Cramer's Rule - Linear Algebra

1. Use Cramer's rule to find the value of $y$ for the following system of linear equations: $\begin{cases} x + y + z = 6 \\ 2x - y + z = 3 \\ x + 2y - z = 2 \end{cases}$
   Answer: 2 When you encounter a system of linear equations and need to find a specific variable using Cramer's rule, you're applying a powerful method that uses determinants to solve for individual variables without finding the entire solution set. Cramer's rule states that for a system $Ax = b$, the value of variable $y$ equals $\frac{\det(A_y)}{\det(A)}$, where $A_y$ is the coefficient matrix with the $y$-column replaced by the constants vector. First, find the determinant of the coefficient matrix: $A = \begin{pmatrix} 1 & 1 & 1 \\ 2 & -1 & 1 \\ 1 & 2 & -1 \end{pmatrix}$ $\det(A) = 1(-1 \cdot (-1) - 1 \cdot 2) - 1(2 \cdot (-1) - 1 \cdot 1) + 1(2 \cdot 2 - (-1) \cdot 1) = 1(1-2) - 1(-2-1) + 1(4+1) = -1 + 3 + 5 = 7$ Next, replace the second column ($y$-coefficients) with the constants vector: $A_y = \begin{pmatrix} 1 & 6 & 1 \\ 2 & 3 & 1 \\ 1 & 2 & -1 \end{pmatrix}$ $\det(A_y) = 1(3 \cdot (-1) - 1 \cdot 2) - 6(2 \cdot (-1) - 1 \cdot 1) + 1(2 \cdot 2 - 3 \cdot 1) = 1(-5) - 6(-3) + 1(1) = -5 + 18 + 1 = 14$ Therefore: $y = \frac{\det(A_y)}{\det(A)} = \frac{14}{7} = 2$ Choice A (1), B (1/2), and C (3) result from computational errors in calculating determinants or incorrectly applying Cramer's rule—perhaps confusing which column to replace or making arithmetic mistakes. Study tip: Always double-check your determinant calculations by expanding along different rows or columns. The most common errors in Cramer's rule stem from determinant computation mistakes, not conceptual misunderstandings.
2. The linear system $\begin{cases} x - 2y + 4z = 0 \\ 2x + y - z = 0 \\ -x - 8y + kz = 0 \end{cases}$ is known to have non-trivial solutions. According to Cramer's Rule, what condition must be met, and what is the value of $k$?
   Answer: $\det(A) = 0$, which implies $k = 14$. When you encounter a homogeneous linear system (where all constants equal zero) that has non-trivial solutions, you're dealing with a fundamental concept about matrix determinants and solution existence. For any homogeneous system $A\mathbf{x} = \mathbf{0}$, non-trivial solutions exist if and only if the coefficient matrix is singular, meaning $\det(A) = 0$. This occurs because when the determinant is zero, the matrix lacks full rank, creating a non-trivial null space. Let's find the determinant of the coefficient matrix: $A = \begin{pmatrix} 1 & -2 & 4 \\ 2 & 1 & -1 \\ -1 & -8 & k \end{pmatrix}$ Expanding along the first row: $\det(A) = 1(k - 8) - (-2)(-2k - 1) + 4(-16 + 1)$ $= k - 8 - 4k - 2 - 60 = -3k - 70$ For non-trivial solutions, we need $\det(A) = 0$: $-3k - 70 = 0$ $k = -\frac{70}{3} = 14$ Therefore, answer C is correct: $\det(A) = 0$ and $k = 14$. Answer A incorrectly states that $\det(A) \neq 0$, which would mean only the trivial solution exists. Answer B has the right determinant condition but calculates $k$ incorrectly. Answer D confuses "inconsistent" with "having non-trivial solutions" — homogeneous systems are always consistent since $\mathbf{x} = \mathbf{0}$ is always a solution. Study tip: Remember that for homogeneous systems, "non-trivial solutions" always means $\det(A) = 0$. This is a direct application of the fundamental theorem connecting determinants to solution uniqueness.
3. In solving a system of two linear equations for variables $x$ and $y$ using Cramer's rule, the following determinants are computed: $D = \det \begin{pmatrix} 2 & -5 \\ 1 & 3 \end{pmatrix}$, $D_x = \det \begin{pmatrix} 8 & -5 \\ -2 & 3 \end{pmatrix}$, and $D_y = \det \begin{pmatrix} 2 & 8 \\ 1 & -2 \end{pmatrix}$. What is the solution $(x, y)$ to the system?
