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 inverse (if it exists): A inverse times AX = A inverse times B. The left side simplifies to (A inverse times A) times X = I times X = X (the identity matrix I acts like the number 1), giving X = A inverse 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 2x2 system, det A = 5 ≠ 0, inverse is (1/5)[[4, -1], [-3, 2]], then X = (1/5)[13, -1] = [13/5, -1/5]. Choice A correctly computes the inverse by hand and performs the multiplication to obtain the solution. Choice B might result from forgetting to negate the off-diagonal elements or a sign error in multiplication, leading to a positive y-value that doesn't satisfy the equations. The matrix inverse solving process: (1) Write system as AX = B, (2) Check det A ≠ 0, (3) Find A inverse using the 2x2 formula, (4) Multiply A inverse times B, (5) Extract x and y. This method is elegant—one multiplication gives all variables simultaneously! Keep practicing these calculations—you're building a strong foundation!

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 $AX = B$ (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 = [[2,1,0],[1,2,1],[0,1,2]], B = [4,6,4], and using the inverse method yields $$X = \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix}$$, which matches choice B. 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. 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. 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 inverse (if it exists): A inverse times AX = A inverse times B. The left side simplifies to (A inverse times A) times X = I times X = X (the identity matrix I acts like the number 1), giving X = A inverse 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. This conceptual question reminds us that to isolate X in AX = B, we multiply both sides on the left by $A^{-1}$, yielding X = $A^{-1}$B—great foundational knowledge! Choice B correctly gives the expression for the solution vector as X = $A^{-1}$B. Choice A multiplies in the wrong order: it computes B times $A^{-1}$ instead of $A^{-1}$ times B. Matrix multiplication order matters—it's not commutative! The matrix inverse solving process: (1) Write system as AX = B (represent as matrix equation), (2) Check if A is invertible, (3) Find A inverse, (4) Multiply: compute A inverse times B to get solution vector X, (5) Extract solutions: read x, y, z values from the result vector. This method is elegant—one multiplication gives all variables simultaneously!

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 2 by 2 system like $2x + y = 5$ and $3x + 4y = 7$, write as matrix equation: $$\begin{pmatrix} 2 & 1 \\ 3 & 4 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 5 \\ 7 \end{pmatrix}$$. Find inverse of A = $$\begin{pmatrix} 2 & 1 \\ 3 & 4 \end{pmatrix}$$: determinant = $24 - 13 = 8 - 3 = 5$. Using the 2 by 2 inverse formula, $A^{-1} = \frac{1}{5} \begin{pmatrix} 4 & -1 \\ -3 & 2 \end{pmatrix}$. Now solve: $X = A^{-1}$ times $B = \frac{1}{5} \begin{pmatrix} 4 & -1 \\ -3 & 2 \end{pmatrix} \begin{pmatrix} 5 \\ 7 \end{pmatrix} = \frac{1}{5} \begin{pmatrix} 20 - 7 \\ -15 + 14 \end{pmatrix} = \frac{1}{5} \begin{pmatrix} 13 \\ -1 \end{pmatrix} = \begin{pmatrix} \frac{13}{5} \\ -\frac{1}{5} \end{pmatrix}$. You're doing great—keep verifying by plugging back in! Choice B correctly finds the matrix inverse and performs the matrix multiplication $A^{-1}$ times $B$ to obtain the solution vector $(13/5, -1/5)$. Choice C makes an error in applying the 2 by 2 inverse formula incorrectly, perhaps negating the wrong elements or miscalculating the determinant. Each step must be precise! The matrix inverse solving process: (1) Write system as $AX = B$ (represent as matrix equation), (2) Check if A is invertible: for 2 by 2, verify determinant not equal 0; for larger, attempt to find inverse with technology, (3) Find $A^{-1}$: for 2 by 2, use formula $(1$ over determinant) times $$\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$$ from $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$; for 3 by 3 or larger, use calculator inverse function, (4) Multiply: compute $A^{-1}$ times $B$ to get solution vector X, (5) Extract solutions: read x, y, z values from the result vector. This method is elegant—one multiplication gives all variables simultaneously!

