Question

1. use the adjoint-matrix formula to calculate the inverse of the following matrix:

$a = \

$$\begin{pmatrix} 2 & 3 & 4 \\\\ 4 & 3 & 1 \\\\ 1 & 2 & 4 \\end{pmatrix}$$

$
$a^{-1} = \frac{\text{adj}(a)}{|a|} = \frac{c}{|a|}$
$|a| = 24 + 32 + 3 - 12 - 4 - 48 = -5$
$c = \

$$\begin{pmatrix} 10 & -15 & 5 \\\\ -4 & 4 & -1 \\\\ -9 & 14 & -6 \\end{pmatrix}$$

$; $c = \

$$\begin{pmatrix} 10 & -4 & -9 \\\\ -15 & 4 & 14 \\\\ 5 & -1 & -6 \\end{pmatrix}$$

$
$a^{-1} = \frac{1}{-5}c = \

$$\begin{pmatrix} -2 & \\frac{4}{5} & \\frac{9}{5} \\\\ 3 & \\frac{-4}{5} & -\\frac{14}{5} \\\\ 1 & \\frac{1}{5} & \\frac{6}{5} \\end{pmatrix}$$

= \

$$\begin{pmatrix} -2 & 0.8 & 1.8 \\\\ 3 & -0.8 & -2.8 \\\\ -1 & 0.2 & 1.2 \\end{pmatrix}$$

$

Explanation:

Step1: Recall the adjoint - matrix formula

The formula for the inverse of a matrix \(A\) using the adjoint matrix is \(A^{-1}=\frac{\text{adj}(A)}{|A|}=\frac{C'}{|A|}\), where \(\text{adj}(A)\) is the adjoint of \(A\) (the transpose of the cofactor matrix \(C\)), and \(|A|\) is the determinant of \(A\).

Step2: Calculate the determinant of \(A\)

For a \(3\times3\) matrix \(A =

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

\), the determinant \(|A|=a_{11}(a_{22}a_{33}-a_{23}a_{32})-a_{12}(a_{21}a_{33}-a_{23}a_{31})+a_{13}(a_{21}a_{32}-a_{22}a_{31})\)

For \(A=

$$\begin{pmatrix}2&3&4\\4&3&1\\1&2&4\end{pmatrix}$$

\)

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Step3: Calculate the cofactor matrix \(C\)

The cofactor \(C_{ij}=(- 1)^{i + j}M_{ij}\), where \(M_{ij}\) is the minor (the determinant of the sub - matrix obtained by deleting the \(i\) - th row and \(j\) - th column)

• For \(C_{11}\): \(i = 1,j = 1\), \(M_{11}=

$$\begin{vmatrix}3&1\\2&4\end{vmatrix}$$

=3\times4-1\times2 = 10\), \(C_{11}=(-1)^{1 + 1}\times10 = 10\)

• For \(C_{12}\): \(i = 1,j = 2\), \(M_{12}=

$$\begin{vmatrix}4&1\\1&4\end{vmatrix}$$

=4\times4-1\times1 = 15\), \(C_{12}=(-1)^{1+2}\times15=- 15\)

• For \(C_{13}\): \(i = 1,j = 3\), \(M_{13}=

$$\begin{vmatrix}4&3\\1&2\end{vmatrix}$$

=4\times2-3\times1 = 5\), \(C_{13}=(-1)^{1 + 3}\times5 = 5\)

• For \(C_{21}\): \(i = 2,j = 1\), \(M_{21}=

$$\begin{vmatrix}3&4\\2&4\end{vmatrix}$$

=3\times4-4\times2 = 4\), \(C_{21}=(-1)^{2 + 1}\times4=-4\)

• For \(C_{22}\): \(i = 2,j = 2\), \(M_{22}=

$$\begin{vmatrix}2&4\\1&4\end{vmatrix}$$

=2\times4-4\times1 = 4\), \(C_{22}=(-1)^{2+2}\times4 = 4\)

• For \(C_{23}\): \(i = 2,j = 3\), \(M_{23}=

$$\begin{vmatrix}2&3\\1&2\end{vmatrix}$$

=2\times2-3\times1 = 1\), \(C_{23}=(-1)^{2 + 3}\times1=-1\)

• For \(C_{31}\): \(i = 3,j = 1\), \(M_{31}=

$$\begin{vmatrix}3&4\\3&1\end{vmatrix}$$

=3\times1-4\times3=-9\), \(C_{31}=(-1)^{3 + 1}\times(-9)=-9\)

• For \(C_{32}\): \(i = 3,j = 2\), \(M_{32}=

$$\begin{vmatrix}2&4\\4&1\end{vmatrix}$$

=2\times1-4\times4=-14\), \(C_{32}=(-1)^{3 + 2}\times(-14)=14\)

• For \(C_{33}\): \(i = 3,j = 3\), \(M_{33}=

$$\begin{vmatrix}2&3\\4&3\end{vmatrix}$$

=2\times3-3\times4=-6\), \(C_{33}=(-1)^{3 + 3}\times(-6)=-6\)

So the cofactor matrix \(C=

$$\begin{pmatrix}10&-15&5\\-4&4&-1\\-9&14&-6\end{pmatrix}$$

\)

Step4: Find the transpose of the cofactor matrix \(C'\)

The transpose of a matrix is obtained by interchanging rows and columns. So \(C'=

$$\begin{pmatrix}10&-4&-9\\-15&4&14\\5&-1&-6\end{pmatrix}$$

\)

Step5: Calculate the inverse of \(A\)

Using the formula \(A^{-1}=\frac{C'}{|A|}\), and \(|A|=-5\)

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Answer:

The inverse of the matrix \(A\) is \(

$$\begin{pmatrix}-2&0.8&1.8\\3&-0.8&-2.8\\-1&0.2&1.2\end{pmatrix}$$

\)