   Answer: ($14/11$, $-12/11$) First, we evaluate each determinant: $D = (2)(3) - (-5)(1) = 6 + 5 = 11$. $D_x = (8)(3) - (-5)(-2) = 24 - 10 = 14$. $D_y = (2)(-2) - (8)(1) = -4 - 8 = -12$. According to Cramer's rule, $x = \frac{D_x}{D}$ and $y = \frac{D_y}{D}$. Thus, $x = \frac{14}{11}$ and $y = \frac{-12}{11}$. The solution is $(14/11, -12/11)$.
4. When using Cramer's rule to solve the system below for $y$, what is the value of the numerator determinant, $\det(A_y)$? $\begin{cases} 3x + y - 2z = 4 \\ x - 5z = 0 \\ 2x - y + z = -1 \end{cases}$
   Answer: -57 First, write the system with all coefficients, including the zero for the missing $y$ term in the second equation: $x + 0y - 5z = 0$. The coefficient matrix is $A = \begin{pmatrix} 3 & 1 & -2 \\ 1 & 0 & -5 \\ 2 & -1 & 1 \end{pmatrix}$ and the constant vector is $\mathbf{b} = \begin{pmatrix} 4 \\ 0 \\ -1 \end{pmatrix}$. To find $\det(A_y)$, we replace the second (y-coefficient) column of $A$ with $\mathbf{b}$: $A_y = \begin{pmatrix} 3 & 4 & -2 \\ 1 & 0 & -5 \\ 2 & -1 & 1 \end{pmatrix}$. We calculate its determinant, for instance by cofactor expansion along the second row: $\det(A_y) = -1 \cdot \det\begin{pmatrix} 4 & -2 \\ -1 & 1 \end{pmatrix} + 0 - (-5) \cdot \det\begin{pmatrix} 3 & 4 \\ 2 & -1 \end{pmatrix}$ $= -1(4(1) - (-2)(-1)) + 5(3(-1) - 4(2))$ $= -1(4 - 2) + 5(-3 - 8) = -1(2) + 5(-11) = -2 - 55 = -57$.
5. Consider the system of equations: $\begin{cases} 2x - 3y + 7z = 9 \\ 5y - 2z = -1 \\ 3z = 6 \end{cases}$ If one were to use Cramer's rule to find $x$, what would be the value of the determinant of the coefficient matrix, $\det(A)$?
   Answer: 30 When you encounter Cramer's rule problems, you need to find the determinant of the coefficient matrix — the matrix formed by the coefficients of the variables on the left side of your system. From this system, the coefficient matrix $A$ is: $A = \begin{pmatrix} 2 & -3 & 7 \\ 0 & 5 & -2 \\ 0 & 0 & 3 \end{pmatrix}$ Notice this is an upper triangular matrix (all entries below the main diagonal are zero). For triangular matrices, the determinant equals the product of the diagonal entries: $\det(A) = 2 \times 5 \times 3 = 30$. Let's examine why the other answers are incorrect: A) 0 — This would occur if the matrix were singular (non-invertible), meaning the system has either no solution or infinitely many solutions. Since we have a triangular system with non-zero diagonal entries, the system has a unique solution. B) -30 — This might result from incorrectly handling signs when expanding the determinant or miscalculating the product of diagonal elements. The diagonal entries are clearly positive: $2 \times 5 \times 3 = 30$, not $-30$. D) 108 — This could come from incorrectly including the constants from the right side of the equations (9, -1, 6) in your calculation, but remember that the coefficient matrix $A$ only contains the coefficients of $x$, $y$, and $z$. Study tip: Always identify the structure of your matrix first. Upper or lower triangular matrices make determinant calculations much easier — just multiply the diagonal entries. Don't let the constants on the right side distract you from isolating the coefficient matrix.
6. For a $3 \times 3$ system where Cramer's rule applies, suppose $\det(A) = 12$ and the determinant obtained by replacing the third column with the constants vector is $-36$. If the system is modified by doubling all coefficients in the third equation only, how does this affect the value of $z$?
   Answer: The new value of $z$ remains $\frac{-36}{24} = -\frac{3}{2}$, unchanged from the original value despite coefficient changes Originally, $z = \frac{-36}{12} = -3$. When we double all coefficients in the third equation, both $\det(A)$ and $D_z$ are affected. Doubling the third row doubles the determinant of the coefficient matrix, so the new $\det(A) = 24$. Similarly, doubling the third row in $D_z$ doubles that determinant, so the new $D_z = -72$. Therefore, the new $z = \frac{-72}{24} = -3$, which is the same as the original value. The doubling effect cancels out in the ratio. Choice A gives the wrong original value. Choice B miscalculates both values. Choice D incorrectly suggests the value doubles.
