2x2 & 3x3 Determinants - Linear Algebra

The determinant of matrix $E = \begin{pmatrix} 2 & 1 & 3 \\ 1 & 2 & 1 \\ 3 & 1 & 2 \end{pmatrix}$ can be computed most efficiently using which approach?
Answer: Row operations to create zeros before expansion, since no row or column has zeros initially Since matrix $E$ has no zeros, cofactor expansion along any row or column requires computing three 2×2 determinants. However, using row operations to create zeros first would reduce this to computing only one 2×2 determinant. Choice A ignores the computational advantage of creating zeros. Choice B incorrectly suggests expanding along the diagonal (not a standard method). Choice D mentions Sarrus's rule, which works but isn't more efficient than expansion with zeros created first.
Let $A$ be a $2 \times 2$ matrix such that $\det(A) = 5$ . What is the value of $\det(3A)$ ?
Answer: 45 For any $n \times n$ matrix $A$ and any scalar $k$ , the determinant of the matrix $kA$ is given by the property $\det(kA) = k^n \det(A)$ . In this problem, $A$ is a $2 \times 2$ matrix, so $n=2$ . The scalar is $k=3$ , and we are given $\det(A)=5$ . Applying the formula: $\det(3A) = 3^2 \det(A)$ $= 9 \times 5$ $= 45$ . Distractor A is the result of the common mistake of thinking $\det(kA) = k \det(A)$ . Distractor C incorrectly assumes the scalar has no effect. Distractor D is the result of incorrectly assuming the matrix is $3 \times 3$ , which would yield $3^3 \det(A) = 27 \times 5 = 135$ .
A student computes $\det\begin{pmatrix} 4 & 7 \\ 2 & 3 \end{pmatrix}$ and gets -2. To check this answer, they compute $\det\begin{pmatrix} 3 & 7 \\ 2 & 4 \end{pmatrix}$ (swapping the diagonal entries). What result should they expect if their original calculation was correct?
Answer: $-2$ The original determinant is $4 \cdot 3 - 7 \cdot 2 = 12 - 14 = -2$ , so the student's answer is correct. For the swapped matrix: $\det\begin{pmatrix} 3 & 7 \\ 2 & 4 \end{pmatrix} = 3 \cdot 4 - 7 \cdot 2 = 12 - 14 = -2$ . The determinants are the same because swapping diagonal entries doesn't change the value $ac - bd$ when the off-diagonal entries remain in the same positions. Choice B assumes the sign changes. Choice C assumes $3 \cdot 4 - 7 \cdot 2 = 12 - 18 = -6$ . Choice D assumes $3 \cdot 4 - 2 \cdot 7 = 12 - 14 = -2$ but with wrong sign.
What is the area of a triangle with vertices at coordinates $(1, 2)$ , $(4, 3)$ , and $(2, 5)$ ?
Answer: 4 The area of a triangle with vertices $(x_1, y_1)$ , $(x_2, y_2)$ , and $(x_3, y_3)$ can be found using the formula: Area $= \frac{1}{2} |\det(M)|$ , where $M = \begin{pmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{pmatrix}$ . Let's set up the matrix with the given vertices: $M = \begin{pmatrix} 1 & 2 & 1 \\ 4 & 3 & 1 \\ 2 & 5 & 1 \end{pmatrix}$ . Now, calculate the determinant of $M$ : $\det(M) = 1(3 \cdot 1 - 5 \cdot 1) - 2(4 \cdot 1 - 2 \cdot 1) + 1(4 \cdot 5 - 2 \cdot 3)$ $= 1(3-5) - 2(4-2) + 1(20-6)$ $= 1(-2) - 2(2) + 1(14)$ $= -2 - 4 + 14 = 8$ . The area is $\frac{1}{2} |8| = 4$ . An alternative method is to translate one vertex to the origin, e.g., $(1,2)$ to $(0,0)$ . The other vertices become $(4-1, 3-2) = (3,1)$ and $(2-1, 5-2) = (1,3)$ . The area is half the absolute value of the determinant of the matrix formed by these new vectors: Area $= \frac{1}{2} |\det\begin{pmatrix} 3 & 1 \\ 1 & 3 \end{pmatrix}| = \frac{1}{2} |(3)(3)-(1)(1)| = \frac{1}{2} |9-1| = \frac{1}{2}(8) = 4$ . Distractor C is the value of the determinant before multiplying by $\frac{1}{2}$ . Distractor B results from an error in the 2x2 determinant calculation, computing $ad+bc$ instead of $ad-bc$ , leading to $\frac{1}{2}|9+1|=5$ .
