Solving Ax=b via Inverses - Linear Algebra

Consider the system of equations $3x + 2y = 7$ and $4x + 3y = 9$ . If this system is represented as $A\vec{x}=\vec{b}$ , and the solution is found using $\vec{x}=A^{-1}\vec{b}$ , what is the matrix $A^{-1}$ ?
Answer: $\begin{pmatrix} 3 & -2 \\ -4 & 3 \end{pmatrix}$ The matrix for the system is $A = \begin{pmatrix} 3 & 2 \\ 4 & 3 \end{pmatrix}$ . To find the inverse, first calculate the determinant: $\det(A) = (3)(3) - (2)(4) = 9 - 8 = 1$ . The formula for the inverse of a 2x2 matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ is $\frac{1}{ad-bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$ . Applying this, $A^{-1} = \frac{1}{1}\begin{pmatrix} 3 & -2 \\ -4 & 3 \end{pmatrix} = \begin{pmatrix} 3 & -2 \\ -4 & 3 \end{pmatrix}$ .
For an invertible matrix $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ , the solution to $A\vec{x} = \begin{pmatrix} p \\ q \end{pmatrix}$ is $\vec{x} = \begin{pmatrix} x_1 \\ x_2 \end{pmatrix}$ . Which expression represents $x_1$ ?
Answer: $x_1 = \frac{dp - bq}{ad-bc}$ First, find the inverse of $A$ . The determinant is $\det(A)=ad-bc$ . The inverse is $A^{-1} = \frac{1}{ad-bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$ . The solution is $\vec{x} = A^{-1}\vec{b}$ . So, $\begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \frac{1}{ad-bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \begin{pmatrix} p \\ q \end{pmatrix} = \frac{1}{ad-bc} \begin{pmatrix} dp - bq \\ -cp + aq \end{pmatrix}$ . The first component of the solution vector is $x_1 = \frac{dp - bq}{ad-bc}$ .
For the system $Ax = b$ where $A = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$ and $b = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$ , the solution using $x = A^{-1}b$ is $x = \begin{pmatrix} \cos\theta \\ \sin\theta \end{pmatrix}$ . What does this solution represent geometrically?
Answer: Rotating the vector $b$ clockwise by angle $\theta$ to obtain the solution vector $x$ Matrix $A$ is a rotation matrix that rotates vectors counterclockwise by $\theta$ . Since $x = A^{-1}b$ , we have $Ax = b$ , meaning rotating $x$ counterclockwise by $\theta$ gives $b = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$ . Therefore, $x$ is obtained by rotating $b$ clockwise by $\theta$ . Indeed, $A^{-1} = A^T = \begin{pmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{pmatrix}$ rotates clockwise by $\theta$ . Choice A incorrectly states counterclockwise rotation. Choice C describes projection, not rotation. Choice D describes reflection, which is incorrect.
The solution to the system $A\vec{x}=\vec{b}$ is $\vec{x}=\begin{pmatrix} 5 \\ -2 \end{pmatrix}$ . If the same matrix $A$ is used in a new system $A\vec{y}=\vec{c}$ , where $\vec{c} = 3\vec{b}$ , what is the solution vector $\vec{y}$ ?
Answer: $\begin{pmatrix} 15 \\ -6 \end{pmatrix}$ From the first system, we know that $\vec{x} = A^{-1}\vec{b}$ . The solution to the new system is $\vec{y} = A^{-1}\vec{c}$ . Since we are given that $\vec{c} = 3\vec{b}$ , we can substitute this into the equation for $\vec{y}$ : $\vec{y} = A^{-1}(3\vec{b})$ . By the properties of matrix multiplication, the scalar can be factored out: $\vec{y} = 3(A^{-1}\vec{b})$ . Since $A^{-1}\vec{b}=\vec{x}$ , we have $\vec{y} = 3\vec{x}$ . Therefore, $\vec{y} = 3\begin{pmatrix} 5 \\ -2 \end{pmatrix} = \begin{pmatrix} 15 \\ -6 \end{pmatrix}$ .
A system of linear equations $A\vec{x}=\vec{b}$ is solved using the inverse matrix $A^{-1}$ . Given $A^{-1} = \begin{pmatrix} 2 & 5 \\ 1 & 3 \end{pmatrix}$ and $\vec{b} = \begin{pmatrix} -1 \\ 2 \end{pmatrix}$ , what is the solution vector $\vec{x}$ ?