7. A system of two linear equations, $a_1x + b_1y = c_1$ and $a_2x + b_2y = c_2$, is represented graphically by two distinct lines. If the determinant of the coefficient matrix $\begin{pmatrix} a_1 & b_1 \\ a_2 & b_2 \end{pmatrix}$ is zero, what must be true about the two lines?
   Answer: The lines are parallel. If the determinant of the coefficient matrix is zero, the system does not have a unique solution. Geometrically, for a system of two linear equations in two variables, this means the lines are either parallel or coincident. Since the problem states that the lines are 'distinct', they cannot be coincident. Therefore, the lines must be parallel.
8. When attempting to solve a linear system $A\mathbf{x} = \mathbf{b}$ using Cramer's rule, a student finds that $\det(A) = 0$ and also that $\det(A_i) = 0$ for every variable $x_i$. What is the correct conclusion to draw about the system?
   Answer: The system has infinitely many solutions. The case where $\det(A) = 0$ and all $\det(A_i) = 0$ corresponds to the indeterminate form $0/0$ for each variable. This indicates that the system is dependent, meaning there are infinitely many solutions. While Cramer's rule itself cannot produce a specific solution in this case, the values of the determinants allow us to conclude that the system is dependent.
9. Consider the linear system: $\begin{cases} 2x - 3y + z = 5 \\ x + y - 2z = -1 \\ -x + 2y + z = 0 \end{cases}$ Using Cramer's Rule, which of the following expressions correctly represents the solution for $y$?
   Answer: $y = \frac{\det \begin{pmatrix} 2 & 5 & 1 \\ 1 & -1 & -2 \\ -1 & 0 & 1 \end{pmatrix}}{\det \begin{pmatrix} 2 & -3 & 1 \\ 1 & 1 & -2 \\ -1 & 2 & 1 \end{pmatrix}}$ To solve for the variable $y$ using Cramer's rule, the numerator is the determinant of the matrix formed by replacing the $y$-coefficient column (the second column) with the constant vector $\mathbf{b} = \begin{pmatrix} 5 \\ -1 \\ 0 \end{pmatrix}$. The denominator is the determinant of the original coefficient matrix. Choice B correctly shows the second column of the coefficient matrix replaced by the constant vector in the numerator's determinant.
10. In applying Cramer's rule to a $3 \times 3$ linear system $A\mathbf{x} = \mathbf{b}$, it is found that the determinant of the coefficient matrix, $\det(A)$, is zero. However, the determinant of the matrix $A_1$, formed by replacing the first column of $A$ with $\mathbf{b}$, is non-zero. What can be concluded about the solution set of the system?
    Answer: The system has no solution. Cramer's rule states that for a system $A\mathbf{x} = \mathbf{b}$, the solution for a variable $x_i$ is given by $x_i = \frac{\det(A_i)}{\det(A)}$. This rule is only applicable if $\det(A) \neq 0$. If $\det(A) = 0$, the system does not have a unique solution. Furthermore, if $\det(A) = 0$ and at least one numerator determinant $\det(A_i)$ is non-zero, the system is inconsistent, meaning it has no solution. This case corresponds to an undefined expression of the form (non-zero)/0.
11. Consider solving the system $x + 2y + z = 6$, $2x - y + 3z = 14$, $3x + y - z = 2$ using Cramer's rule. If a computational error results in calculating $D_y = 28$ instead of the correct value $D_y = -28$, and $D = 14$, what would be the incorrect value obtained for $y$, and how does it relate to the correct answer?
    Answer: The incorrect value is $y = 2$, which is the negative of the correct value $y = -2$ With the computational error, $y = \frac{D_y}{D} = \frac{28}{14} = 2$. With the correct calculation, $y = \frac{-28}{14} = -2$. The error in sign of $D_y$ leads to an answer that is the negative of the correct value. This demonstrates how sign errors in determinant calculations directly propagate to the final answer in Cramer's rule. Choices B, C, and D all give incorrect relationships between the erroneous and correct values.
12. When applying Cramer's rule to a $2 \times 2$ system, a student finds that swapping two equations changes $D$ from $15$ to $-15$ and $D_x$ from $30$ to $-30$. The student concludes that $x$ remains unchanged at $x = 2$. Is this reasoning correct?
