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 Adjoint Matrix Formula

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

Step2: Calculate Determinant \( |A| \)

For matrix \( A=

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

\), use the rule for 3x3 determinants: \( |A| = a(ei - fh)-b(di - fg)+c(dh - eg) \).
Substituting \( a = 2, b = 3, c = 4, d = 4, e = 3, f = 1, g = 1, h = 2, i = 4 \):

$$
LATEXBLOCK1
$$

(Or as given: \( |A| = 24 + 32 + 3 - 12 - 4 - 48=-5 \))

Step3: Find Cofactor Matrix \( C \)

Cofactor \( C_{ij}=(-1)^{i + j}M_{ij} \), where \( M_{ij} \) is the minor.

• For \( C_{11} \): \( (-1)^{1 + 1}

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

=1\times(12 - 2)=10 \)

• \( C_{12} \): \( (-1)^{1 + 2}

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

=-1\times(16 - 1)=-15 \)

• \( C_{13} \): \( (-1)^{1 + 3}

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

=1\times(8 - 3)=5 \)

• \( C_{21} \): \( (-1)^{2 + 1}

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

=-1\times(12 - 8)=-4 \)

• \( C_{22} \): \( (-1)^{2 + 2}

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

=1\times(8 - 4)=4 \)

• \( C_{23} \): \( (-1)^{2 + 3}

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

=-1\times(4 - 3)=-1 \)

• \( C_{31} \): \( (-1)^{3 + 1}

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

=1\times(3 - 12)=-9 \)

• \( C_{32} \): \( (-1)^{3 + 2}

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

=-1\times(2 - 16)=14 \)

• \( C_{33} \): \( (-1)^{3 + 3}

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

=1\times(6 - 12)=-6 \)
So \( C=

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

\)

Step4: Transpose Cofactor Matrix (\( C' \))

Transpose \( C \) to get \( C' \): \( C'=

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

\)

Step5: Compute Inverse \( A^{-1} \)

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

$$ A^{-1}=\frac{1}{-5}
LATEXBLOCK13
=
LATEXBLOCK14
=
LATEXBLOCK15
$$

Answer:

The inverse of matrix \( A \) is \( \boldsymbol{

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

} \) (or in fraction form \(

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

\))