Linear Algebra Quiz: Solving Ax Equals B Via Inverses

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?

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 $A^{-1}$?

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?

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 $\vec{x}=A^{-1}\vec{b}$?

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 $X$?

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?

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 $x_1$?

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?

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 $\vec{x}$?

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 $\vec{y}$?

A student is solving the system $A\vec{x}=\vec{b}$ where $A=\begin{pmatrix} 4 & 2 \\ 5 & 3 \end{pmatrix}$ and $\vec{b}=\begin{pmatrix} 8 \\ 10 \end{pmatrix}$. Their work is shown below.

Step 1: Calculate determinant: $\det(A) = 4(3) - 2(5) = 12 - 10 = 2$. Step 2: Find the adjugate matrix: $\text{adj}(A) = \begin{pmatrix} 3 & -2 \\ -5 & 4 \end{pmatrix}$. Step 3: Determine the inverse: $A^{-1} = \begin{pmatrix} 3 & -2 \\ -5 & 4 \end{pmatrix}$. Step 4: Calculate the solution: $\vec{x} = A^{-1}\vec{b} = \begin{pmatrix} 3 & -2 \\ -5 & 4 \end{pmatrix}\begin{pmatrix} 8 \\ 10 \end{pmatrix} = \begin{pmatrix} 4 \\ 0 \end{pmatrix}$.

In which step did the student first make an error?

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 $(2A)y = b$?

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?

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?

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}

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 $x_1$?

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}

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 $x = M^{-1}c$?

Matrix $A$ satisfies $A^2 = I + 2A$. To solve $Ax = b$ using the inverse method, what is the correct expression for $A^{-1}$?

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?