Question

which equation represents the solution to this system?
x + y + z = 350
x + y - z = 130
2x + 3y + 2z = 790
a. $x = \

$$\begin{bmatrix}2.5 & -0.5 & -1\\\\ 2 & 0 & -1\\\\ -0.5 & 0.5 & 0\\end{bmatrix}$$

\

$$\begin{bmatrix}790\\\\ 130\\\\ 795\\end{bmatrix}$$

$
b. $x = \

$$\begin{bmatrix}2.5 & 0.5 & -1\\\\ -2 & 0 & 1\\\\ 0.5 & -0.5 & 0\\end{bmatrix}$$

\

$$\begin{bmatrix}350\\\\ 130\\\\ 790\\end{bmatrix}$$

$
c. $x = \

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

\

$$\begin{bmatrix}350\\\\ 130\\\\ 790\\end{bmatrix}$$

$
d. $x = \

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

\

$$\begin{bmatrix}790\\\\ 130\\\\ 350\\end{bmatrix}$$

$

Explanation:

Step1: Recall System of Equations Matrix Form

A system of linear equations \( a_1x + b_1y + c_1z = d_1 \), \( a_2x + b_2y + c_2z = d_2 \), \( a_3x + b_3y + c_3z = d_3 \) can be written as \( AX = B \), where \( A=\begin{bmatrix}a_1&b_1&c_1\\a_2&b_2&c_2\\a_3&b_3&c_3\end{bmatrix} \), \( X=\begin{bmatrix}x\\y\\z\end{bmatrix} \), and \( B=\begin{bmatrix}d_1\\d_2\\d_3\end{bmatrix} \). The solution is \( X = A^{-1}B \) (if \( A \) is invertible).

Step2: Identify Matrix \( A \) and \( B \) for Given System

The given system is:
\( x + y + z = 350 \)
\( x + y - z = 130 \)
\( 2x + 3y + 2z = 790 \)

So, \( A=\begin{bmatrix}1&1&1\\1&1&-1\\2&3&2\end{bmatrix} \), \( B=\begin{bmatrix}350\\130\\790\end{bmatrix} \), and the solution should be \( X = A^{-1}B \) (or in the form where the matrix multiplied by \( B \) is \( A^{-1} \)).

Step3: Analyze Each Option

- Option A: The matrix and \( B \)-vector do not match the coefficients or constants of the system.
- Option B: The matrix and \( B \)-vector do not match the coefficients or constants of the system.
- Option C: The matrix \( A \) is the coefficient matrix of the system, and \( B \) is the constant vector. So \( X = A^{-1}B \) (here the form shows \( X \) equal to the coefficient matrix times the constant vector? Wait, no—wait, actually, the form \( X = \text{matrix} \times \text{vector} \) should have the matrix as \( A^{-1} \), but wait, no—wait, the system is \( AX = B \), so \( X = A^{-1}B \). But option C shows \( X = A \times B \)? No, wait, no—wait, the option C is \( X = \begin{bmatrix}1&1&1\\1&1&-1\\2&3&2\end{bmatrix}\begin{bmatrix}350\\130\\790\end{bmatrix} \)? No, that's not correct. Wait, no—wait, maybe I made a mistake. Wait, no—wait, the options are in the form \( X = \text{some matrix} \times \text{some vector} \). Wait, actually, the correct form for the solution of \( AX = B \) is \( X = A^{-1}B \). But let's check the coefficient matrix. The coefficient matrix \( A \) for the system is \( \begin{bmatrix}1&1&1\\1&1&-1\\2&3&2\end{bmatrix} \), and the constant vector \( B \) is \( \begin{bmatrix}350\\130\\790\end{bmatrix} \). Option C shows \( X \) equal to the coefficient matrix multiplied by the constant vector? No, that would be \( AX \), not \( X \). Wait, no—wait, maybe the options are written as \( X = \text{inverse of } A \times B \), but option C's matrix is \( A \), not \( A^{-1} \). Wait, no—wait, maybe I messed up. Wait, let's re-express the system.

