 \end{array}]$ . Finally, do $(-3) R_{2} + R_{1}$ (negative three times row two added into row 1). From that we get $[\begin{array}{r} 1 & 0 | & -2 \\ 0 & 1 | & 1 \end{array}]$ . The solution is $x = -2, y = 1$ .

[15.] Solve ${\begin{array}{r} x + 2 y = 1 \\ -2 x + 3 y = 5 \end{array}$ by Gaussian elimination.

The augmented matrix is $[\begin{array}{r} 1 & 2 | & 1 \\ -2 & 3 | & 5 \end{array}]$ . $(2) R_{1} + R_{2}$ gives $[\begin{array}{r} 1 & 2 | & 1 \\ 0 & 7 | & 7 \end{array}]$ . Next, $(\frac{1}{7}) R_{2}$ gives $[\begin{array}{r} 1 & 2 | & 1 \\ 0 & 1 | & 1 \end{array}]$ . Finally, $(-2) R_{2} + R_{1}$ gives $[\begin{array}{r} 1 & 0 | & -1 \\ 0 & 1 | & 1 \end{array}]$ . The solution is $x = -1, y = 1$ .

[16.] Solve ${\begin{array}{r} x + 3 z = 2 \\ - x + y + 5 z = -4 \\ y + 3 z = 0 \end{array}$ by Gaussian elimination.

The augmented matrix is $[\begin{array}{r} 1 & 0 & 3 | & 2 \\ -1 & 1 & 5 | & -4 \\ 0 & 1 & 3 | & 0 \end{array}]$ .

$(1) R_{1} + R_{2} ⇒ [\begin{array}{r} 1 & 0 & 3 | & 2 \\ 0 & 1 & 8 | & -2 \\ 0 & 1 & 3 | & 0 \end{array}]$

$(-1) R_{2} + R_{3} ⇒ [\begin{array}{r} 1 & 0 & 3 | & 2 \\ 0 & 1 & 8 | & -2 \\ 0 & 0 & -5 | & 2 \end{array}]$

$(- \frac{1}{5}) R_{3} ⇒ [\begin{array}{r} 1 & 0 & 3 | & 2 \\ 0 & 1 & 8 | & -2 \\ 0 & 0 & 1 | & - \frac{2}{5} \end{array}]$

$(-8) R_{3} + R_{2} ⇒ [\begin{array}{r} 1 & 0 & 3 | & 2 \\ 0 & 1 & 0 | & \frac{6}{5} \\ 0 & 0 & 1 | & - \frac{2}{5} \end{array}]$

$(-3) R_{3} + R_{1} ⇒ [\begin{array}{r} 1 & 0 & 0 | & \frac{16}{5} \\ 0 & 1 & 0 | & \frac{6}{5} \\ 0 & 0 & 1 | & - \frac{2}{5} \end{array}]$

The solution is $x = \frac{16}{5}, y = \frac{6}{5}, z = - \frac{2}{5}$ .

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