Solving Systems Using Matrix Inverses - Algebra 2

Use technology to compute $A^{-1}$ and solve $AX=B$ using $X=A^{-1}B$ . $ A=\begin{bmatrix}4&0&1\\2&1&0\\0&3&1\end{bmatrix},\quad B=\begin{bmatrix}9\\5\\9\end{bmatrix} $ What is $X$ ?
Answer: $\begin{bmatrix}2\\1\\1\end{bmatrix}$ 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. The left side simplifies to ( $A^{-1}$ × A) × X = I × X = X (the identity matrix I acts like the number 1), giving X = $A^{-1}$ × 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. Given A = [[4,0,1],[2,1,0],[0,3,1]] and B = [9,5,9], we use technology to find $A^{-1}$ , then compute X = $A^{-1}$ × B. Enter matrix A into your calculator, find its inverse using the INV or $x^{-1}$ function, then multiply by vector B. The calculation yields X = [2,1,1], meaning x = 2, y = 1, z = 1. Choice A correctly identifies this solution vector. Choice B would give x = 1, y = 2, z = 1, but checking the first equation: 4(1) + 0(2) + 1(1) = 5 ≠ 9, so this is incorrect. The matrix inverse solving process: (1) Write system as AX = B (organize coefficients into matrix A, constants into vector B), (2) Use technology to find $A^{-1}$ , (3) Multiply: compute $A^{-1}$ × B to get solution vector X, (4) Extract solutions: read x, y, z values from the result vector. Technology makes the inverse method practical—for a 3×3 matrix, the calculator efficiently computes the inverse that would take many error-prone steps by hand, letting you focus on the problem setup and interpretation!
Solve the system using the inverse of the coefficient matrix. $\begin{cases} x+3y=7\\ 2x+5y=12 \end{cases}$ Let $A=\begin{bmatrix}1&3\\2&5\end{bmatrix}$ , $X=\begin{bmatrix}x\\y\end{bmatrix}$ , and $B=\begin{bmatrix}7\\12\end{bmatrix}$ . Find $A^{-1}$ and compute $X=A^{-1}B$ .
Answer: $A^{-1}=\begin{bmatrix}-5&3\\2&-1\end{bmatrix},\; X=\begin{bmatrix}1\\2\end{bmatrix}$ 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 coefficient matrix A = [[1,3],[2,5]], we first find the determinant: det(A) = 1(5) - 3(2) = 5 - 6 = -1. Using the 2×2 inverse formula, A⁻¹ = (1/-1)[[5,-3],[-2,1]] = [[-5,3],[2,-1]] (swap diagonal elements, negate off-diagonal elements, divide by determinant). Now compute X = A⁻¹B = [[-5,3],[2,-1]][[7],[12]] = [[-5(7)+3(12)],[2(7)+(-1)(12)]] = [[-35+36],[14-12]] = [[1],[2]]. Choice A correctly shows both A⁻¹ = [[-5,3],[2,-1]] and X = [1,2], meaning x = 1, y = 2. Choices B and C have errors in either the inverse calculation or the final multiplication, while Choice D incorrectly computes the determinant as 11 instead of -1. The matrix inverse method elegantly solves for all variables with one multiplication—but every arithmetic step must be precise!
Use technology to solve the system by the inverse-matrix method. Write it as $AX=B$ and compute $X=A^{-1}B$ . $\begin{cases} 2x+y-z=3\\ x-y+2z=5\\ 3x+2y+z=7 \end{cases}$ What is $X=\begin{bmatrix}x\\y\\z\end{bmatrix}$ ?
Answer: $\begin{bmatrix}1\\2\\1\end{bmatrix}$ 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!
A matrix $A$ is invertible if there exists $A^{-1}$ such that $AA^{-1}=I$ . For the system below, determine whether the coefficient matrix is invertible and, if it is, solve using $X=A^{-1}B$ (use technology). $\begin{cases} x+2y+3z=1 \\ 2x+4y+6z=2 \\ 3x+6y+9z=3 \end{cases}$ Which statement is correct?
