This question tests your understanding of using matrix inverses to solve linear systems—a powerful method that reduces solving a system to a single matrix multiplication once the inverse is found. When a system is written as matrix equation $AX = B$ (where $A$ is coefficient matrix, $X$ is variable vector, $B$ is constant vector), we can solve for $X$ by multiplying both sides by the inverse matrix $A^{-1}$ (if it exists): $A^{-1} \times AX = A^{-1} \times B$. The left side simplifies to $(A^{-1} \times A) \times X = I \times X = X$ (the identity matrix $I$ acts like the number 1), giving $X = A^{-1} \times B$. This single matrix multiplication produces the solution vector! The method works if and only if $A$ has an inverse, which requires the determinant of $A$ to be nonzero. For 3×3 or larger matrices, we use technology (graphing calculator or computer) to find the inverse because hand calculation is extremely tedious—the standard explicitly acknowledges technology use for these sizes. The system $3x + y + z = 9$, $x + 4y + z = 9$, $x + y + 5z = 17$ gives us $$A = \begin{bmatrix} 3 & 1 & 1 \\ 1 & 4 & 1 \\ 1 & 1 & 5 \end{bmatrix}$$ and $$B = \begin{bmatrix} 9 \\ 9 \\ 17 \end{bmatrix}$$. Using technology to find $A^{-1}$ and compute $X = A^{-1} \times B$, we get $X = [1,2,3]$, so $(x,y,z) = (1,2,3)$. Choice A correctly identifies this solution. Choice B would give $(2,1,3)$, but checking: $3(2) + 1 + 3 = 10 \neq 9$, so this is incorrect. The matrix inverse solving process: (1) Write system as $AX = B$ (identify coefficient matrix $A$ and constant vector $B$), (2) Use technology to find $A^{-1}$, (3) Multiply: compute $A^{-1} \times B$ to get solution vector $X$, (4) Extract solutions: read x, y, z values as the ordered triple. Technology is essential here—manually finding the inverse of a 3×3 matrix requires computing 9 cofactors and dividing by the determinant, a process prone to arithmetic errors. Let the calculator handle the computation while you focus on setting up the problem correctly!

This question tests your understanding of when a matrix is invertible for using the inverse method to solve systems— a key prerequisite before applying $X = A^{-1} B$. When a system is written as $AX = B$, the inverse method requires $A$ to have an inverse, which exists if and only if the determinant of $A$ is nonzero; if $det A = 0$, $A$ is singular, has no inverse, and the method cannot be used (though the system might still have solutions via other means). For 3x3 matrices, computing the determinant by hand is feasible using the formula: expand along a row or use the general method, watching for linear dependence in rows/columns that makes $det=0$. In this case, rows 1 and 2 are linearly dependent ($row2 = 2*row1$), so $det A = 0$, confirming it's singular. Choice B correctly identifies that $A$ is not invertible, so the inverse method cannot be used. Choice C mistakenly claims invertibility because $det=0$, but actually $det=0$ means no inverse—remember, nonzero $det$ is required! To check invertibility: (1) Compute $det A$ using expansion or technology, (2) If $det ≠ 0$, proceed to find inverse; if $=0$, switch to row reduction for the augmented matrix to analyze solutions. This check prevents wasted effort on non-invertible matrices! You're building important skills—keep practicing determinant calculations for confidence.

This question tests your understanding of using matrix inverses to solve linear systems—a powerful method that reduces solving a system to a single matrix multiplication once the inverse is found. When a system is written as matrix equation $AX = B$, we can solve for $X$ by computing $X = A^{-1}B$, where $A^{-1}$ is the inverse of the coefficient matrix. For the 2×2 system $2x + y = 5$ and $3x + 4y = 7$, we have $$A = \begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix}$$, $$X = \begin{bmatrix} x \\ y \end{bmatrix}$$, and $$B = \begin{bmatrix} 5 \\ 7 \end{bmatrix}$$. To find $A^{-1}$, we first calculate $\det(A) = 2(4) - 1(3) = 8 - 3 = 5$, then $$A^{-1} = \frac{1}{5} \begin{bmatrix} 4 & -1 \\ -3 & 2 \end{bmatrix}$$. Computing $$X = A^{-1}B = \frac{1}{5} \begin{bmatrix} 4 & -1 \\ -3 & 2 \end{bmatrix} \begin{bmatrix} 5 \\ 7 \end{bmatrix} = \frac{1}{5} \begin{bmatrix} 4(5) - 1(7) \\ -3(5) + 2(7) \end{bmatrix} = \frac{1}{5} \begin{bmatrix} 13 \\ -1 \end{bmatrix} = \begin{bmatrix} \frac{13}{5} \\ -\frac{1}{5} \end{bmatrix}$$. Choice D correctly shows $(\frac{13}{5}, -\frac{1}{5})$ as the solution. Choices A, B, and C represent common errors like forgetting to divide by the determinant or making arithmetic mistakes in the matrix multiplication. The inverse method transforms the system-solving process into a single matrix multiplication—elegant and efficient!

