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\begin{tabular}{ccc|c}
30 & 48 & 18 & 2 \\
\hline
15 & 24 & 9 & 3 \\
5 & 8 & 3 &
\end{tabular}

Sagot :

To solve the given problem, we first need to recognize that we are dealing with a system of linear equations. Our objective is typically to determine whether the system has a unique solution, infinitely many solutions, or no solution. This can be done by associating the system with its augmented matrix and then calculating the determinant of the coefficient matrix.

Let’s start by writing out the augmented matrix and the coefficient matrix:

[tex]\[ \begin{pmatrix} 30 & 48 & 18 & \vert & 2 \\ 15 & 24 & 9 & \vert & 3 \\ 5 & 8 & 3 & \vert & \end{pmatrix} \][/tex]

The coefficient matrix [tex]\( A \)[/tex] is:

[tex]\[ A = \begin{pmatrix} 30 & 48 & 18 \\ 15 & 24 & 9 \\ 5 & 8 & 3 \\ \end{pmatrix} \][/tex]

And the constant vector [tex]\( B \)[/tex] is:

[tex]\[ B = \begin{pmatrix} 2 \\ 3 \\ \end{pmatrix} \][/tex]

To determine if there is a unique solution to this system, we need to compute the determinant of the matrix [tex]\( A \)[/tex].

The determinant of a 3x3 matrix [tex]\( A \)[/tex], given by the elements [tex]\( a_{ij} \)[/tex], can be calculated using the formula:

[tex]\[ \text{det}(A) = a_{11}(a_{22}a_{33} - a_{32}a_{23}) - a_{12}(a_{21}a_{33} - a_{31}a_{23}) + a_{13}(a_{21}a_{32} - a_{31}a_{22}) \][/tex]

Plugging in our values from matrix [tex]\( A \)[/tex]:

[tex]\[ \text{det}(A) = 30(24 \cdot 3 - 8 \cdot 9) - 48(15 \cdot 3 - 5 \cdot 9) + 18(15 \cdot 8 - 5 \cdot 24) \][/tex]

Let's simplify each term step by step:

1. Compute the minor determinants:
- [tex]\( 24 \cdot 3 - 8 \cdot 9 = 72 - 72 = 0 \)[/tex]
- [tex]\( 15 \cdot 3 - 5 \cdot 9 = 45 - 45 = 0 \)[/tex]
- [tex]\( 15 \cdot 8 - 5 \cdot 24 = 120 - 120 = 0 \)[/tex]

2. Multiply by the corresponding element in matrix [tex]\( A \)[/tex]:
- [tex]\( 30 \cdot 0 = 0 \)[/tex]
- [tex]\( -48 \cdot 0 = 0 \)[/tex]
- [tex]\( 18 \cdot 0 = 0 \)[/tex]

Therefore:

[tex]\[ \text{det}(A) = 0 + 0 + 0 = 0 \][/tex]

So, the determinant of the matrix [tex]\( A \)[/tex] is [tex]\( 0 \)[/tex].

The determinant being zero indicates that the matrix [tex]\( A \)[/tex] is singular, which implies that the system of linear equations either has no solution or has infinitely many solutions. It does not have a unique solution.

To summarize, the determinant of the given coefficient matrix is [tex]\( 0 \)[/tex], which means the system of equations does not have a unique solution.