To solve the matrix equation [tex]\( AX = B \)[/tex] for [tex]\( X \)[/tex], we need to determine the values of the vector [tex]\( X \)[/tex] that satisfy the equation given [tex]\( A \)[/tex] and [tex]\( B \)[/tex]. Here, we have:
[tex]\[ A = \begin{pmatrix} 1 & 3 \\ 0 & 1 \end{pmatrix}, \][/tex]
[tex]\[ B = \begin{pmatrix} -17 \\ -5 \end{pmatrix}. \][/tex]
Let [tex]\( X \)[/tex] be a vector with components [tex]\( x_1 \)[/tex] and [tex]\( x_2 \)[/tex]:
[tex]\[ X = \begin{pmatrix} x_1 \\ x_2 \end{pmatrix}. \][/tex]
The equation [tex]\( AX = B \)[/tex] can be written as:
[tex]\[ \begin{pmatrix} 1 & 3 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \begin{pmatrix} -17 \\ -5 \end{pmatrix}. \][/tex]
Performing the matrix multiplication on the left-hand side, we get two equations:
1. From the first row:
[tex]\[ 1 \cdot x_1 + 3 \cdot x_2 = -17 \][/tex]
[tex]\[ x_1 + 3x_2 = -17 \][/tex]
2. From the second row:
[tex]\[ 0 \cdot x_1 + 1 \cdot x_2 = -5 \][/tex]
[tex]\[ x_2 = -5 \][/tex]
Having found [tex]\( x_2 \)[/tex], we can substitute [tex]\( x_2 = -5 \)[/tex] back into the first equation to find [tex]\( x_1 \)[/tex]:
[tex]\[ x_1 + 3(-5) = -17 \][/tex]
[tex]\[ x_1 - 15 = -17 \][/tex]
[tex]\[ x_1 = -17 + 15 \][/tex]
[tex]\[ x_1 = -2 \][/tex]
Thus, the solution for [tex]\( X \)[/tex] is:
[tex]\[ X = \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \begin{pmatrix} -2 \\ -5 \end{pmatrix}. \][/tex]
Therefore, the solution for the matrix equation [tex]\( AX = B \)[/tex] is:
[tex]\[ X = \begin{pmatrix} -2 \\ -5 \end{pmatrix}. \][/tex]
