Determine if a 2x2 matrix is invertible and find its inverse if it exists Algebra II

Let A be a matrix such that:

A=\begin {pmatrix} a_{11} & a_{12} \cr \cr a_{21} & a_{22} \end {pmatrix}

Then A is invertible if:

det \left(A\right) = a_ {11} a_ {22} -a_ {12} a_ {21} \neq 0

And if A is invertible its inverse is:

A^{- 1} = \dfrac{1}{det\left(A\right)} \begin {pmatrix} a_ {22} & -a_ {12} \cr \ -a_ {21} & a_ {11} \end {pmatrix}

Here, we have:

A=\begin {pmatrix} 8 & 3 \cr \cr 4 & 2 \end {pmatrix}

Let's compute:

a_{11}a_{22}-a_{12}a_{21} = 8 \times 2 -3 \times 4 = 4

a_{11}a_{22}-a_{12}a_{21} \neq 0

Then A is invertible. We can determine its inverse matrix:

A^{-1}=\dfrac{1}{4} \begin {pmatrix} 2 & -3 \cr \cr -4 & 8 \end {pmatrix}

A^{- 1} = \begin {pmatrix} \dfrac{1}{2} & \dfrac{-3}{4} \cr \cr -1 & 2 \end {pmatrix}

Let A be a matrix such that:

A=\begin {pmatrix} a_{11} & a_{12} \cr \cr a_{21} & a_{22} \end {pmatrix}

Then A is invertible if:

a_{11}a_{22}-a_{12}a_{21} \neq 0

And if A is invertible its inverse is:

A^{- 1} = \dfrac{1}{det\left(A\right)} \begin {pmatrix} a_ {22} & -a_ {12} \cr \ -a_ {21} & a_ {11} \end {pmatrix}

Here, we have:

A = \begin {pmatrix} 4 & 8 \cr \cr 1 & 2 \end {pmatrix}

Let's compute:

a_ {11} a_ {22} -a_ {12} a_ {21} = 8-8=0

Let A be a matrix such that:

A=\begin {pmatrix} a_{11} & a_{12} \cr \cr a_{21} & a_{22} \end {pmatrix}

Then A is invertible if:

det \left(A\right) = a_ {11} a_ {22} -a_ {12} a_ {21} \neq 0

And if A is invertible its inverse is:

A^{- 1} = \dfrac{1}{det\left(A\right)} \begin {pmatrix} a_ {22} & -a_ {12} \cr \ -a_ {21} & a_ {11} \end {pmatrix}

Here, we have:

A = \begin {pmatrix} 2 & 1 \cr \cr 3 & 2 \end {pmatrix}

det\left(A\right)=2\times 2-3 \times 1=1

Therefore A is invertible and the inverse is given by:

A^{-1} = \begin {pmatrix} 2 & -1 \cr \cr -3 & 2 \end {pmatrix}

Let A be a matrix such that:

A=\begin {pmatrix} a_{11} & a_{12} \cr \cr a_{21} & a_{22} \end {pmatrix}

Then A is invertible if:

a_{11}a_{22}-a_{12}a_{21} \neq 0

And if A is invertible its inverse is:

A^{- 1} = \dfrac{1}{det\left(A\right)} \begin {pmatrix} a_ {22} & -a_ {12} \cr \ -a_ {21} & a_ {11} \end {pmatrix}

Here we have :

A = \begin {pmatrix} 4 & -6 \cr \cr -2 & 3 \end {pmatrix}

Let's compute :

a_ {11} a_ {22} -a_ {12} a_ {21} = 12-12 = 0

Let A be a matrix such that:

A=\begin {pmatrix} a_{11} & a_{12} \cr \cr a_{21} & a_{22} \end {pmatrix}

Then A is invertible if:

det \left(A\right) = a_ {11} a_ {22} -a_ {12} a_ {21} \neq 0

And if A is invertible its inverse is:

A^{- 1} = \dfrac{1}{det\left(A\right)} \begin {pmatrix} a_ {22} & -a_ {12} \cr \ -a_ {21} & a_ {11} \end {pmatrix}

Here, we have:

A = \begin {pmatrix} 1& 2 \cr \cr -3 & -4 \end {pmatrix}

Let's compute:

a_ {11} a_ {22} -a_ {12} a_ {21} = 1 \times -4 -3 \times 2 = 2

a_{11}a_{22}-a_{12}a_{21} \neq 0

Then A is invertible. We can determine its inverse matrix:

A^{-1} =\dfrac{1}{2} \begin {pmatrix} -4 & -2 \cr \cr 3 & 1 \end {pmatrix}

A^{-1}= \begin {pmatrix} -2 & -1 \cr \cr \dfrac {3} {2} & \dfrac {1} {2} \end {pmatrix}

Let A be a matrix such that:

A=\begin {pmatrix} a_{11} & a_{12} \cr \cr a_{21} & a_{22} \end {pmatrix}

Then A is invertible if:

det \left(A\right) = a_ {11} a_ {22} -a_ {12} a_ {21} \neq 0

And if A is invertible its inverse is:

A^{- 1} = \dfrac{1}{det\left(A\right)} \begin {pmatrix} a_ {22} & -a_ {12} \cr \ -a_ {21} & a_ {11} \end {pmatrix}

Here, we have:

A = \begin {pmatrix} 2 & 3 \cr \cr 3 & 5 \end {pmatrix}

Let's compute:

a_ {11} a_ {22} -a_ {12} a_ {21} = 2 \times 5 -3 \times 3 = 1

a_{11}a_{22}-a_{12}a_{21} \neq 0

Then A is invertible. We can determine its inverse matrix:

A^{-1} =\begin {pmatrix} 5 & -3 \cr \cr -3 & 2 \end {pmatrix}

Let A be a matrix such that:

A=\begin {pmatrix} a_{11} & a_{12} \cr \cr a_{21} & a_{22} \end {pmatrix}

Then A is invertible if:

a_{11}a_{22}-a_{12}a_{21} \neq 0

And if A is invertible its inverse is:

A^{- 1} = \dfrac{1}{det\left(A\right)} \begin {pmatrix} a_ {22} & -a_ {12} \cr \ -a_ {21} & a_ {11} \end {pmatrix}

Here we have:

A=\begin{pmatrix} 8 & 2 \cr\cr 12 & 3 \end{pmatrix}

Let's compute :

a_ {11} a_ {22} -a_ {12} a_ {21} = 24-24 = 0