2x2 Matrix Inverses - Finite Mathematics

Given the matrix equation $XA = B$ , where $A = \begin{pmatrix} 3 & -2 \\ -4 & 3 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & 1 \\ 2 & 3 \end{pmatrix}$ , the solution for $X$ is found by computing $X = BA^{-1}$ . Which of the following matrices is $X$ ?
Answer: $\begin{pmatrix} 7 & 5 \\ 18 & 13 \end{pmatrix}$ To solve for $X$ , we must first calculate $A^{-1}$ . The determinant is $\det(A) = (3)(3) - (-2)(-4) = 9 - 8 = 1$ . The inverse is $A^{-1} = \frac{1}{1} \begin{pmatrix} 3 & 2 \\ 4 & 3 \end{pmatrix} = \begin{pmatrix} 3 & 2 \\ 4 & 3 \end{pmatrix}$ . Now, we compute the product $X = BA^{-1}$ : $X = \begin{pmatrix} 1 & 1 \\ 2 & 3 \end{pmatrix} \begin{pmatrix} 3 & 2 \\ 4 & 3 \end{pmatrix} = \begin{pmatrix} 1(3)+1(4) & 1(2)+1(3) \\ 2(3)+3(4) & 2(2)+3(3) \end{pmatrix} = \begin{pmatrix} 3+4 & 2+3 \\ 6+12 & 4+9 \end{pmatrix} = \begin{pmatrix} 7 & 5 \\ 18 & 13 \end{pmatrix}$ .
For what value of $c$ does the linear system represented by the matrix equation $\begin{pmatrix} 2 & -6 \\ -3 & c \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 5 \\ 1 \end{pmatrix}$ fail to have a unique solution?
Answer: $9$ A system of linear equations $A\mathbf{x}=\mathbf{b}$ has a unique solution if and only if the coefficient matrix $A$ is invertible. The system fails to have a unique solution if $A$ is singular, which means its determinant is zero. The coefficient matrix is $A = \begin{pmatrix} 2 & -6 \\ -3 & c \end{pmatrix}$ . We calculate its determinant: $\det(A) = (2)(c) - (-6)(-3) = 2c - 18$ . To find the value of $c$ for which there is no unique solution, we set the determinant to zero: $2c - 18 = 0$ . Solving for $c$ gives $2c = 18$ , so $c = 9$ .
Matrix $A = \begin{pmatrix} 2 & 3 \\ 1 & 4 \end{pmatrix}$ represents the coefficients of a system of equations. If the system has solution $\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 5 \\ -1 \end{pmatrix}$ , what is the constant vector $\mathbf{b}$ in the system $A\mathbf{x} = \mathbf{b}$ ?
Answer: $\begin{pmatrix} 7 \\ 1 \end{pmatrix}$ To find $\mathbf{b}$ , we multiply $A$ by the solution vector: $A\mathbf{x} = \begin{pmatrix} 2 & 3 \\ 1 & 4 \end{pmatrix} \begin{pmatrix} 5 \\ -1 \end{pmatrix} = \begin{pmatrix} 2(5) + 3(-1) \\ 1(5) + 4(-1) \end{pmatrix} = \begin{pmatrix} 10 - 3 \\ 5 - 4 \end{pmatrix} = \begin{pmatrix} 7 \\ 1 \end{pmatrix}$ . Choice B incorrectly uses the inverse formula components. Choice C uses only the first calculation row. Choice D forgets to compute the second component correctly.
Let $A = \begin{pmatrix} k & 5 \\ -1 & -k \end{pmatrix}$ . If the inverse of matrix $A$ is equal to matrix $A$ itself (i.e., $A^{-1} = A$ ), what is the value of $k^2$ ?
Answer: $6$ If $A^{-1} = A$ , then multiplying by $A$ on the left gives $A A^{-1} = A^2$ , which means $I = A^2$ , where $I$ is the identity matrix. Let's compute $A^2$ : $A^2 = \begin{pmatrix} k & 5 \\ -1 & -k \end{pmatrix} \begin{pmatrix} k & 5 \\ -1 & -k \end{pmatrix} = \begin{pmatrix} k(k) + 5(-1) & k(5) + 5(-k) \\ -1(k) + (-k)(-1) & -1(5) + (-k)(-k) \end{pmatrix} = \begin{pmatrix} k^2 - 5 & 0 \\ 0 & k^2 - 5 \end{pmatrix}$ . We set this equal to the identity matrix $I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ . Equating the diagonal entries gives the equation $k^2 - 5 = 1$ . Solving for $k^2$ yields $k^2 = 6$ .