Answer: The coefficient matrix is singular (no inverse), so the inverse method cannot be used to find a unique solution. 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!
Use technology to solve the system by the inverse-matrix method. Write $AX=B$ and compute $X=A^{-1}B$ . $\begin{cases} 2x+y+z=4 \\ x+3y+2z=5 \\ 3x+y+4z=6 \end{cases}$ What is $X=\begin{bmatrix} x \\ y \\ z \end{bmatrix}$ ?
Answer: $\begin{bmatrix} \tfrac{5}{6} \\ \tfrac{5}{6} \\ \tfrac{7}{6} \end{bmatrix}$ 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!
Solve the $2\times 2$ system using the inverse-matrix method. First find the inverse of the coefficient matrix, then compute $X=A^{-1}B$ . $\begin{cases} 2x+y=5\\ 3x+4y=7 \end{cases}$ What is $X=\begin{bmatrix}x\\y\end{bmatrix}$ ?
Answer: $\begin{bmatrix}\frac{13}{5}\\-\frac{1}{5}\end{bmatrix}$ 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.
Use technology to solve $AX=B$ by computing $X=A^{-1}B$ . $A=\begin{bmatrix}2&1&0\\1&3&1\\0&2&4\end{bmatrix},\quad B=\begin{bmatrix}5\\10\\16\end{bmatrix}$ What is $X$ ?
Answer: $\begin{bmatrix}1\\2\\3\end{bmatrix}$ 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. Given $A = \begin{bmatrix} 2 & 1 & 0 \\ 1 & 3 & 1 \\ 0 & 2 & 4 \end{bmatrix}$ and $B = \begin{bmatrix} 5 \\ 10 \\ 16 \end{bmatrix}$ , we use technology to find $A^{-1}$ , then compute $X = A^{-1} \times B$ . Enter matrix $A$ into your calculator, find its inverse using the INV or $x^{-1}$ function, then multiply by vector $B$ . The calculation yields $X = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$ , meaning $x = 1$ , $y = 2$ , $z = 3$ . Choice A correctly identifies this solution vector. Choice B would give $x = 2$ , $y = 1$ , $z = 3$ , which doesn't satisfy the original system—always verify your answer! The matrix inverse solving process: (1) Write system as $AX = B$ (represent as matrix equation), (2) Check if $A$ is invertible: for 3×3, use technology to attempt finding inverse, (3) Find $A^{-1}$ : 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. Technology makes matrix inverse method practical: for 3×3 systems, the calculator handles finding the inverse (which involves computing 9 cofactors and dividing by the determinant). This lets you focus on understanding the method rather than getting bogged down in arithmetic!
The inverse of matrix $A = \begin{pmatrix} 2 & 1 & 0 \\ 1 & 2 & 1 \\ 0 & 1 & 2 \end{pmatrix}$ contains the entry $\frac{3}{4}$ in position $(1,1)$ . If this matrix is used to solve the system $A\vec{x} = \begin{pmatrix} 8 \\ 6 \\ 4 \end{pmatrix}$ , and the solution has integer coordinates, what is the sum of all coordinates in the solution vector?
Answer: $6$ from adding the solution components The determinant of A is 4, and $A^{-1} = \frac{1}{4}\begin{pmatrix} 3 & -2 & 1 \\ -2 & 4 & -2 \\ 1 & -2 & 3 \end{pmatrix}$ . Computing $\vec{x} = A^{-1}\vec{b} = \frac{1}{4}\begin{pmatrix} 3 & -2 & 1 \\ -2 & 4 & -2 \\ 1 & -2 & 3 \end{pmatrix}\begin{pmatrix} 8 \\ 6 \\ 4 \end{pmatrix} = \begin{pmatrix} 2 \\ 0 \\ 4 \end{pmatrix}$ . The sum is $2 + 0 + 4 = 6$ . Choice B results from computational errors in matrix multiplication. Choice C comes from doubling the correct answer. Choice D results from major calculation mistakes.