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): $A^{-1} \cdot AX = A^{-1} \cdot B$. The left side simplifies to $(A^{-1} \cdot A) \cdot X = I \cdot X = X$ (the identity matrix $I$ acts like the number 1), giving $X = A^{-1} \cdot B$. 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, first check the determinant of $A = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 3 & 6 & 9 \end{bmatrix}$, which is 0 since rows are proportional, making $A$ singular with no inverse—excellent observation! Choice C correctly identifies that the coefficient matrix is singular (no inverse), so the inverse method cannot be used to find a unique solution. Choice A attempts to use the inverse method on a singular matrix, which fails because there's no inverse. Check determinant first: if it's 0, the method can't be used. The matrix inverse solving process: (1) Write system as $AX = B$, (2) Check if $A$ is invertible: verify determinant not equal 0, (3) If invertible, find $A^{-1}$ with technology, (4) Multiply $A^{-1}$ times $B$ to get $X$, (5) If singular, use other methods like row reduction to analyze solutions. This method is elegant when applicable—keep practicing!

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!

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 inverse (if it exists): A inverse times AX = A inverse times B. The left side simplifies to (A inverse times A) times X = I times X = X (the identity matrix I acts like the number 1), giving X = A inverse 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. Given $A^{-1}$ = (1/5)[[4,-1],[-3,2]] and B = [5,7], compute X = $A^{-1}$ B = (1/5)[13,-1] = [13/5, -1/5]—excellent matrix multiplication! Choice A correctly performs the matrix multiplication A inverse times B to obtain the solution vector. Choice C makes an error in the multiplication, perhaps forgetting the negative signs in the inverse matrix. Each step must be precise! The matrix inverse solving process: (1) Write system as AX = B (represent as matrix equation), (2) Check if A is invertible: for 2 by 2, verify determinant not equal 0; for larger, attempt to find inverse with technology, (3) Find A inverse: for 2 by 2, use formula (1 over determinant) times [[d, -b], [-c, a]] from A = [[a, b], [c, d]]; for 3 by 3 or larger, use calculator inverse function, (4) Multiply: compute A inverse times B to get solution vector X, (5) Extract solutions: read x, y, z values from the result vector. This method is elegant—one multiplication gives all variables simultaneously!

This question tests your understanding of using matrix inverses to solve a 2x2 linear system by hand—a powerful method that reduces solving the 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 2x2 system, $A = \begin{bmatrix} 3 & -2 \\ 5 & 1 \end{bmatrix}$, $\det(A) = 13$, $A^{-1} = \frac{1}{13} \begin{bmatrix} 1 & 2 \\ -5 & 3 \end{bmatrix}$, so $X = \frac{1}{13} \begin{bmatrix} 18 \\ 1 \end{bmatrix} = \begin{bmatrix} \frac{18}{13} \\ \frac{1}{13} \end{bmatrix}$, which is choice A. Choice A correctly finds the inverse using the 2x2 formula and multiplies $A^{-1}$ times $B$—excellent precision! Choice B makes an error by negating the off-diagonal incorrectly, leading to a negative y value. The matrix inverse solving process: (1) Write system as $AX = B$ (represent as matrix equation), (2) Check if $A$ is invertible: for 2 by 2, verify determinant not equal 0; for larger, attempt to find inverse with technology, (3) Find $A^{-1}$: for 2 by 2, use formula (1 over determinant) times $[[d, -b], [-c, a]]$ from $A = [[a, b], [c, d]]$; for 3 by 3 or larger, use calculator inverse function, (4) Multiply: compute $A^{-1}$ times $B$ to get solution vector $X$, (5) Extract solutions: read x, y, z values from the result vector. This method is elegant—one multiplication gives all variables simultaneously! Technology makes matrix inverse method practical: for 3 by 3 systems, hand-calculating the inverse requires finding 9 cofactors and dividing by the determinant (extremely tedious and error-prone). Instead, enter the coefficient matrix into your calculator, press the inverse button (usually x to the -1 or INV), then multiply by the constant vector. The calculator handles the messy arithmetic, letting you focus on understanding the method. The standard explicitly allows technology for 3 by 3 or greater—use it!