This question tests your understanding of using matrix inverses to solve linear systems—a powerful method that reduces solving a system to a single matrix multiplication once the inverse is found. To solve the system x+y+z=6, 2x-y+z=3, x+2y-z=2, we first write it as AX = B where A = [[1,1,1], [2,-1,1], [1,2,-1]], X = [x,y,z], and B = [6,3,2]. Using technology to compute A^(-1) and then X = A^(-1) × B, we find X = [1,2,3], meaning x=1, y=2, z=3. Choice B correctly identifies this solution as the ordered triple (1,2,3). The other choices represent common errors: Choice A reverses two values, Choice C completely reverses the order, and Choice D swaps the last two values—these errors often occur when reading the solution vector carelessly or mixing up variable order. Always verify your solution by substituting back: 1+2+3=6 ✓, 2(1)-2+3=3 ✓, 1+2(2)-3=2 ✓. The matrix inverse method elegantly finds all three unknowns simultaneously through one matrix multiplication!

This question tests your understanding of using matrix inverses to solve linear systems—a powerful method that reduces solving a system to a single matrix multiplication once the inverse is found. When a system is written as matrix equation $AX = B$ (where A is coefficient matrix, X is variable vector, B is constant vector), we can solve for X by multiplying both sides by the inverse matrix $A^{-1}$ (if it exists): $A^{-1} \times AX = A^{-1} \times B$. The left side simplifies to $(A^{-1} \times A) \times X = I \times X = X$ (the identity matrix I acts like the number 1), giving $X = A^{-1} \times B$. This single matrix multiplication produces the solution vector! The method works if and only if A has an inverse, which requires the determinant of A to be nonzero. For 3×3 or larger matrices, we use technology (graphing calculator or computer) to find the inverse because hand calculation is extremely tedious—the standard explicitly acknowledges technology use for these sizes. The system 2x + y - z = 3, x - y + 2z = 5, 3x + 2y + z = 7 gives us A = [[2,1,-1],[1,-1,2],[3,2,1]] and B = [3,5,7]. Using technology to find $A^{-1}$ and compute $X = A^{-1} \times B$, we get X = [1,2,1], meaning x = 1, y = 2, z = 1. Choice A correctly identifies this solution vector. Choice B would give x = 2, y = 1, z = 1, but checking the first equation: 2(2) + 1 - 1 = 4 ≠ 3, so this is incorrect. The matrix inverse solving process: (1) Write system as $AX = B$ (organize equations into matrix form), (2) Use technology to find $A^{-1}$, (3) Multiply: compute $A^{-1} \times B$ to get solution vector X, (4) Extract solutions: read x, y, z values from the result. Technology is crucial for 3×3 systems—the calculator efficiently handles the inverse calculation that would otherwise require computing 9 cofactors and extensive arithmetic!

This question tests your understanding of using matrix inverses to solve linear systems—a powerful method that reduces solving a system to a single matrix multiplication once the inverse is found. For the 3×3 system given, we write it as $AX = B$ where $A = \begin{bmatrix} 2 & 1 & 1 \\ 1 & 3 & 2 \\ 3 & 1 & 4 \end{bmatrix}$, $X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$, and $B = \begin{bmatrix} 4 \\ 5 \\ 6 \end{bmatrix}$. Using technology to find $A^{-1}$ and compute $X = A^{-1}B$, we obtain the solution vector $X = \begin{bmatrix} \frac{5}{6} \\ \frac{5}{6} \\ \frac{7}{6} \end{bmatrix}$, which corresponds to choice D. Choices B and C show different arrangements of these fraction values, likely from misreading the solution vector order, while choice A suggests all variables equal 1, which doesn't satisfy the system. The matrix inverse method is particularly powerful for systems with fractional solutions—technology handles the arithmetic precisely where hand calculation might introduce rounding errors. For 3×3 systems, always use technology as the standard recommends—it ensures accuracy and saves tremendous time!