    Answer: Yes, swapping equations changes signs of both numerator and denominator equally, leaving $x = \frac{30}{15} = \frac{-30}{-15} = 2$ unchanged This question tests your understanding of how row operations affect determinants in Cramer's rule. When you swap two rows in a matrix, the determinant changes sign - this is a fundamental property of determinants. In Cramer's rule for a $2 \times 2$ system, $x = \frac{D_x}{D}$ where $D$ is the determinant of the coefficient matrix and $D_x$ is the determinant when the first column is replaced by the constants. When you swap two equations in the system, you're effectively swapping rows in both the coefficient matrix and the augmented matrix used to find $D_x$. Since swapping rows changes the sign of any determinant, both $D$ and $D_x$ will have their signs flipped. The student's reasoning is correct: $x = \frac{30}{15} = 2$ originally, and $x = \frac{-30}{-15} = 2$ after swapping. The negative signs cancel out, leaving $x$ unchanged. Option A is wrong because this result doesn't depend on the special relationship $D_x = 2D$ - it's true for any system where you swap equations. Option B misunderstands what happens when you swap equations: you're changing the entire system, which affects both the coefficient matrix and the constants vector. Option C incorrectly assumes that only $D$ changes sign while $D_x$ stays the same, but swapping equations affects both determinants equally. Remember: swapping rows in any matrix changes the determinant's sign. In Cramer's rule, this affects both numerator and denominator equally, so the solution remains unchanged.
13. A system of linear equations has coefficient matrix $A$ with $\det(A) = 0$. Which statement about using Cramer's rule for this system is most accurate?
    Answer: Cramer's rule cannot be applied because division by zero would occur, regardless of system consistency Cramer's rule requires $\det(A) \neq 0$ to be applicable. When $\det(A) = 0$, the formula $x_i = \frac{\det(A_i)}{\det(A)}$ involves division by zero, making Cramer's rule undefined regardless of whether the system has no solutions, infinitely many solutions, or is inconsistent. Choice A is wrong because Cramer's rule cannot be applied at all when $\det(A) = 0$. Choice C is wrong because even if numerator determinants are zero, we still have $\frac{0}{0}$ which is indeterminate. Choice D is wrong because we cannot conclude that variables equal zero when the rule is inapplicable.
14. A system of equations $Ax = b$ has the property that when Cramer's rule is applied, $D_x = 2D$, $D_y = -D$, and $D_z = \frac{D}{2}$ where $D = \det(A) \neq 0$. What can be concluded about the solution?
    Answer: The solution is $(x, y, z) = (2, -1, \frac{1}{2})$ and represents the unique solution to the system When you encounter Cramer's rule problems, remember that this method provides a direct way to solve systems of linear equations by using determinants. For a 3×3 system $Ax = b$, each variable is found using the formula: variable = (corresponding numerator determinant) ÷ (coefficient matrix determinant). Given the relationships $D_x = 2D$, $D_y = -D$, and $D_z = \frac{D}{2}$ where $D = \det(A) \neq 0$, you can apply Cramer's rule directly: $x = \frac{D_x}{D} = \frac{2D}{D} = 2$ $y = \frac{D_y}{D} = \frac{-D}{D} = -1$ $z = \frac{D_z}{D} = \frac{D/2}{D} = \frac{1}{2}$ This gives the solution $(x, y, z) = (2, -1, \frac{1}{2})$, confirming answer D. Let's examine why the other options fail: A is incorrect because Cramer's rule gives you exact values, not just ratios—you divide each numerator determinant by the coefficient determinant to get precise solutions. B contains a calculation error; the student likely confused the determinant ratios with the actual variable values. C reflects a fundamental misunderstanding—there's no mathematical requirement that $D_x + D_y + D_z = D$; these determinants are formed by replacing different columns of the coefficient matrix and have no such additive relationship. Study tip: With Cramer's rule, always remember the formula is simply "variable = numerator determinant ÷ coefficient determinant." Don't overthink the relationships between different determinants—focus on the straightforward division for each variable.
15. Let the linear system $A\mathbf{x} = \mathbf{b}$ have a unique solution given by $x_i = \frac{\det(A_i)}{\det(A)}$. If the system is changed to $A\mathbf{x} = k\mathbf{b}$ where $k$ is a non-zero scalar, how does the new solution for $x_i$, denoted $x_i'$, relate to the original solution $x_i$?