A system of linear equations is given by $5x - 3y = 15$ and $3x - 2y = 8$ . The solution to the system can be found using the matrix equation $A\mathbf{x}=\mathbf{b}$ . What is the value of $y$ in the solution?
Answer: $5$ The system can be written as $A\mathbf{x}=\mathbf{b}$ with $A = \begin{pmatrix} 5 & -3 \\ 3 & -2 \end{pmatrix}$ , $\mathbf{x} = \begin{pmatrix} x \\ y \end{pmatrix}$ , and $\mathbf{b} = \begin{pmatrix} 15 \\ 8 \end{pmatrix}$ . The solution is $\mathbf{x} = A^{-1}\mathbf{b}$ . First, find $A^{-1}$ . The determinant is $\det(A) = (5)(-2) - (-3)(3) = -10 - (-9) = -1$ . The inverse is $A^{-1} = \frac{1}{-1} \begin{pmatrix} -2 & 3 \\ -3 & 5 \end{pmatrix} = \begin{pmatrix} 2 & -3 \\ 3 & -5 \end{pmatrix}$ . Now, calculate the solution vector: $\mathbf{x} = \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 2 & -3 \\ 3 & -5 \end{pmatrix} \begin{pmatrix} 15 \\ 8 \end{pmatrix} = \begin{pmatrix} 2(15) + (-3)(8) \\ 3(15) + (-5)(8) \end{pmatrix} = \begin{pmatrix} 30 - 24 \\ 45 - 40 \end{pmatrix} = \begin{pmatrix} 6 \\ 5 \end{pmatrix}$ . The value of $y$ is the second entry in the solution vector, which is 5.
If $A = \begin{pmatrix} 3 & 1 \\ 4 & 2 \end{pmatrix}$ , what is the sum of all elements of the matrix $A + A^{-1}$ ?
Answer: $10.5$ First, we need to find the inverse of matrix $A$ . The determinant of $A$ is $\det(A) = (3)(2) - (1)(4) = 6 - 4 = 2$ . The inverse is $A^{-1} = \frac{1}{2} \begin{pmatrix} 2 & -1 \\ -4 & 3 \end{pmatrix} = \begin{pmatrix} 1 & -0.5 \\ -2 & 1.5 \end{pmatrix}$ . Next, we compute the sum of the matrices $A + A^{-1}$ : $A + A^{-1} = \begin{pmatrix} 3 & 1 \\ 4 & 2 \end{pmatrix} + \begin{pmatrix} 1 & -0.5 \\ -2 & 1.5 \end{pmatrix} = \begin{pmatrix} 3+1 & 1-0.5 \\ 4-2 & 2+1.5 \end{pmatrix} = \begin{pmatrix} 4 & 0.5 \\ 2 & 3.5 \end{pmatrix}$ . Finally, we find the sum of all elements of this resulting matrix: $4 + 0.5 + 2 + 3.5 = 10.5$ .
The inverse of the matrix $A = \begin{pmatrix} 3 & x \\ 4 & 6 \end{pmatrix}$ is $A^{-1} = \begin{pmatrix} -3 & 2.5 \\ 2 & -1.5 \end{pmatrix}$ . What is the value of $x$ ?
Answer: $5$ The product of a matrix and its inverse is the identity matrix, i.e., $AA^{-1} = I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ . We can use this property to solve for $x$ . Let's compute the entry in the first row, second column of the product $AA^{-1}$ : $\begin{pmatrix} 3 & x \\ 4 & 6 \end{pmatrix} \begin{pmatrix} -3 & 2.5 \\ 2 & -1.5 \end{pmatrix}$ . The entry is $(3)(2.5) + (x)(-1.5) = 7.5 - 1.5x$ . This must be equal to the corresponding entry in the identity matrix, which is 0. So, we set up the equation $7.5 - 1.5x = 0$ . Solving for $x$ gives $1.5x = 7.5$ , which means $x = 5$ .
