Matrices – Elementary Row Operations

Hi, and welcome to this video about matrix manipulation!

In this video, we’re going to look at:

• Elementary row operations
• Row echelon form of a matrix
• Reduced row echelon form of a matrix
• Finding the inverse of a matrix

Elementary Row Operations

Matrices of any dimension can be manipulated using three elementary row operations:

1. Any two rows can be swapped.
2. Any row can be multiplied by a non-zero scalar.
3. A multiple of a row can be added to another row.

The process of using row operations on a matrix is referred to as row reduction.

Row Echelon Form

The ultimate goal of row reduction is to end up with an upper triangular matrix—a matrix with all zero entries below the main diagonal. When this is achieved, the matrix is said to be in row echelon form.

Specifically, to be in row echelon form:

1. Rows with all zero entries must be at the bottom of the matrix
2. The first non-zero element of a row must be to the right of the first non-zero element in the row above.

Let’s see some row reduction in action! Keep in mind that the series of row operations is not unique. Also, the row echelon form of a matrix is not unique. Different people might use different series of operations to end up with different row echelon forms of the same matrix.

Here’s [X]:

$$X=\left[\begin{array}{ccc}2& 3& 5\\ 4& 7& 1\\ 3& 0& 9\end{array}\right]$$

My first operation will be to use the 2 to turn the 4 below it into a 0. I want to subtract 2 times row 1 from row 2 and replace row 2.

In shorthand, this looks like $R_2–2R_1=R_2$.

$$X=\left[\begin{array}{ccc}2& 3& 5\\ 4–2(2)& 7–2(3)& 1–2(5)\\ 3& 0& 9\end{array}\right]$$

Now, I want to use the 2 again to turn the 3 in the bottom left into a 0. To make this easier, I’ll use two steps:

1. $2R_3=R_3$

2. $R_3–3R_1=R_3$

$$X=\left[\begin{array}{ccc}2& 3& 5\\ 0& 1& -9\\ 6–3(2)& 0–3(3)& 18–3(5)\end{array}\right]$$

Now there’s only one more move to get this matrix to row echelon form. I need to use the 1 to make the negative 9 in row three into a 0. So $R_3+9R_2=R_3$.

This is one version of [X] in row echelon form:

$$X=\left[\begin{array}{ccc}2& 3& 5\\ 0& 1& -9\\ 0& 0& -78\end{array}\right]$$

Another version might look like this, if we had decided, for example, to divide row 3 by row 2:

$$X=\left[\begin{array}{ccc}2& 3& 5\\ 0& 1& -9\\ 0& 0& -39\end{array}\right]$$

As you can see, we ended with an upper triangular matrix. The first nonzero entry in each row is to the right of the nonzero entry in the row above. In other words, the 1 is to the right of the 2 and the negative 78 is to the right of the 1.

Reduced Row Echelon Form

We could go further and get [X] into reduced row echelon form. Reduced row echelon form can be obtained using different sequences of row operations, but this form is unique.

A matrix has only one reduced row echelon form. In order to be in reduced row echelon form, a matrix must meet these conditions:

• Be in row echelon form.
• The leading entry in each nonzero row must be a 1.
• Wherever there is a leading 1, the entries in the rest of the column must be zeros.

Let’s pick up with [X] where we left off:

$$X=\left[\begin{array}{ccc}2& 3& 5\\ 0& 1& -9\\ 0& 0& -78\end{array}\right]$$

First, I need to turn that 2 into a 1: $\frac{1}{2}{R}_1=R_1$.

$$X=\left[\begin{array}{ccc}\frac{1}{2}(2)& \frac{1}{2}(3)& \frac{1}{2}(5)\\ 0& 1& -9\\ 0& 0& -78\end{array}\right]$$

The first column looks good. Now to column 2. I need to use the 1 to turn the three halves above it into a 0: $R_1–\frac{3}{2}{R}_2=R_1$.

Now columns 1 and 2 look good. Time to turn that -78 into a 1: $\frac{-1}{78}{R}_3=R_3$.

Row 3 looks good. All that’s left is to use row 3 to change the 16 and -9 to zeroes: $-16R_3+R_1=R_1$ and $R_2+9R_3=R_2$.

$$X=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$$

In this case, the reduced row echelon form of the matrix is $I_3$. With square matrices, this will happen frequently, but it doesn’t need to happen.