The inverse of a matrix is given. Use the inverse-matrix method to solve $AX=B$ by computing $X=A^{-1}B$ . $A^{-1}=\begin{bmatrix}1&-2\0&1\end{bmatrix},\quad B=\begin{bmatrix}5\3\end{bmatrix}$ What is $X$ ?
Answer: $\begin{bmatrix}-1\\3\end{bmatrix}$ 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 given $A^{-1}$ directly, we simply compute $X = A^{-1}B$ without needing to find the inverse ourselves. Here, $A^{-1} = \begin{bmatrix} 1 & -2 \\ 0 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} 5 \\ 3 \end{bmatrix}$ , so $X = A^{-1}B = \begin{bmatrix} 1 & -2 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 5 \\ 3 \end{bmatrix} = \begin{bmatrix} 1(5) + (-2)(3) \\ 0(5) + 1(3) \end{bmatrix} = \begin{bmatrix} 5 - 6 \\ 0 + 3 \end{bmatrix} = \begin{bmatrix} -1 \\ 3 \end{bmatrix}$ . Choice A correctly shows $[-1,3]$ as the solution. Choice B might result from sign errors, choice C from incorrect matrix multiplication, and choice D from reversing the components. The beauty of having $A^{-1}$ given is that we skip the most computationally intensive step—finding the inverse—and go straight to the multiplication. This problem emphasizes that once you have the inverse matrix, solving the system is just one matrix multiplication away!
A system is written in matrix form as $AX=B$ . Use technology to find $A^{-1}$ and then compute $X=A^{-1}B$ . $A=\begin{bmatrix}1&1&1\\0&2&1\\2&-1&1\end{bmatrix},\quad B=\begin{bmatrix}6\\7\\3\end{bmatrix}$ What is the solution vector $X$ ?
Answer: $\begin{bmatrix}1\\2\\3\end{bmatrix}$ 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. With $A = \begin{bmatrix} 1 & 1 & 1 \\ 0 & 2 & 1 \\ 2 & -1 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} 6 \\ 7 \\ 3 \end{bmatrix}$ , we need to find $A^{-1}$ using technology, then multiply $A^{-1} \times B$ . Using a calculator: enter matrix $A$ , find its inverse (INV or $x^{-1}$ button), then multiply by $B$ . The calculation yields $X = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$ , meaning $x = 1$ , $y = 2$ , $z = 3$ . Choice B correctly identifies this solution vector. Choice A would give $x = 2$ , $y = 1$ , $z = 3$ , which doesn't satisfy the equations—substituting into the first equation: $2 + 1 + 3 = 6 \checkmark$ , but into the second: $0(2) + 2(1) + 3 = 5 \neq 7 \times$ . The matrix inverse solving process: (1) Write system as $AX = B$ (represent as matrix equation), (2) Check if $A$ is invertible: use technology to verify inverse exists, (3) Find $A^{-1}$ : 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. Technology makes this method practical for 3×3 systems—the calculator efficiently handles the complex inverse calculation, allowing you to focus on the conceptual understanding of how matrix inverses solve systems!
Use technology (for a $3\times 3$ matrix) to solve the system by writing it as $AX=B$ and using $X=A^{-1}B$ . \begin{align*} x+y+z&=6\\ 2x-y+z&=3\\ x+2y-z&=2 \end{align*} What is $(x,y,z)$ ?
Answer: $(1,2,3)$ 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!
Solve the system using the inverse of the coefficient matrix. $\begin{cases} 2x+y=5\\ 3x+4y=7 \end{cases}$ Let $A=\begin{bmatrix}2&1\\3&4\end{bmatrix}$ , $X=\begin{bmatrix}x\\y\end{bmatrix}$ , and $B=\begin{bmatrix}5\\7\end{bmatrix}$ . Find $A^{-1}$ and then compute $X=A^{-1}B$ .