Answer: $\begin{pmatrix} 8 \\ 5 \end{pmatrix}$ The solution is given by the formula $\vec{x} = A^{-1}\vec{b}$ . We compute the matrix-vector product: $\vec{x} = \begin{pmatrix} 2 & 5 \\ 1 & 3 \end{pmatrix} \begin{pmatrix} -1 \\ 2 \end{pmatrix} = \begin{pmatrix} (2)(-1) + (5)(2) \\ (1)(-1) + (3)(2) \end{pmatrix} = \begin{pmatrix} -2 + 10 \\ -1 + 6 \end{pmatrix} = \begin{pmatrix} 8 \\ 5 \end{pmatrix}$
The matrix equation $AXB = C$ can be solved for $X$ when $A$ and $B$ are invertible. If this equation is rewritten as a system of the form $Mx = c$ where $x$ contains the entries of $X$ , which matrix $M$ allows solution via $x = M^{-1}c$ ?
Answer: $M = B^T \otimes A$ where $\otimes$ denotes the Kronecker product When you encounter matrix equations like $AXB = C$ , the key is understanding how to vectorize them using the Kronecker product. This transformation allows you to convert a matrix equation into a standard linear system that can be solved using matrix inversion. The vectorization of $AXB = C$ follows a specific rule: $\text{vec}(AXB) = (B^T \otimes A)\text{vec}(X)$ . This identity comes from the fundamental property that $\text{vec}(PQR) = (R^T \otimes P)\text{vec}(Q)$ . Applying this to our equation, we get $(B^T \otimes A)\text{vec}(X) = \text{vec}(C)$ , which gives us the system $Mx = c$ where $M = B^T \otimes A$ . Answer D is correct because it provides the matrix $M = B^T \otimes A$ that emerges from proper vectorization of the original equation. Answer A is wrong because $A \otimes B$ doesn't account for the transpose needed in the vectorization formula. Answer B is incorrect because $AB$ uses standard multiplication, which doesn't preserve the structure needed for vectorization—the dimensions won't even match in most cases. Answer C is close but has the Kronecker product terms reversed; $A \otimes B^T$ doesn't follow the correct vectorization rule. Remember this pattern: when vectorizing $AXB = C$ , the coefficient matrix is always $(B^T \otimes A)$ —the rightmost matrix gets transposed and appears first in the Kronecker product. This reversal and transpose rule is essential for solving matrix equations via vectorization.
Matrix $A$ satisfies $A^2 = I + 2A$ . To solve $Ax = b$ using the inverse method, what is the correct expression for $A^{-1}$ ?
Answer: $A^{-1} = A - 2I$ When you encounter a matrix equation like $A^2 = I + 2A$ , you're dealing with finding a matrix inverse through algebraic manipulation. The key insight is to rearrange this equation to isolate terms that will help you express $A^{-1}$ . Starting with $A^2 = I + 2A$ , subtract $2A$ from both sides: $A^2 - 2A = I$ . Now factor the left side: $A(A - 2I) = I$ . This is the crucial step—you've found that $A$ multiplied by $(A - 2I)$ equals the identity matrix $I$ . By definition of matrix inverses, if $AB = I$ , then $B = A^{-1}$ . Therefore, $A^{-1} = A - 2I$ . Choice A ( $A^{-1} = I + 2A$ ) comes from incorrectly assuming you can just move terms around without proper algebraic manipulation. If this were correct, then $A(I + 2A) = I$ , which would mean $A + 2A^2 = I$ —but substituting our original equation shows this doesn't work. Choice B ( $A^{-1} = \frac{1}{2}(A - I)$ ) likely results from incorrect factoring or arithmetic errors in the algebraic manipulation. Choice D ( $A^{-1} = 2I - A$ ) is close to the correct answer but has the wrong sign. This might come from incorrectly rearranging $A(A - 2I) = I$ as $(2I - A)A = I$ . Study tip: When finding matrix inverses from polynomial equations, always factor to get the form $A \cdot (\text{expression}) = I$ . The expression in parentheses is your inverse. Double-check by verifying that your candidate inverse actually satisfies $AA^{-1} = I$ .
The system $Ax = b$ has solution $x^*$ . A student claims that if $\|A^{-1}\|$ is large, then small changes in $b$ will cause large changes in the solution $x^*$ . To test this claim using the inverse method $x = A^{-1}b$ , which approach is most appropriate?