The system of linear equations $3x + 5y = -1$ and $2x + 4y = 2$ is represented by the matrix equation $A\mathbf{x} = \mathbf{b}$ . The solution is given by $\mathbf{x} = A^{-1}\mathbf{b}$ . What is the entry in the first row of the solution vector $\mathbf{x}$ ?
Answer: $-7$ First, identify the coefficient matrix $A = \begin{pmatrix} 3 & 5 \\ 2 & 4 \end{pmatrix}$ and the constant vector $\mathbf{b} = \begin{pmatrix} -1 \\ 2 \end{pmatrix}$ . To find the solution $\mathbf{x} = A^{-1}\mathbf{b}$ , we must first compute $A^{-1}$ . The determinant is $\det(A) = (3)(4) - (5)(2) = 12 - 10 = 2$ . The inverse is $A^{-1} = \frac{1}{2} \begin{pmatrix} 4 & -5 \\ -2 & 3 \end{pmatrix}$ . The solution is $\mathbf{x} = \frac{1}{2} \begin{pmatrix} 4 & -5 \\ -2 & 3 \end{pmatrix} \begin{pmatrix} -1 \\ 2 \end{pmatrix} = \frac{1}{2} \begin{pmatrix} 4(-1) + (-5)(2) \\ -2(-1) + 3(2) \end{pmatrix} = \frac{1}{2} \begin{pmatrix} -14 \\ 8 \end{pmatrix} = \begin{pmatrix} -7 \\ 4 \end{pmatrix}$ . The entry in the first row of the solution vector is $x = -7$ .
The matrix equation $\begin{pmatrix} 3 & k \\ 1 & 2 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 5 \\ 3 \end{pmatrix}$ has solution $x = 1$ . For what value of $k$ does $y = 1$ as well?
Answer: $k = 2$ If $x = 1$ and $y = 1$ , then substituting into the matrix equation: $\begin{pmatrix} 3 & k \\ 1 & 2 \end{pmatrix}\begin{pmatrix} 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 5 \\ 3 \end{pmatrix}$ . This gives us $\begin{pmatrix} 3 + k \\ 1 + 2 \end{pmatrix} = \begin{pmatrix} 5 \\ 3 \end{pmatrix}$ . From the first component: $3 + k = 5$ , so $k = 2$ . We can verify with the second component: $1 + 2 = 3$ ✓. The other choices would give different values for the first component that don't equal 5.
Let $P = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}$ and $Q = \begin{pmatrix} 3 & 0 \\ 1 & 1 \end{pmatrix}$ . What is the entry in the first row, second column of the matrix $(PQ)^{-1}$ ?
Answer: $-2/3$ To find $(PQ)^{-1}$ , we first compute the product $PQ$ . $PQ = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 3 & 0 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} 1(3)+2(1) & 1(0)+2(1) \\ 0(3)+1(1) & 0(0)+1(1) \end{pmatrix} = \begin{pmatrix} 5 & 2 \\ 1 & 1 \end{pmatrix}$ . Next, we find the inverse of this resulting matrix. The determinant is $\det(PQ) = (5)(1) - (2)(1) = 3$ . The inverse is $(PQ)^{-1} = \frac{1}{3} \begin{pmatrix} 1 & -2 \\ -1 & 5 \end{pmatrix} = \begin{pmatrix} 1/3 & -2/3 \\ -1/3 & 5/3 \end{pmatrix}$ . The entry in the first row, second column is $-2/3$ .
Let $A$ be a $2 \times 2$ matrix such that $A^{-1} = \begin{pmatrix} -1/2 & 1/2 \\ 3/2 & -1/2 \end{pmatrix}$ . If the system of equations $A\mathbf{x} = \begin{pmatrix} 4 \\ 6 \end{pmatrix}$ is solved for $\mathbf{x} = \begin{pmatrix} x \\ y \end{pmatrix}$ , what is the value of $x+y$ ?
