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Cofactors Determinant and inverse of a 3x3 matrix.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

Inverse of a $3 \\times 3$ Matrix

\n

Follow the steps in the questions to find the inverse of a $3 \\times 3$ matrix

", "advice": "

a)

\n

For simplicity, we will use the expansion of the first row to find the determinant

\n

$$
\\begin{aligned}
\\det{A} &= a_1A_1 + b_1B_1+ c_1C_1 \\\\
&= \\var{matrixA[0][0]}
\\begin{vmatrix}
\\var{matrixA[1][1]} & \\var{matrixA[1][2]} \\\\
\\var{matrixA[2][1]} & \\var{matrixA[2][2]} \\\\
\\end{vmatrix} - \\var{matrixA[0][1]} \\begin{vmatrix}
\\var{matrixA[1][0]} & \\var{matrixA[1][2]} \\\\
\\var{matrixA[2][0]} & \\var{matrixA[2][2]} \\\\
\\end{vmatrix} + \\var{matrixA[0][2]} \\begin{vmatrix}
\\var{matrixA[1][0]} & \\var{matrixA[1][1]} \\\\
\\var{matrixA[2][0]} & \\var{matrixA[2][1]} \\\\
\\end{vmatrix} \\\\
&= \\var{matrixA[0][0]} \\times \\var{cof11} - \\left(\\var{matrixA[0][1]} \\times \\var{cof12}\\right) + \\var{matrixA[0][2]} \\times \\var{cof13} \\\\
&= \\var{deta}
\\end{aligned}
$$

\n

b)

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Given arbitrary matrix

\n

$$
A = \\begin{pmatrix}
a & b & c \\\\
d & e & f \\\\
g & h & j \\\\
\\end{pmatrix}
$$

\n

It's cofactors are given by

\n

$$
\\begin{aligned}
A_{11} &=  +\\begin{vmatrix}
e & f \\\\
h & j \\\\
\\end{vmatrix} \\\\
&= +\\begin{vmatrix}
\\var{a22} & \\var{a23} \\\\
\\var{a32} & \\var{a33} \\\\
\\end{vmatrix} \\\\
&= \\var{cof11}
\\end{aligned}
$$

\n

$$
\\begin{aligned}
A_{12} &=  -\\begin{vmatrix}
d & f \\\\
g & j \\\\
\\end{vmatrix} \\\\
&= -\\begin{vmatrix}
\\var{a21} & \\var{a23} \\\\
\\var{a31} & \\var{a33} \\\\
\\end{vmatrix} \\\\
&= \\var{cof12}
\\end{aligned}
$$

\n

$$
\\begin{aligned}
A_{13} &=  +\\begin{vmatrix}
d & e \\\\
g & h \\\\
\\end{vmatrix} \\\\
&= +\\begin{vmatrix}
\\var{a21} & \\var{a22} \\\\
\\var{a31} & \\var{a32} \\\\
\\end{vmatrix} \\\\
&= \\var{cof13}
\\end{aligned}
$$

\n

$$
\\begin{aligned}
A_{21} &=  -\\begin{vmatrix}
b & c \\\\
h & j \\\\
\\end{vmatrix} \\\\
&= -\\begin{vmatrix}
\\var{a12} & \\var{a13} \\\\
\\var{a32} & \\var{a33} \\\\
\\end{vmatrix} \\\\
&= \\var{cof21}
\\end{aligned}
$$

\n

$$
\\begin{aligned}
A_{22} &=  +\\begin{vmatrix}
a & c \\\\
g & j \\\\
\\end{vmatrix} \\\\
&= +\\begin{vmatrix}
\\var{a11} & \\var{a13} \\\\
\\var{a31} & \\var{a33} \\\\
\\end{vmatrix} \\\\
&= \\var{cof22}
\\end{aligned}
$$

\n

$$
\\begin{aligned}
A_{23} &=  -\\begin{vmatrix}
a & b \\\\
g & h \\\\
\\end{vmatrix} \\\\
&= -\\begin{vmatrix}
\\var{a11} & \\var{a12} \\\\
\\var{a31} & \\var{a32} \\\\
\\end{vmatrix} \\\\
&= \\var{cof23}
\\end{aligned}
$$