Answer: Compute $\|A^{-1}(b + \delta b) - A^{-1}b\|$ for small perturbations $\delta b$ and compare to $\|A^{-1}\|\|\delta b\|$ The change in solution is $\Delta x = A^{-1}(b + \delta b) - A^{-1}b = A^{-1}\delta b$ . So $\|\Delta x\| = \|A^{-1}\delta b\| \leq \|A^{-1}\|\|\delta b\|$ . Testing this relationship directly validates the student's claim about sensitivity. Choice A looks at changes in norms rather than changes in vectors. Choice C incorrectly relates determinant to matrix norm and sensitivity. Choice D confuses existence of solution with stability of solution.
Let $A$ be an invertible $3 \times 3$ matrix. If one were to solve the system $A\vec{x} = \vec{e_2}$ , where $\vec{e_2} = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}$ is the second standard basis vector, what would the solution vector $\vec{x}$ represent?
Answer: The second column of $A^{-1}$ . The solution to the system is given by $\vec{x} = A^{-1}\vec{b}$ . In this case, $\vec{b} = \vec{e_2}$ . So, $\vec{x} = A^{-1}\vec{e_2}$ . The product of a matrix and the $j$ -th standard basis vector, $\vec{e_j}$ , results in the $j$ -th column of that matrix. Therefore, $A^{-1}\vec{e_2}$ is precisely the second column of the matrix $A^{-1}$ .
Consider two systems: $Ax = b_1$ with solution $x_1 = A^{-1}b_1$ and $Ax = b_2$ with solution $x_2 = A^{-1}b_2$ . If $b_2 = b_1 + \alpha Ax_1$ for some scalar $\alpha$ , what is $x_2$ in terms of $x_1$ ?
Answer: $x_2 = x_1 + \alpha x_1 = (1 + \alpha)x_1$ When you encounter linear systems with related right-hand sides, the key insight is recognizing how changes in the input vector affect the solution. This question tests your understanding of how matrix inverses distribute over vector addition. Let's work through this systematically. We know that $x_2 = A^{-1}b_2$ , and we're given that $b_2 = b_1 + \alpha Ax_1$ . Substituting this expression: $x_2 = A^{-1}(b_1 + \alpha Ax_1) = A^{-1}b_1 + A^{-1}(\alpha Ax_1) = A^{-1}b_1 + \alpha A^{-1}Ax_1$ Since $A^{-1}A = I$ (the identity matrix), we have $A^{-1}Ax_1 = x_1$ . Also, since $x_1 = A^{-1}b_1$ , our expression becomes: $x_2 = x_1 + \alpha x_1 = (1 + \alpha)x_1$ This confirms answer C is correct. Looking at the wrong answers: A) $x_2 = (1 + \alpha)A^{-1}x_1$ incorrectly applies the inverse to $x_1$ instead of recognizing that $A^{-1}b_1 = x_1$ . B) $x_2 = x_1 + \alpha A^{-1}x_1$ stops one step short of the simplification, failing to recognize that $A^{-1}Ax_1 = x_1$ . D) $x_2 = x_1 + \alpha A^{-2}x_1$ incorrectly squares the inverse, showing a fundamental misunderstanding of matrix multiplication. Study tip: When working with linear systems and matrix inverses, always look for opportunities to use the fact that $A^{-1}A = I$ . This identity is your key to simplifying expressions involving both a matrix and its inverse.
A system $Ax = b$ has a unique solution that can be expressed as $x = A^{-1}b$ . If matrix $A$ is replaced by $2A$ , what expression gives the solution to the new system $(2A)y = b$ ?
Answer: $y = \frac{1}{2}A^{-1}b$ For the system $(2A)y = b$ , we need $(2A)^{-1}$ . Since $(2A)^{-1} = \frac{1}{2}A^{-1}$ , the solution is $y = (2A)^{-1}b = \frac{1}{2}A^{-1}b$ . Choice B incorrectly multiplies by 2 instead of $\frac{1}{2}$ . Choice C ignores the scalar multiplication entirely. Choice D is technically correct notation but doesn't simplify the expression as requested.
Consider the block matrix $M = \begin{pmatrix} A & 0 \\ 0 & B \end{pmatrix}$ where $A$ and $B$ are invertible $2 \times 2$ matrices. To solve $M\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} u \\ v \end{pmatrix}$ using $M^{-1}$ , which statement is correct?
