Question

1. write the system as a single matrix equation and solve:

4a - 2b + c = 4

a + b + c = 1

9a + 3b + c = -21

Explanation:

Step1: Write the matrix equation

A system of linear equations \(

$$\begin{cases} 4a - 2b + c = 4 \\ a + b + c = 1 \\ 9a + 3b + c = -21 \end{cases}$$

\) can be written in the form \( AX = B \), where \( A \) is the coefficient matrix, \( X \) is the variable matrix, and \( B \) is the constant matrix.

So, \( A=

$$\begin{bmatrix}4&-2&1\\1&1&1\\9&3&1\end{bmatrix}$$

\), \( X =

$$\begin{bmatrix}a\\b\\c\end{bmatrix}$$

\), and \( B=

$$\begin{bmatrix}4\\1\\-21\end{bmatrix}$$

\). The matrix equation is \(

$$\begin{bmatrix}4&-2&1\\1&1&1\\9&3&1\end{bmatrix}$$

$$\begin{bmatrix}a\\b\\c\end{bmatrix}$$

=

$$\begin{bmatrix}4\\1\\-21\end{bmatrix}$$

\).

Step2: Find the inverse of matrix \( A \) (using the formula for 3x3 matrix or row operations)

First, we can find the determinant of \( A \), \( \det(A)=4

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

-(-2)

$$\begin{vmatrix}1&1\\9&1\end{vmatrix}$$

+1

$$\begin{vmatrix}1&1\\9&3\end{vmatrix}$$

\)

\( = 4(1 - 3)+ 2(1 - 9)+1(3 - 9) \)

\( = 4(-2)+2(-8)+1(-6) \)

\( = -8 - 16 - 6=-30 \)

Now, find the adjugate matrix of \( A \). The cofactor matrix \( C \) is:

\( C_{11}=

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

=1 - 3=-2 \), \( C_{12}=-

$$\begin{vmatrix}1&1\\9&1\end{vmatrix}$$

=-(1 - 9)=8 \), \( C_{13}=

$$\begin{vmatrix}1&1\\9&3\end{vmatrix}$$

=3 - 9=-6 \)

\( C_{21}=-

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

=-(-2 - 3)=5 \), \( C_{22}=

$$\begin{vmatrix}4&1\\9&1\end{vmatrix}$$

=4 - 9=-5 \), \( C_{23}=-

$$\begin{vmatrix}4&-2\\9&3\end{vmatrix}$$

=-(12 + 18)=-30 \)

\( C_{31}=

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

=-2 - 1=-3 \), \( C_{32}=-

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

=-(4 - 1)=-3 \), \( C_{33}=

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

=4 + 2=6 \)

So, the cofactor matrix \( C=

$$\begin{bmatrix}-2&8&-6\\5&-5&-30\\-3&-3&6\end{bmatrix}$$

\)

The adjugate matrix \( \text{adj}(A)=C^T=

$$\begin{bmatrix}-2&5&-3\\8&-5&-3\\-6&-30&6\end{bmatrix}$$

\)

Then, the inverse of \( A \) is \( A^{-1}=\frac{1}{\det(A)}\text{adj}(A)=-\frac{1}{30}

$$\begin{bmatrix}-2&5&-3\\8&-5&-3\\-6&-30&6\end{bmatrix}$$

=

$$\begin{bmatrix}\frac{1}{15}&-\frac{1}{6}&\frac{1}{10}\\-\frac{4}{15}&\frac{1}{6}&\frac{1}{10}\\\frac{1}{5}&1&-\frac{1}{5}\end{bmatrix}$$

\)

Step3: Solve for \( X \)

Since \( AX = B \), then \( X = A^{-1}B \)

\( X=

$$\begin{bmatrix}\frac{1}{15}&-\frac{1}{6}&\frac{1}{10}\\-\frac{4}{15}&\frac{1}{6}&\frac{1}{10}\\\frac{1}{5}&1&-\frac{1}{5}\end{bmatrix}$$

$$\begin{bmatrix}4\\1\\-21\end{bmatrix}$$

\)

Calculate each component:

First component (for \( a \)): \( \frac{1}{15}\times4-\frac{1}{6}\times1+\frac{1}{10}\times(-21)=\frac{4}{15}-\frac{1}{6}-\frac{21}{10}=\frac{8 - 5 - 63}{30}=\frac{-60}{30}=-2 \)

Second component (for \( b \)): \( -\frac{4}{15}\times4+\frac{1}{6}\times1+\frac{1}{10}\times(-21)=-\frac{16}{15}+\frac{1}{6}-\frac{21}{10}=\frac{-32 + 5 - 63}{30}=\frac{-90}{30}=-3 \)

Third component (for \( c \)): \( \frac{1}{5}\times4+1\times1-\frac{1}{5}\times(-21)=\frac{4}{5}+1+\frac{21}{5}=\frac{4 + 5 + 21}{5}=\frac{30}{5}=6 \)

Answer:

The matrix equation is \( \boldsymbol{

$$\begin{bmatrix}4&-2&1\\1&1&1\\9&3&1\end{bmatrix}$$

$$\begin{bmatrix}a\\b\\c\end{bmatrix}$$

=

$$\begin{bmatrix}4\\1\\-21\end{bmatrix}$$

} \), and the solution is \( a = -2 \), \( b = -3 \), \( c = 6 \) (or in matrix form \(

$$\begin{bmatrix}a\\b\\c\end{bmatrix}$$

=

$$\begin{bmatrix}-2\\-3\\6\end{bmatrix}$$

\))