Wait, no—wait, the options are in the form \( X = M \times V \), where \( M \) is a matrix and \( V \) is a vector. For the system \( AX = B \), \( X = A^{-1}B \). But let's check the coefficient matrix \( A \) and the constant vector \( B \). The given system's \( A \) is \( \begin{bmatrix}1&1&1\\1&1&-1\\2&3&2\end{bmatrix} \) and \( B \) is \( \begin{bmatrix}350\\130\\790\end{bmatrix} \). Option C has \( M = A \) and \( V = B \), so \( X = A \times B \)? No, that's not the solution. Wait, no—wait, maybe the options are miswritten, or I misinterpret. Wait, no—wait, the system is linear, and the solution using matrix inversion is \( X = A^{-1}B \). But let's check the other options. Option A and B have non-integer matrices, but option C has the coefficient matrix. Wait, maybe the question is about representing the system in matrix form \( AX = B \), so \( X \) is the vector of variables, \( A \) is the coefficient matrix, and \( B \) is the constant vector. So the equation \( AX = B \) can be written as \( X = A^{-1}B \), but if we are just writing the system as \( AX = B \), then \( A \) is the coefficient matrix, \( X \) is \( \begin{bmatrix}x\\y\\z\end{bmatrix} \), and \( B \) is the constant vector. But the options are in the form \( X = M \times V \), which would be \( X = A^{-1}B \). Wait, but option C's matrix is \( A \), not \( A^{-1} \). Wait, maybe the question is actually asking for the matrix equation \( AX = B \), so \( X \) is the variable vector, \( A \) is the coefficient matrix, and \( B \) is the constant vector. So the equation is \( \begin{bmatrix}1&1&1\\1&1&-1\\2&3&2\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}350\\130\\790\end{bmatrix} \), which can be written as \( X = A^{-1}B \), but if we consider the form \( X = M \times V \), where \( M \) is \( A^{-1} \), but option C has \( M = A \). Wait, no—maybe the question is a trick, and option C is the matrix equation \( AX = B \), so the solution is represented as \( X \) being equal to... No, wait, the options are in the form \( X = M \times V \), so we need to see which \( M \) and \( V \) correspond to \( A^{-1}B \). But let's check the coefficients. The given system:

First equation: \( x + y + z = 350 \)

Second: \( x + y - z = 130 \)

Third: \( 2x + 3y + 2z = 790 \)

Let's try to solve the system. Add first and second equations: \( 2x + 2y = 480 \implies x + y = 240 \). Subtract second from first: \( 2z = 220 \implies z = 110 \). Then from \( x + y = 240 \), we have \( x = 240 - y \). Substitute into third equation: \( 2(240 - y) + 3y + 2(110) = 790 \implies 480 - 2y + 3y + 220 = 790 \implies y + 700 = 790 \implies y = 90 \), then \( x = 240 - 90 = 150 \). So \( x = 150 \), \( y = 90 \), \( z = 110 \).

Now let's check option C: \( X = \begin{bmatrix}1&1&1\\1&1&-1\\2&3&2\end{bmatrix}\begin{bmatrix}350\\130\\790\end{bmatrix} \). Let's compute the first component: \( 1*350 + 1*130 + 1*790 = 350 + 130 + 790 = 1270 \), which is not \( x = 150 \). So that's wrong. Wait, so my initial thought was wrong. So the options are \( X = M \times V \), where \( M \) is \( A^{-1} \). Let's find \( A^{-1} \).

Matrix \( A = \begin{bmatrix}1&1&1\\1&1&-1\\2&3&2\end{bmatrix} \). Let's compute its inverse. First, find the determinant of \( A \).

\( \det(A) = 1*(1*2 - (-1)*3) - 1*(1*2 - (-1)*2) + 1*(1*3 - 1*2) \)

\( = 1*(2 + 3) - 1*(2 + 2) + 1*(3 - 2) \)

\( = 5 - 4 + 1 = 2 \).

Now, find the adjugate matrix. The cofactor matrix:

Cofactor of (1,1): \( \det\begin{bmatrix}1&-1\\3&2\end{bmatrix} = 2 + 3 = 5 \)

Cofactor of (1,2): \( -\det\begin{bmatrix}1&-1\\2&2\end{bmatrix} = -(2 + 2) = -4 \)

Cofactor of (1,3): \( \det\begin{bmatrix}1&1\\2&3\end{bmatrix} = 3 - 2 = 1 \)

Cofactor of (2,1): \( -\det\begin{bmatrix}1&1\\3&2\end{bmatrix} = -(2 - 3) = 1 \)

Cofactor of (2,2): \( \det\begin{bmatrix}1&1\\2&2\end{bmatrix} = 2 - 2 = 0 \)

Cofactor of (2,3): \( -\det\begin{bmatrix}1&1\\2&3\end{bmatrix} = -(3 - 2) = -1 \)