Answer: $M^{-1} = \begin{pmatrix} A^{-1} & 0 \\ 0 & B^{-1} \end{pmatrix}$ and the system decouples into $Ax = u$ and $By = v$ For a block diagonal matrix $M = \begin{pmatrix} A & 0 \\ 0 & B \end{pmatrix}$ , the inverse is $M^{-1} = \begin{pmatrix} A^{-1} & 0 \\ 0 & B^{-1} \end{pmatrix}$ . The system $M\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} u \\ v \end{pmatrix}$ decouples into two independent systems: $Ax = u$ and $By = v$ . Choice B gives wrong form for block diagonal inverse. Choice C confuses block diagonal with general $2 \times 2$ matrix inverse formula. Choice D correctly identifies $M^{-1}$ but incorrectly claims the system cannot decouple.
Let $A = \begin{pmatrix} 1 & 2 \\ 2 & 4 \end{pmatrix}$ . An attempt to solve the system $A\vec{x}=\vec{b}$ using the inverse method fails. What is the fundamental reason for this failure?
Answer: The matrix $A$ is singular because its columns are linearly dependent. The inverse method requires the matrix $A$ to be invertible. A matrix is invertible if and only if its determinant is non-zero. The determinant of $A$ is $\det(A) = (1)(4) - (2)(2) = 4 - 4 = 0$ . Since the determinant is zero, $A$ is singular and not invertible. This singularity is a result of the columns being linearly dependent (the second column is twice the first column). This is the fundamental reason the inverse method fails.
To solve the matrix equation $XA = B$ for the matrix $X$ , where $A$ and $B$ are known $n \times n$ matrices and $A$ is invertible, one must perform a matrix multiplication involving $A^{-1}$ and $B$ . Which expression correctly calculates $X$ ?
Answer: $X = BA^{-1}$ To isolate the matrix $X$ in the equation $XA = B$ , we need to multiply by $A^{-1}$ . Since matrix multiplication is not commutative, the side on which we multiply matters. To cancel $A$ , we must multiply by $A^{-1}$ on the right side of both sides of the equation: $XAA^{-1} = BA^{-1}$ . This simplifies to $XI = BA^{-1}$ , so $X = BA^{-1}$ . Choice A is the solution to $AX=B$ .
A $3 \times 3$ matrix $A$ has the property that $A^{-1} = \frac{1}{2}A^T$ . If the system $Ax = e_1$ (where $e_1 = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$ ) has solution $x = \begin{pmatrix} a \\ b \\ c \end{pmatrix}$ , what constraint must hold?
Answer: $a^2 + b^2 + c^2 = \frac{1}{2}$ This question tests your understanding of orthogonal matrices and their properties. When you see that $A^{-1} = \frac{1}{2}A^T$ , you're dealing with a scaled orthogonal matrix, which means the columns (and rows) of $A$ have special geometric relationships. To find the constraint, start with the given relationship $A^{-1} = \frac{1}{2}A^T$ . Since $AA^{-1} = I$ , we have: $A \cdot \frac{1}{2}A^T = I$ $\frac{1}{2}AA^T = I$ $AA^T = 2I$ Now, since $Ax = e_1$ where $x = \begin{pmatrix} a \\ b \\ c \end{pmatrix}$ , we can use the relationship $AA^T = 2I$ . The first column of $AA^T$ equals $A$ times the first column of $A^T$ (which is the first row of $A$ ). But we also know that the first column of $AA^T$ is the first column of $2I$ , which is $\begin{pmatrix} 2 \\ 0 \\ 0 \end{pmatrix}$ . Since $Ax = e_1$ , we have $x = A^{-1}e_1 = \frac{1}{2}A^T e_1$ , which means $x$ is half the first column of $A^T$ (or half the first row of $A$ ). The constraint $AA^T = 2I$ tells us that the dot product of the first row of $A$ with itself equals 2. Since $x$ is half this row, we get $a^2 + b^2 + c^2 = \frac{1}{2}$ . Choice A gives $\frac{1}{4}$ (dividing by 4 instead of 2), choice B gives 2 (forgetting the scaling), and choice C gives 1 (assuming standard orthogonal matrix). Choice D correctly accounts for the $\frac{1}{2}$ scaling factor. Study tip: When working with scaled orthogonal matrices, always track how the scaling affects inner products and norms throughout your calculations.
Consider the system of equations defined by $A\vec{x}=\vec{b}$ where $A = \begin{pmatrix} k & 4 \\ 9 & k \end{pmatrix}$ . For which value(s) of $k$ is it NOT possible to find a unique solution using the formula $\vec{x}=A^{-1}\vec{b}$ ?