This question tests your understanding of using matrix inverses to solve linear systems—a powerful method that reduces solving a system to a single matrix multiplication once the inverse is found. When a system is written as matrix equation $AX = B$ (where A is coefficient matrix, X is variable vector, B is constant vector), we can solve for X by multiplying both sides by the inverse matrix $A^{-1}$ (if it exists): $A^{-1} AX = A^{-1} B$, simplifying to $X = A^{-1} B$. The method works if and only if A has an inverse, which requires the determinant of A to be nonzero; for 3x3 or larger matrices, we use technology to find the inverse. For this specific system, $$A = \begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \end{bmatrix}$$, $$B = \begin{bmatrix} 3 \\ 5 \\ 4 \end{bmatrix}$$, and using the inverse method yields $X = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$, which matches choice A. Remember to verify by plugging back into the original equations to ensure accuracy—you've got this! A common mistake is multiplying in the wrong order, like $B A^{-1}$ instead of $A^{-1} B$, since matrix multiplication isn't commutative. Keep practicing with technology, and this method will become second nature!

This question tests your understanding of using matrix inverses to solve linear systems—a powerful method that reduces solving a system to a single matrix multiplication once the inverse is found. For the 2×2 system with coefficient matrix A = [[2,1], [3,4]] and constant vector B = [5,7], we first find A^(-1). The determinant is 2(4)-1(3)=8-3=5. Using the 2×2 inverse formula, A^(-1) = (1/5)[[4,-1], [-3,2]]. Now we compute X = A^(-1) × B = (1/5)[[4,-1], [-3,2]] × [5,7] = (1/5)[4(5)-1(7), -3(5)+2(7)] = (1/5)[20-7, -15+14] = (1/5)[13,-1] = [13/5, -1/5]. Choice A correctly shows this solution vector. The other choices represent common calculation errors: forgetting to divide by the determinant, using the wrong inverse formula, or making arithmetic mistakes in the matrix multiplication.

This question tests your understanding of using matrix inverses to solve linear systems—a powerful method that reduces solving a system to a single matrix multiplication once the inverse is found. When a system is written as matrix equation AX = B (where A is coefficient matrix, X is variable vector, B is constant vector), we can solve for X by multiplying both sides by the inverse matrix A^(-1) (if it exists): A^(-1) × AX = A^(-1) × B, which simplifies to X = A^(-1) × B since A^(-1) × A = I (the identity matrix). For this system, we form A = [[2,1,-1], [1,-1,2], [3,2,1]] and B = [1,8,7]. Using technology to find A^(-1) and multiply by B, we get X = [2,1,3], meaning x=2, y=1, z=3. Choice D correctly identifies this solution vector. The other choices represent calculation errors or incorrect matrix setups—always double-check your coefficient matrix matches the system exactly, including signs! Technology makes this method practical for 3×3 systems: enter matrix A, find its inverse (usually INV or x^(-1) button), then multiply by B to get the solution instantly.

This question tests your understanding of using matrix inverses to solve linear systems—a powerful method that reduces solving a system to a single matrix multiplication once the inverse is found. When a system is written as matrix equation $AX = B$ (where $A$ is coefficient matrix, $X$ is variable vector, $B$ is constant vector), we can solve for $X$ by multiplying both sides by the inverse matrix $A^{-1}$ (if it exists): $A^{-1} \times AX = A^{-1} \times B$. The left side simplifies to $(A^{-1} \times A) \times X = I \times X = X$ (the identity matrix $I$ acts like the number 1), giving $X = A^{-1} \times B$. This single matrix multiplication produces the solution vector! The method works if and only if $A$ has an inverse, which requires the determinant of $A$ to be nonzero. For 3 by 3 or larger matrices, we use technology (graphing calculator or computer) to find the inverse because hand calculation is extremely tedious—the standard explicitly acknowledges technology use for these sizes. For this 3x3 system, use technology to compute $A^{-1}$ (where $A = \begin{bmatrix} 2 & 1 & 1 \\ 1 & -1 & 1 \\ 3 & 1 & -1 \end{bmatrix}$) and multiply by $B = \begin{bmatrix} 9 \\ 5 \\ 3 \end{bmatrix}$ to get $X = \begin{bmatrix} 2 \\ 1 \\ 4 \end{bmatrix}$—wonderful job! Choice A correctly finds the matrix inverse and performs the matrix multiplication $A^{-1} \times B$ to obtain the solution vector. Choice B makes an error, perhaps in transposing values during computation or misreading calculator output. Each step must be precise! The matrix inverse solving process: (1) Write system as $AX = B$, (2) Check if $A$ is invertible by attempting to find the inverse with technology, (3) Compute $A^{-1}$ using calculator, (4) Multiply $A^{-1} \times B$ to get $X$, (5) Extract solutions from the vector. Technology makes this method practical—use it confidently!