2: Determinants and Inverses

Determinants

Consider row reducing the standard 2x2 matrix. Suppose that \(a\) is nonzero.

\[\begin{pmatrix} a &b \\ c &d \end{pmatrix}\nonumber \]

\[\frac{1}{a} R_1 \rightarrow R_1, \;\;\; R_2-cR_1 \rightarrow R_2\nonumber \]

\[\begin{pmatrix} 1 &\frac{b}{a} \\ c &d \end{pmatrix}\nonumber \]

\[\begin{pmatrix} 1 & \frac{b}{a} \\ 0 & d-\frac{cb}{a}\end{pmatrix}\nonumber \]

Now notice that we cannot make the lower right corner a 1 if

\[d - \frac{cb}{a} = 0\nonumber \]

or

\[ad - bc = 0.\nonumber \]

Definition: The Determinant

We call \(ad - bc\) the determinant of the 2 by 2 matrix

\[\begin{pmatrix} a &b \\ c &d \end{pmatrix}\nonumber \]

it tells us when it is possible to row reduce the matrix and find a solution to the linear system.

Example \(\PageIndex{1}\)

The determinant of the matrix

\[\begin{pmatrix} 3 & 1\\ 5 & 2 \end{pmatrix}\nonumber \]

is

\[3(2) - 1(5) = 6 - 5 = 1.\nonumber \]

Determinants of 3 x 3 Matrices

We define the determinant of a triangular matrix

\[\begin{pmatrix} a &d &e \\ 0 &b &f \\ 0 &0 &c \end{pmatrix}\nonumber \]

by

\[\text{det} = abc.\nonumber \]

Notice that if we multiply a row by a constant \(k\) then the new determinant is \(k\) times the old one. We list the effect of all three row operations below.

Theorem

The effect of the the three basic row operations on the determinant are as follows

1. Multiplication of a row by a constant multiplies the determinant by that constant.
2. Switching two rows changes the sign of the determinant.
3. Replacing one row by that row + a multiply of another row has no effect on the determinant.

To find the determinant of a matrix we use the operations to make the matrix triangular and then work backwards.

Example \(\PageIndex{2}\)

Find the determinant of

\[\begin{pmatrix} 2 & 6 &10 \\ 2 &4 &-3 \\ 0 &4 &2 \end{pmatrix}\nonumber \]

We use row operations until the matrix is triangular.

\[\dfrac{1}{2}R_1 \rightarrow R_1 \text{(Multiplies the determinant by } \dfrac{1}{2})\nonumber \]

\[\begin{pmatrix} 1 & 3 &5 \\ 2 &4 &-3 \\ 0 &4 &2 \end{pmatrix}\nonumber \]

\[R_2 - 2R_1 \rightarrow R_2 \text{ (No effect on the determinant)}\nonumber \]

\[\begin{pmatrix} 1 & 3 &5 \\ 0 &-2 &-13 \\ 0 &4 &2 \end{pmatrix}\nonumber \]

Note that we do not need to zero out the upper middle number. We only need to zero out the bottom left numbers.

\[R_3 + 2R_2 \rightarrow R_3 \text{ (No effect on the determinant)}.\nonumber \]

\[\begin{pmatrix} 1 & 3 &5 \\ 0 &-2 &-13 \\ 0 &0 &-24 \end{pmatrix}\nonumber \]

Note that we do not need to make the middle number a 1.

The determinant of this matrix is 48. Since this matrix has \(\frac{1}{2}\) the determinant of the original matrix, the determinant of the original matrix has

\[\text{determinant} = 48(2) = 96.\nonumber \]

Inverses

We call the square matrix I with all 1's down the diagonal and zeros everywhere else the identity matrix. It has the unique property that if \(A\) is a square matrix with the same dimensions then

\[AI = IA = A.\nonumber \]

Definition

If \(A\) is a square matrix then the inverse \(A^{-1}\) of \(A\) is the unique matrix such that

\[AA^{-1}=A^{-1}A=I.\nonumber \]

Example \(\PageIndex{3}\)

Let

\[A=\begin{pmatrix} 2 &5 \\ 1 &3 \end{pmatrix}\nonumber \]

then

\[A^{-1}= \begin{pmatrix} 3 &-5 \\ -1 &2 \end{pmatrix} \nonumber \]

Verify this!

Theorem: ExistEnce

The inverse of a matrix exists if and only if the determinant is nonzero.

To find the inverse of a matrix, we write a new extended matrix with the identity on the right. Then we completely row reduce, the resulting matrix on the right will be the inverse matrix.

Example \(\PageIndex{4}\)

\[\begin{pmatrix} 2 &-1 \\ 1 &-1 \end{pmatrix}\nonumber \]

First note that the determinant of this matrix is

\[-2 + 1 = -1\nonumber \]

hence the inverse exists. Now we set the augmented matrix as

\[\begin{pmatrix}\begin{array}{cc|cc}2&-1&1&0 \\1&-1&0&1\end{array}\end{pmatrix}\nonumber \]

\[R_1 {\leftrightarrow} R_2\nonumber \]

\[\begin{pmatrix}\begin{array}{cc|cc}1&-1&0&1 \\2&-1&1&0\end{array}\end{pmatrix}\nonumber \]

\[ R_2 - 2R_1 {\rightarrow} R_2\nonumber \]

\[\begin{pmatrix}\begin{array}{cc|cc}1&-1&0&1 \\0&1&1&-2\end{array}\end{pmatrix}\nonumber \]

\[ R_1 + R_2 {\rightarrow} R_1 \nonumber \]

\[\begin{pmatrix}\begin{array}{cc|cc}1&0&1&-1 \\0&1&1&-2\end{array}\end{pmatrix}\nonumber \]

Notice that the left hand part is now the identity. The right hand side is the inverse. Hence

\[A^{-1}= \begin{pmatrix} 1&-1 \\ 1&-2 \end{pmatrix} \nonumber \]

Solving Equations Using Matrices

Example \(\PageIndex{5}\)

Suppose we have the system

\[2x - y = 3\nonumber \]

\[ x - y = 4\nonumber \]

Then we can write this in matrix form

\[Ax = b\nonumber \]

where

\[A=\begin{pmatrix} 2&-1 \\ 1&-1 \end{pmatrix}, \;\;\; x= \begin{pmatrix} x \\ y \end{pmatrix}, \;\;\; \text{and} \; b=\begin{pmatrix} 3\\4 \end{pmatrix}\nonumber \]

We can multiply both sides by \(A^{-1}\):

\[A^{-1}A x = A^{-1}b\nonumber \]

or

\[x = A^{-1}b\nonumber \]

From before,

\[A^{-1}=\begin{pmatrix} 1&-1 \\ 1&-2 \end{pmatrix} \nonumber \]

Hence our solution is

\[\begin{pmatrix} -1&-5 \end{pmatrix}\nonumber \]

or

\[x = -1 \text{ and } y = 5\nonumber \]

Contributors

• Larry Green (Lake Tahoe Community College)

• Integrated by Justin Marshall.