Answer: $k=6$ and $k=-6$ The inverse method is not possible when the matrix $A$ is singular, which occurs when its determinant is zero. The determinant of $A$ is $\det(A) = (k)(k) - (4)(9) = k^2 - 36$ . Setting the determinant to zero gives the equation $k^2 - 36 = 0$ . This factors as $(k-6)(k+6)=0$ , which has solutions $k=6$ and $k=-6$ . For these values of $k$ , the matrix $A$ is not invertible.
In which step did the student first make an error?
Answer: Step 3 The formula for the inverse matrix is $A^{-1} = \frac{1}{\det(A)}\text{adj}(A)$ . The student correctly calculated the determinant in Step 1 and the adjugate matrix in Step 2. However, in Step 3, the student failed to divide the adjugate matrix by the determinant (which is 2). The correct inverse should be $A^{-1} = \frac{1}{2}\begin{pmatrix} 3 & -2 \\ -5 & 4 \end{pmatrix}$ . This is the first error in the process.
The system of linear equations $A\vec{x}=\vec{b}$ is to be solved using the matrix inverse method, yielding the unique solution $\vec{x}=A^{-1}\vec{b}$ . Which of the following conditions is necessary for this method to be applicable?
Answer: The determinant of matrix $A$ must be non-zero. The formula $\vec{x}=A^{-1}\vec{b}$ requires the existence of the inverse matrix $A^{-1}$ . A square matrix $A$ has an inverse if and only if it is non-singular, which is equivalent to its determinant being non-zero. The other conditions are not necessary. Symmetry is not required for invertibility. The method works for $\vec{b}=\vec{0}$ (giving $\vec{x}=\vec{0}$ ). For a unique solution via matrix inversion, $A$ must be a square matrix (number of equations equals number of variables).
Let $A$ and $B$ be invertible $n \times n$ matrices. To solve the matrix equation $AB\vec{x} = \vec{c}$ for the vector $\vec{x}$ , which of the following computations is correct?
Answer: $\vec{x} = B^{-1}A^{-1}\vec{c}$ To isolate $\vec{x}$ , we need to multiply both sides by the inverse of the matrix product $AB$ . The inverse of a product of matrices is the product of their inverses in reverse order: $(AB)^{-1} = B^{-1}A^{-1}$ . We must left-multiply both sides of the equation by this inverse: $(AB)^{-1}(AB)\vec{x} = (AB)^{-1}\vec{c}$ , which simplifies to $I\vec{x} = B^{-1}A^{-1}\vec{c}$ , so $\vec{x} = B^{-1}A^{-1}\vec{c}$ .
A bakery sells two types of boxes: a pastry box with 3 croissants and 4 muffins, and a breakfast box with 5 croissants and 2 muffins. On a given day, they use a total of 160 croissants and 100 muffins to fill an order. Let $p$ be the number of pastry boxes and $b$ be the number of breakfast boxes. The situation is modeled by the matrix equation $A\begin{pmatrix} p \\ b \end{pmatrix} = \begin{pmatrix} 160 \\ 100 \end{pmatrix}$ . Which of the following expressions correctly solves for the number of boxes?
Answer: $\frac{1}{-14} \begin{pmatrix} 2 & -5 \\ -4 & 3 \end{pmatrix} \begin{pmatrix} 160 \\ 100 \end{pmatrix}$ The system of equations is $3p + 5b = 160$ (croissants) and $4p + 2b = 100$ (muffins). The coefficient matrix is $A = \begin{pmatrix} 3 & 5 \\ 4 & 2 \end{pmatrix}$ . The solution is given by $\begin{pmatrix} p \\ b \end{pmatrix} = A^{-1}\begin{pmatrix} 160 \\ 100 \end{pmatrix}$ . We must calculate $A^{-1}$ . The determinant is $\det(A) = (3)(2) - (5)(4) = 6 - 20 = -14$ . The inverse is $A^{-1} = \frac{1}{-14}\begin{pmatrix} 2 & -5 \\ -4 & 3 \end{pmatrix}$ . Therefore, the correct expression for the solution is $\frac{1}{-14} \begin{pmatrix} 2 & -5 \\ -4 & 3 \end{pmatrix} \begin{pmatrix} 160 \\ 100 \end{pmatrix}$ .

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