Answer: $4$ The solution to the system $A\mathbf{x} = \mathbf{b}$ is given by $\mathbf{x} = A^{-1}\mathbf{b}$ . We compute: $\mathbf{x} = \begin{pmatrix} -1/2 & 1/2 \\ 3/2 & -1/2 \end{pmatrix} \begin{pmatrix} 4 \\ 6 \end{pmatrix} = \begin{pmatrix} (-1/2)(4) + (1/2)(6) \\ (3/2)(4) + (-1/2)(6) \end{pmatrix} = \begin{pmatrix} -2 + 3 \\ 6 - 3 \end{pmatrix} = \begin{pmatrix} 1 \\ 3 \end{pmatrix}$ . Therefore, $x = 1$ and $y = 3$ , so $x + y = 1 + 3 = 4$ .
Let $A = \begin{pmatrix} 1 & -2 \\ -3 & 7 \end{pmatrix}$ . If its inverse is $A^{-1} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ , what is the value of the expression $a - d$ ?
Answer: $6$ First, compute the inverse of matrix $A$ . The determinant is $\det(A) = (1)(7) - (-2)(-3) = 7 - 6 = 1$ . The formula for the inverse of a $2 \times 2$ matrix $\begin{pmatrix} w & x \\ y & z \end{pmatrix}$ is $\frac{1}{wz-xy} \begin{pmatrix} z & -x \\ -y & w \end{pmatrix}$ . Applying this to $A$ , we get $A^{-1} = \frac{1}{1} \begin{pmatrix} 7 & -(-2) \\ -(-3) & 1 \end{pmatrix} = \begin{pmatrix} 7 & 2 \\ 3 & 1 \end{pmatrix}$ . By comparing this to $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ , we identify $a=7$ , $b=2$ , $c=3$ , and $d=1$ . The required expression is $a - d = 7 - 1 = 6$ .
If $P = \begin{pmatrix} 2 & 1 \\ 3 & 2 \end{pmatrix}$ and $Q = \begin{pmatrix} 5 & 4 \\ 1 & 2 \end{pmatrix}$ , what is the determinant of $(PQ)^{-1}$ ?
Answer: $\frac{1}{6}$ Using the property that $\det((PQ)^{-1}) = \frac{1}{\det(PQ)} = \frac{1}{\det(P)\det(Q)}$ . We have $\det(P) = 2(2) - 1(3) = 4 - 3 = 1$ and $\det(Q) = 5(2) - 4(1) = 10 - 4 = 6$ . Therefore $\det(PQ) = \det(P)\det(Q) = 1 \cdot 6 = 6$ , so $\det((PQ)^{-1}) = \frac{1}{6}$ . Choice B gives $\det(PQ)$ instead of its reciprocal. Choice C incorrectly squares the reciprocal. Choice D incorrectly squares the determinant.
Let matrices $A = \begin{pmatrix} 2 & 1 \\ 5 & 3 \end{pmatrix}$ and $C = \begin{pmatrix} 1 & 0 \\ -1 & 2 \end{pmatrix}$ . If the matrix equation $AX = C$ holds, what is the sum of the elements in the first column of matrix $X$ ?
Answer: $-3$ To solve for matrix $X$ in the equation $AX = C$ , we multiply both sides on the left by $A^{-1}$ , yielding $X = A^{-1}C$ . First, we must find the inverse of $A$ . The determinant of $A$ is $\det(A) = (2)(3) - (1)(5) = 6 - 5 = 1$ . The inverse is $A^{-1} = \frac{1}{1} \begin{pmatrix} 3 & -1 \\ -5 & 2 \end{pmatrix} = \begin{pmatrix} 3 & -1 \\ -5 & 2 \end{pmatrix}$ . Next, we compute the product $X = A^{-1}C = \begin{pmatrix} 3 & -1 \\ -5 & 2 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ -1 & 2 \end{pmatrix} = \begin{pmatrix} 3(1)+(-1)(-1) & 3(0)+(-1)(2) \\ -5(1)+2(-1) & -5(0)+2(2) \end{pmatrix} = \begin{pmatrix} 4 & -2 \\ -7 & 4 \end{pmatrix}$ . The elements in the first column of $X$ are 4 and -7. Their sum is $4 + (-7) = -3$ .

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