    Answer: $x_i' = kx_i$ In the new system $A\mathbf{x} = k\mathbf{b}$, the coefficient matrix $A$ remains the same, so $\det(A)$ is unchanged. The new numerator determinant, $\det(A_i')$, is calculated from the matrix $A_i'$ formed by replacing the $i$-th column of $A$ with the vector $k\mathbf{b}$. A property of determinants states that if a single column of a matrix is multiplied by a scalar $k$, the determinant of the new matrix is $k$ times the determinant of the original. Thus, $\det(A_i') = k \cdot \det(A_i)$. The new solution is $x_i' = \frac{\det(A_i')}{\det(A)} = \frac{k \cdot \det(A_i)}{\det(A)} = k \cdot x_i$.
16. A student applies Cramer's rule to solve $3x + 2y = 8$ and $6x + 4y = 16$. The student calculates $D = \begin{vmatrix} 3 & 2 \\ 6 & 4 \end{vmatrix} = 0$ but continues with $D_x = \begin{vmatrix} 8 & 2 \\ 16 & 4 \end{vmatrix} = 0$ and concludes $x = \frac{0}{0} = 0$. What is wrong with this reasoning?
    Answer: While both determinants are correctly calculated as zero, $\frac{0}{0}$ is indeterminate, not zero, and Cramer's rule doesn't apply here The student correctly calculated both determinants as zero: $D = 3(4) - 2(6) = 0$ and $D_x = 8(4) - 2(16) = 0$. However, $\frac{0}{0}$ is mathematically indeterminate, not equal to zero. When $D = 0$, Cramer's rule cannot be applied regardless of what $D_x$ equals. Choice A is wrong because $D_x = 0$ is correct. Choice C is partially correct about infinite solutions but wrong about using Cramer's rule. Choice D is wrong because this system actually has infinite solutions (the second equation is twice the first), not no solutions.
17. The system of equations $kx + 3y = -1$ and $4x + 2y = 5$ has a unique solution for all values of $k$ except one. What is this exceptional value of $k$?
    Answer: 6 When you encounter a system of linear equations and need to find when it lacks a unique solution, you're dealing with the fundamental concept of linear dependence. A system has a unique solution except when the equations become linearly dependent (proportional to each other) but have different constant terms, creating an inconsistent system. For the system $kx + 3y = -1$ and $4x + 2y = 5$ to lack a unique solution, the left sides must be proportional while the right sides are not. This means the coefficients must satisfy: $\frac{k}{4} = \frac{3}{2}$ Solving for $k$: $k = 4 \cdot \frac{3}{2} = 6$ When $k = 6$, the first equation becomes $6x + 3y = -1$, which is equivalent to $2x + y = -\frac{1}{3}$. The second equation $4x + 2y = 5$ is equivalent to $2x + y = \frac{5}{2}$. These equations demand the same expression ($2x + y$) equal two different values, making the system inconsistent with no solution. Choice A) $k = -6$ would make the coefficients $\frac{-6}{4} = -\frac{3}{2}$, which doesn't equal $\frac{3}{2}$. Choice B) $k = 12$ gives $\frac{12}{4} = 3 \neq \frac{3}{2}$. Choice C) $k = \frac{2}{3}$ yields $\frac{2/3}{4} = \frac{1}{6} \neq \frac{3}{2}$. All these values preserve the system's unique solution. Remember: A system loses its unique solution when coefficient ratios match but constant term ratios don't. Set up the proportion $\frac{a_1}{a_2} = \frac{b_1}{b_2}$ and solve for the parameter.
18. A student is solving for $x$ in the system $\begin{cases} 3x + y = 5 \\ 2x - 4y = 1 \end{cases}$ using Cramer's rule. Their work is shown below. Step 1: $D = \det \begin{pmatrix} 3 & 1 \\ 2 & -4 \end{pmatrix} = -12 - 2 = -14$. Step 2: $D_x = \det \begin{pmatrix} 3 & 5 \\ 2 & 1 \end{pmatrix} = 3 - 10 = -7$. Step 3: $x = \frac{D_x}{D} = \frac{-7}{-14} = \frac{1}{2}$. In which step did the student make their first mistake?
    Answer: Step 2 The student's calculation of the main determinant $D$ in Step 1 is correct. However, in Step 2, the student makes an error in setting up the determinant $D_x$. To find $D_x$, the column of x-coefficients (the first column) should be replaced by the constant vector $\begin{pmatrix} 5 \\ 1 \end{pmatrix}$. The student incorrectly replaced the y-coefficient column. The correct setup for $D_x$ is $\det \begin{pmatrix} 5 & 1 \\ 1 & -4 \end{pmatrix} = -20 - 1 = -21$. The matrix the student used in Step 2 is actually the correct setup for $D_y$.

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