Search in book...
Toggle Font Controls
Create new playlist

Name your new playlist

Playlist description (optional)
Sign In

Email address

Password

Forgot Password?

or

Continue with Facebook

Continue with Google
Sign Up

Full Name

Email address

Confirm Email Address

Password

or

Continue with Facebook

Continue with Google

M5.3 Finding the Inverse of a Matrix

The inverse of a matrix is a unique matrix of the same dimensions that, when multiplied by the original matrix, produces a unit or identity matrix. For example, if A is any $2 \times 2$ $2 \times 2$ matrix and its inverse is denoted $A^{- 1},$ $A^{- 1},$ then

A \times A^{- 1} = (\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}) = Identity matrix

$A \times A^{- 1} = (\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}) = Identity matrix$ (M5-3)

The adjoint of a matrix is extremely helpful in forming the inverse of the original matrix. We simply compute the value of the determinant of the original matrix and divide each term of the adjoint by this value.

To find the inverse of the matrix just presented, we need to know the adjoint (already computed) and the value of the determinant of the original matrix:

(\begin{matrix} 3 & 7 & 5 \\ 2 & 0 & 3 \\ 4 & 1 & 8 \end{matrix}) = Original matrix

$(\begin{matrix} 3 & 7 & 5 \\ 2 & 0 & 3 \\ 4 & 1 & 8 \end{matrix}) = Original matrix$

Value of determinant:

An image shows the primary and secondary diagonals of the matrix and the calculation shows the value of the determinant.

5.1-11 Full Alternative Text

The inverse is found by dividing each element in the adjoint by $- 27 :$ $- 27 :$

\begin{array}{l} Inverse & = & (\begin{array}{l} \frac{- 3}{- 27} & \frac{- 51}{- 27} & \frac{21}{- 27} \\ \frac{- 4}{- 27} & \frac{4}{- 27} & \frac{1}{- 27} \\ \frac{2}{- 27} & \frac{25}{- 27} & \frac{- 14}{- 27} \end{array}) \\ = & (\begin{array}{l} \frac{3}{27} & \frac{51}{27} & \frac{- 21}{27} \\ \frac{4}{27} & \frac{- 4}{27} & \frac{- 1}{27} \\ \frac{- 2}{27} & \frac{- 25}{27} & \frac{14}{27} \end{array}) \end{array}

$\begin{array}{l} Inverse & = & (\begin{array}{l} \frac{- 3}{- 27} & \frac{- 51}{- 27} & \frac{21}{- 27} \\ \frac{- 4}{- 27} & \frac{4}{- 27} & \frac{1}{- 27} \\ \frac{2}{- 27} & \frac{25}{- 27} & \frac{- 14}{- 27} \end{array}) \\ = & (\begin{array}{l} \frac{3}{27} & \frac{51}{27} & \frac{- 21}{27} \\ \frac{4}{27} & \frac{- 4}{27} & \frac{- 1}{27} \\ \frac{- 2}{27} & \frac{- 25}{27} & \frac{14}{27} \end{array}) \end{array}$

We can verify that this is indeed the correct inverse of the original matrix by multiplying the original matrix by the inverse:

\begin{array}{l} Original & Identity \\ matrix & \times & Inverse & = & matrix \\ (\begin{array}{l} 3 & 7 & 5 \\ 2 & 0 & 3 \\ 4 & 1 & 8 \end{array}) & \times & (\begin{array}{l} \frac{3}{27} & \frac{51}{27} & \frac{- 21}{27} \\ \frac{4}{27} & \frac{- 4}{27} & \frac{- 1}{27} \\ \frac{- 2}{27} & \frac{- 25}{27} & \frac{14}{27} \end{array}) & = & (\begin{array}{l} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}) \end{array}

$\begin{array}{l} Original & Identity \\ matrix & \times & Inverse & = & matrix \\ (\begin{array}{l} 3 & 7 & 5 \\ 2 & 0 & 3 \\ 4 & 1 & 8 \end{array}) & \times & (\begin{array}{l} \frac{3}{27} & \frac{51}{27} & \frac{- 21}{27} \\ \frac{4}{27} & \frac{- 4}{27} & \frac{- 1}{27} \\ \frac{- 2}{27} & \frac{- 25}{27} & \frac{14}{27} \end{array}) & = & (\begin{array}{l} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}) \end{array}$

If this process is applied to a $2 \times 2$ $2 \times 2$ matrix, the inverse is easily found, as shown with the following matrix.

\begin{array}{l} Original matrix & = & (\begin{array}{l} a & b \\ c & d \end{array}) \\ Determinant value of original matrix & = & a d - c b \\ Matrix of cofactors & = & (\begin{array}{l} d & - c \\ - b & a \end{array}) \\ Adjoint of the matrix & = & (\begin{array}{l} d & - b \\ - c & a \end{array}) \end{array}

$\begin{array}{l} Original matrix & = & (\begin{array}{l} a & b \\ c & d \end{array}) \\ Determinant value of original matrix & = & a d - c b \\ Matrix of cofactors & = & (\begin{array}{l} d & - c \\ - b & a \end{array}) \\ Adjoint of the matrix & = & (\begin{array}{l} d & - b \\ - c & a \end{array}) \end{array}$

Since the inverse of the matrix is equal to the adjoint divided by the determinant, we have

{(\begin{matrix} a & b \\ c & d \end{matrix})}^{- 1} = (\begin{matrix} \frac{d}{a d - c b} & \frac{- b}{a d - c b} \\ \frac{- c}{a d - c b} & \frac{a}{a d - c b} \end{matrix})

${(\begin{matrix} a & b \\ c & d \end{matrix})}^{- 1} = (\begin{matrix} \frac{d}{a d - c b} & \frac{- b}{a d - c b} \\ \frac{- c}{a d - c b} & \frac{a}{a d - c b} \end{matrix})$ (M5-4)

For example, if

\begin{array}{l} Original matrix & = & (\begin{array}{l} 1 & 2 \\ 3 & 8 \end{array}) \\ Determinant value & = & 1 (8) - 3 (2) = 2 \end{array}

$\begin{array}{l} Original matrix & = & (\begin{array}{l} 1 & 2 \\ 3 & 8 \end{array}) \\ Determinant value & = & 1 (8) - 3 (2) = 2 \end{array}$

then

Inverse = {(\begin{matrix} 1 & 2 \\ 3 & 8 \end{matrix})}^{- 1} = (\begin{matrix} 8 / 2 & - 2 / 2 \\ - 3 / 2 & 1 / 2 \end{matrix}) = (\begin{matrix} 4 & - 1 \\ - 1.5 & 0.5 \end{matrix})

$Inverse = {(\begin{matrix} 1 & 2 \\ 3 & 8 \end{matrix})}^{- 1} = (\begin{matrix} 8 / 2 & - 2 / 2 \\ - 3 / 2 & 1 / 2 \end{matrix}) = (\begin{matrix} 4 & - 1 \\ - 1.5 & 0.5 \end{matrix})$

..................Content has been hidden....................

You can't read the all page of ebook, please click here login for view all page.

Table of Contents for M5.3 Finding the Inverse of a Matrix

Create new playlist

Sign In

Sign Up

Table of Contents for
M5.3 Finding the Inverse of a Matrix