Answer: $A^{-1}=\frac{1}{5}\begin{bmatrix}4&-1\\-3&2\end{bmatrix},\; X=\begin{bmatrix}\frac{13}{5}\\-\frac{1}{5}\end{bmatrix}$ 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 solve for X by computing X = A⁻¹B, where A⁻¹ is the inverse of the coefficient matrix. For the 2×2 matrix A = [[2,1],[3,4]], we first find the determinant: det(A) = 2(4) - 1(3) = 8 - 3 = 5. Using the 2×2 inverse formula, A⁻¹ = (1/5)[[4,-1],[-3,2]] (swap diagonal, negate off-diagonal, divide by determinant). Now compute X = A⁻¹B = (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 both the inverse matrix and the solution vector. Choice B has the wrong inverse (didn't negate the off-diagonal elements), Choice C has the wrong determinant in the scalar factor, and Choice D has arithmetic errors in the final multiplication—each step must be precise!
A $3 \times 3$ system is given in matrix form. Use technology to find $A^{-1}$ and then compute $X = A^{-1} B$ . $A = \begin{bmatrix} 2 & -1 & 0 \\ 1 & 2 & 1 \\ 3 & 0 & 1 \end{bmatrix}, \quad B = \begin{bmatrix} 1 \\ 8 \\ 11 \end{bmatrix}$ What is $X$ ?
Answer: $\begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$ 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 $3 \times 3$ 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 $A = \begin{bmatrix} 2 & -1 & 0 \\ 1 & 2 & 1 \\ 3 & 0 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} 1 \\ 8 \\ 11 \end{bmatrix}$ , we use technology to find $A^{-1}$ , then compute $X = A^{-1} \times B$ . Enter matrix $A$ into your calculator, find its inverse using the INV or $x^{-1}$ function, then multiply by vector $B$ . The calculation yields $X = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$ , meaning $x = 1$ , $y = 2$ , $z = 3$ . Choice B correctly identifies this solution vector. Choice A would give the same values but in wrong order, while choices C and D have different permutations—the order matters in a solution vector! The matrix inverse solving process: (1) Write system as $AX = B$ (set up the matrix equation), (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 in order from the result vector. Technology makes this method efficient—the calculator handles the complex inverse calculation, allowing you to focus on proper problem setup and solution interpretation!
Solve the system using the inverse of the coefficient matrix. Write $AX=B$ and compute $X=A^{-1}B$ . $\begin{cases} 4x - y = 9 \\ 2x + y = 7 \end{cases}$ What is $(x,y)$ ?
Answer: $(\tfrac{8}{3},\tfrac{5}{3})$ 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 $2 \times 2$ system $4x - y = 9$ and $2x + y = 7$ , we write it as $AX = B$ where $A = [[4,-1],[2,1]]$ , $X = [x,y]$ , and $B = [9,7]$ . First, we find $\det(A) = 4(1) - (-1)(2) = 4 + 2 = 6$ , so A is invertible. Using the 2×2 inverse formula, $A^{-1} = (1/6)[[1,1],[-2,4]]$ . Computing $X = A^{-1}B = (1/6)[[1,1],[-2,4]][[9],[7]] = (1/6)[[1(9)+1(7)],[-2(9)+4(7)]] = (1/6)[[16],[10]] = [16/6, 10/6] = [8/3, 5/3]$ . Choice B correctly shows $(8/3, 5/3)$ as the solution. Choices A and C show integer solutions that don't satisfy the system, while choice D reverses the fraction values. The inverse method transforms solving the system into a straightforward matrix multiplication—once you have $A^{-1}$ , the solution follows immediately!