Cofactor of (3,1): \( \det\begin{bmatrix}1&1\\1&-1\end{bmatrix} = -1 - 1 = -2 \)

Cofactor of (3,2): \( -\det\begin{bmatrix}1&1\\1&-1\end{bmatrix} = -(-1 - 1) = 2 \)

Cofactor of (3,3): \( \det\begin{bmatrix}1&1\\1&1\end{bmatrix} = 1 - 1 = 0 \)

So cofactor matrix is \( \begin{bmatrix}5&-4&1\\1&0&-1\\-2&2&0\end{bmatrix} \). The adjugate is the transpose of the cofactor matrix: \( \begin{bmatrix}5&1&-2\\-4&0&2\\1&-1&0\end{bmatrix} \). Then \( A^{-1} = \frac{1}{\det(A)} \times \text{adjugate} = \frac{1}{2}\begin{bmatrix}5&1&-2\\-4&0&2\\1&-1&0\end{bmatrix} = \begin{bmatrix}2.5&0.5&-1\\-2&0&1\\0.5&-0.5&0\end{bmatrix} \). Wait, but none of the options have this. Wait, maybe I made a mistake in the cofactor calculation. Wait, let's recalculate the cofactor of (3,1): the minor for (3,1) is the determinant of the submatrix when removing row 3, column 1: \( \begin{bmatrix}1&1\\1&-1\end{bmatrix} \), determinant is \( (1)(-1) - (1)(1) = -1 -1 = -2 \), so cofactor is \( (-1)^{3+1}*(-2) = 1*(-2) = -2 \), correct. Cofactor of (3,2): \( (-1)^{3+2}*\det\begin{bmatrix}1&1\\1&-1\end{bmatrix} = -1*(-2) = 2 \), correct. Cofactor of (3,3): \( (-1)^{3+3}*\det\begin{bmatrix}1&1\\1&1\end{bmatrix} = 1*0 = 0 \), correct. Adjugate is transpose: \( \begin{bmatrix}5&1&-2\\-4&0&2\\1&-1&0\end{bmatrix} \), so \( A^{-1} = \frac{1}{2}\begin{bmatrix}5&1&-2\\-4&0&2\\1&-1&0\end{bmatrix} = \begin{bmatrix}2.5&0.5&-1\\-2&0&1\\0.5&-0.5&0\end{bmatrix} \). Now, the solution is \( X = A^{-1}B \), where \( B = \begin{bmatrix}350\\130\\790\end{bmatrix} \). Wait, but option B has matrix \( \begin{bmatrix}2.5&0.5&-1\\-2&0&1\\0.5&-0.5&0\end{bmatrix} \) (wait, no—option B's matrix is \( \begin{bmatrix}2.5&0.5&-1\\-2&0&1\\0.5&-0.5&0\end{bmatrix} \)? Wait, option B is:

\( X = \begin{bmatrix}2.5&0.5&-1\\-2&0&1\\0.5&-0.5&0\end{bmatrix}\begin{bmatrix}350\\130\\790\end{bmatrix} \)

Let's compute that. First component: \( 2.5*350 + 0.5*130 -1*790 = 875 + 65 - 790 = 150 \) (which is \( x = 150 \)). Second component: \( -2*350 + 0*130 + 1*790 = -700 + 0 + 790 = 90 \) (which is \( y = 90 \)). Third component: \( 0.5*350 -0.5*130 + 0*790 = 175 - 65 + 0 = 110 \) (which is \( z = 110 \)). Yes! So option B's matrix is \( A^{-1} \) and \( V = B \). Wait, option B's matrix is \( \begin{bmatrix}2.5&0.5&-1\\-2&0&1\\0.5&-0.5&0\end{bmatrix} \) and \( V = \begin{bmatrix}350\\130\\790\end{bmatrix} \), which matches \( A^{-1}B \). So the correct option is B.

Wait, earlier I misread option B's matrix. Let's recheck option B:

Option B: \( X = \begin{bmatrix}2.5&0.5&-1\\-2&0&1\\0.5&-0.5&0\end{bmatrix}\begin{bmatrix}350\\130\\790\end{bmatrix} \)

Yes, that's \( A^{-1}B \), which gives the correct solution \( x = 150 \), \( y = 90 \), \( z = 110 \). So the correct answer is B.

Answer:

B. \( X =

$$\begin{bmatrix}2.5&0.5&-1\\-2&0&1\\0.5&-0.5&0\end{bmatrix}$$

$$\begin{bmatrix}350\\130\\790\end{bmatrix}$$

\)