These matrices are in reduced row echelon form as well:

$$\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right]$$

$$\left[\begin{array}{ccc}1& 0& 2\\ 0& 1& 3\end{array}\right]$$

$$\left[\begin{array}{ccccc}1& 3& 0& 4& 0\\ 0& 0& 1& 6& 0\\ 0& 0& 0& 1& 1\end{array}\right]$$

Inverse Matrices

Some matrices have inverse matrices, denoted by the superscript -1. Inverse matrices satisfy the equation $A{A}^{-1}=A^{-1}A=I$.

For instance, here we have [B] and the [B^-1]:

$$B=\left[\begin{array}{cc}4& 3\\ 5& 4\end{array}\right]$$

$$B^{-1}=\left[\begin{array}{cc}4& -3\\ -5& 4\end{array}\right]$$

Let’s try [BB^-1]. To do this, we do $4\times 4=16$, then $3\times -5=-15$. For our next element, we’ll do $4\times -3=-12$, then $3\times 4=12$. Then, we’re going to do $5\times 4=20$, then $4\times -5=-20$. Then, we’ll do $5\times -3=-15$, then $4\times 4=16$.

$$B{B}^{-1}=\left[\begin{array}{cc}16+(-15)& (-12)+12\\ 20+(-20)& (-15)+16\end{array}\right]$$

If we simplify this, we get $16+(-15)=1$, $(-12)+12=0$, $20+(-20)=0$, and $(-15)+16=1$.

$$B{B}^{-1}=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$$

This is our identity matrix for a 2×2 matrix. If you go ahead and multiply [B times [B], you’ll get the same thing.

In order for a matrix to have an inverse, or to be invertible, two conditions must be met:

1. The matrix must be a square matrix
2. The determinant of the matrix must not be 0.

Let’s work with a 3×3 matrix:

$$Y=\left[\begin{array}{ccc}2& 3& -2\\ -4& 4& 1\\ 3& 0& -1\end{array}\right]$$

Is [Y] a square matrix? Check! Is the determinant of [Y] not equal to 0? Let’s see.

 
Yes, the determinant of [Y] does not equal 0. So now let’s figure out what the inverse of [Y] is.

Here are the two steps to finding the inverse of a matrix:

1. Augment the matrix with the identity matrix of the appropriate size.
   (Y augment I looks like this: [Y|I])
2. Row reduce the augmented matrix. When it is in reduced row echelon form, the structure will be [I|Y^-1] and the inverse can be identified.

Here’s what that process looks like:

$$\left[\begin{array}{cccccc}2& 3& -2& 1& 0& 0\\ -4& 4& 1& 0& 1& 0\\ 3& 0& -1& 0& 0& 1\end{array}\right]$$

First, let’s turn the 2 into a 1: $\frac{1}{2}{R}_1=R_1$.

Next, let’s get zeroes below the 1 in column 1: $R_2+4R_1=R_2$ and $R_3–3R_1=R_3$.

Now we need to make the 10 in column 2 into a 1: $\frac{1}{10}{R}_2=R_2$.

The rest of the entries in column 2 need to be 0: $R_1–\frac{3}{2}{R}_2=R_1$ and $R_3+\frac{9}{2}{R}_2=R_3$.

Now the 13 over 20 in column 3 needs to change to 1: $\frac{20}{13}{R}_3=R_3$.

Lastly, the two remaining entries in column 3 need to be 0: $R_1+\frac{11}{20}{R}_3=R_1$ and $R_2+\frac{3}{10}{R}_3=R_2$.

Now we have an augmented matrix in the form $[I|Y^{-1}$].

$$I|Y^{-1}=\left[\begin{array}{cccccc}1& 0& 0& -\frac{4}{13}& \frac{3}{13}& \frac{11}{13}\\ 0& 1& 0& -\frac{1}{13}& \frac{4}{13}& \frac{6}{13}\\ 0& 0& 1& -\frac{12}{13}& \frac{9}{13}& \frac{20}{13}\end{array}\right]$$

Thanks for watching, and happy studying!

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