\n

$$
\\begin{aligned}
A_{31} &=  +\\begin{vmatrix}
b & c \\\\
e & f \\\\
\\end{vmatrix} \\\\
&= +\\begin{vmatrix}
\\var{a12} & \\var{a13} \\\\
\\var{a22} & \\var{a23} \\\\
\\end{vmatrix} \\\\
&= \\var{cof31}
\\end{aligned}
$$

\n

$$
\\begin{aligned}
A_{32} &=  -\\begin{vmatrix}
a & c \\\\
d & f \\\\
\\end{vmatrix} \\\\
&= -\\begin{vmatrix}
\\var{a11} & \\var{a13} \\\\
\\var{a21} & \\var{a23} \\\\
\\end{vmatrix} \\\\
&= \\var{cof32}
\\end{aligned}
$$

\n

$$
\\begin{aligned}
A_{33} &=  +\\begin{vmatrix}
a & b \\\\
d & e \\\\
\\end{vmatrix} \\\\
&= +\\begin{vmatrix}
\\var{a11} & \\var{a12} \\\\
\\var{a21} & \\var{a22} \\\\
\\end{vmatrix} \\\\
&= \\var{cof33}
\\end{aligned}
$$

\n

c)

\n

Using our answer from the previous question, we simply write the cofactors in the form

\n

$$
\\begin{pmatrix}
A_{11} & A_{12} & A_{13} \\\\
A_{21} & A_{22} & A_{23} \\\\
A_{31} & A_{32} & A_{33} \\\\
\\end{pmatrix}
$$

\n

Giving us our matrix of cofactors

\n

$$
\\begin{pmatrix}
\\var{cof11} & \\var{cof12} & \\var{cof13} \\\\
\\var{cof21} & \\var{cof22} & \\var{cof23} \\\\
\\var{cof31} & \\var{cof32} & \\var{cof33} \\\\
\\end{pmatrix}
$$

\n

d)

\n

The transposition process turns rows into columns and columns into rows

\n

Carrying out this process on our matrix of cofactors gives us the adjugate

\n

$$
\\begin{pmatrix}
\\var{cof11} & \\var{cof21} & \\var{cof31} \\\\
\\var{cof12} & \\var{cof22} & \\var{cof32} \\\\
\\var{cof13} & \\var{cof23} & \\var{cof33} \\\\
\\end{pmatrix}
$$

\n

e)

\n

We can find the inverse of $A$ using our determinant and adjugate, using the formula

\n

$$
A^{-1} = \\frac{1}{\\det A}(adj \\; A)
$$

\n

Therefore, we can calculate $A^{-1}$ by

\n

$$
\\begin{aligned}
A^{-1} &= \\frac{1}{\\var{deta}} \\begin{pmatrix}
\\var{cof11} & \\var{cof21} & \\var{cof31} \\\\
\\var{cof12} & \\var{cof22} & \\var{cof32} \\\\
\\var{cof13} & \\var{cof23} & \\var{cof33} \\\\
\\end{pmatrix} \\\\
&= \\var{inverseA}
\\end{aligned}
$$

\n

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cof23

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Let:

\n

$$
A=\\var{matrixA}
$$

\n

Find the determinant of $A$

\n

$\\det A =$ [[0]]

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Calculate the nine cofactors of $A$

\n

The cofactor $A_{ij}$ denotes the cofactor in row $i$ and column $j$

\n

$A _{11}=$ [[0]]

\n

$A_{12}=$ [[1]]

\n

$A_{13}=$ [[2]]

\n

$A_{21}=$ [[3]]

\n

$A_{22}=$ [[4]]

\n

$A_{23}=$ [[5]]

\n

$A_{31}=$ [[6]]

\n

$A_{32}=$ [[7]]

\n

$A_{33}=$ [[8]]

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Using your answer to part $b)$, state the matrix of cofactors

\n

[[0]]

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Find the transpose of your matrix from part $c)$, giving us the adjugate 

\n

[[0]]

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Using your answers from all the previous parts, find the inverse of $A$

\n

Elements will be accepted as fractions or correct to 2 decimal places

\n

$A^{-1}=$ [[0]]

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