Use technology to find $A^{-1}$ and solve $AX=B$ by computing $X=A^{-1}B$ . $A=\begin{bmatrix}1&1&1\\0&1&1\\0&0&1\end{bmatrix},\quad B=\begin{bmatrix}6\\5\\3\end{bmatrix}$ What is $X$ ?
Answer: $X=\begin{bmatrix}1\\2\\3\end{bmatrix}$ This question tests your understanding of using matrix inverses to solve a system given the coefficient matrix and B vector with technology—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 $X = A^{-1} B$ . 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 system, A is upper triangular with $\det A = 1$ , so invertible; computing $A^{-1}$ times B gives $X = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$ , which matches B when multiplied back by A. Choice A correctly performs the matrix multiplication $A^{-1}$ times B to obtain the solution vector—nice work! Choice B might result from confusing X with B or not multiplying at all; always remember to apply the inverse. 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 $\frac{1}{\det} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$ from A = $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ ; 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!
A student writes $AX=B$ for a system and claims the solution is found by multiplying both sides on the left by $A^{-1}$ . Which expression correctly gives the solution vector (assuming $A$ is invertible and $AA^{-1}=I$ )?
Answer: $X=A^{-1}B$ 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!
Use technology to solve the system by computing $X=A^{-1}B$ . $\begin{cases} 3x+y+z=9\\ x+4y+z=9\\ x+y+5z=17 \end{cases}$ In matrix form, $AX=B$ with $A=\begin{bmatrix}3&1&1\\1&4&1\\1&1&5\end{bmatrix},\quad B=\begin{bmatrix}9\\9\\17\end{bmatrix}.$ What is $(x,y,z)$ ?
Answer: $(1,2,3)$ 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!
A system is written in matrix form $AX=B$ . Determine whether $A$ is invertible (has an inverse). If it is invertible, you may solve using $X=A^{-1}B$ . $A=\begin{bmatrix}1&2&3\\2&4&6\\1&0&1\end{bmatrix},\quad X=\begin{bmatrix}x\\y\\z\end{bmatrix},\quad B=\begin{bmatrix}1\\2\\3\end{bmatrix}.$ Which statement is true?
Answer: $A$ is not invertible (singular), so $A^{-1}$ does not exist and the inverse method cannot be used. 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.
Solve using matrix inverses. Find $A^{-1}$ (by hand, $2\times2$ formula) and compute $X=A^{-1}B$ . $\begin{cases} 3x-2y=4 \\ 5x+y=7 \end{cases}$ What is $(x,y)$ ?
Answer: $(x,y)=\left(\frac{18}{13},\frac{1}{13}\right)$ 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!
Use technology to solve the $3 \times 3$ system by the inverse-matrix method: write $AX=B$ and compute $X=A^{-1}B$ . $\begin{cases} 2x-y+z=2\\ x+y+z=6\\ 3x-2y+4z=8 \end{cases}$ What is $X=\begin{bmatrix}x\\y\\z\end{bmatrix}$ ?
Answer: $\begin{bmatrix}1\\2\\3\end{bmatrix}$ 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 $3 \times 3$ system given, we write it as $AX = B$ where $A = \begin{bmatrix} 2 & -1 & 1 \\ 1 & 1 & 1 \\ 3 & -2 & 4 \end{bmatrix}$ , $X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$ , and $B = \begin{bmatrix} 2 \\ 6 \\ 8 \end{bmatrix}$ . Using technology (as required for $3 \times 3$ systems), we find $A^{-1}$ and compute $X = A^{-1} B$ to get the solution vector. The calculation yields $X = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$ , which matches choice B. Choice A reverses the first two components, while choices C and D show different permutations of the values—these likely result from matrix entry errors or calculation mistakes. The matrix inverse method is particularly powerful for $3 \times 3$ systems because it avoids the tedious elimination process. Always use technology for $3 \times 3$ or larger matrices—the standard explicitly acknowledges this necessity since hand calculation of a $3 \times 3$ inverse requires finding 9 cofactors and is extremely error-prone!

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