Matrix $F = \begin{pmatrix} k & 2 & -1 \\ 1 & k & 2 \\ -1 & 1 & k \end{pmatrix}$ has the property that $\det(F) = 0$ for exactly three distinct values of $k$ . What is the sum of these three values?
Answer: $0$ When you encounter a matrix with a parameter and need to find when the determinant equals zero, you're looking for the characteristic polynomial's roots. The determinant being zero means the matrix is singular (non-invertible). To find $\det(F) = 0$ , calculate the determinant using cofactor expansion along the first row: $\det(F) = k \begin{vmatrix} k & 2 \\ 1 & k \end{vmatrix} - 2 \begin{vmatrix} 1 & 2 \\ -1 & k \end{vmatrix} + (-1) \begin{vmatrix} 1 & k \\ -1 & 1 \end{vmatrix}$ $= k(k^2 - 2) - 2(k + 2) - (1 + k)$ $= k^3 - 2k - 2k - 4 - 1 - k$ $= k^3 - 5k - 5$ Setting this equal to zero: $k^3 - 5k - 5 = 0$ This is a cubic equation, which can have at most 3 real roots. Since the problem states there are exactly 3 distinct values of $k$ , all three roots are real and distinct. By Vieta's formulas, for a cubic $ak^3 + bk^2 + ck + d = 0$ , the sum of roots equals $-\frac{b}{a}$ . In our equation $k^3 + 0k^2 - 5k - 5 = 0$ , we have $a = 1$ and $b = 0$ , so the sum of roots is $-\frac{0}{1} = 0$ . Choice (A) $6$ and (B) $3$ are positive sums that might tempt you if you misapplied Vieta's formulas or made arithmetic errors. Choice (C) $-3$ could result from incorrectly thinking the sum relates to the constant term. Choice (D) $0$ is correct because there's no $k^2$ term in our cubic. Study tip: Remember Vieta's formulas—the sum of roots of $ax^n + bx^{n-1} + ... = 0$ is $-\frac{b}{a}$ . When the second-highest degree term has coefficient zero, the sum of roots is zero.
For the matrix $B = \begin{pmatrix} \cos\theta & -\sin\theta & 0 \\ \sin\theta & \cos\theta & 0 \\ 0 & 0 & 2 \end{pmatrix}$ , what is $\det(B)$ ?
Answer: $2$ Using cofactor expansion along the third row: $\det(B) = 0 \cdot C_{31} + 0 \cdot C_{32} + 2 \cdot C_{33} = 2 \cdot \det\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix} = 2(\cos^2\theta - (-\sin\theta)(\sin\theta)) = 2(\cos^2\theta + \sin^2\theta) = 2(1) = 2$ . Choice A incorrectly assumes only the $\cos\theta$ term matters. Choice C forgets the factor of 2 from the (3,3) entry. Choice D incorrectly computes the 2×2 determinant as $\cos^2\theta - \sin^2\theta$ .
Given matrices $A = \begin{pmatrix} 3 & 1 \\ 2 & 2 \end{pmatrix}$ and $B = \begin{pmatrix} x & 4 \\ 1 & -1 \end{pmatrix}$ , if $\det(AB) = -24$ , what is the value of $x$ ?
Answer: 2 This problem is most efficiently solved using the property that for any two square matrices $A$ and $B$ of the same size, $\det(AB) = \det(A)\det(B)$ . First, calculate the determinant of matrix $A$ : $\det(A) = (3)(2) - (1)(2) = 6 - 2 = 4$ . Next, calculate the determinant of matrix $B$ in terms of $x$ : $\det(B) = (x)(-1) - (4)(1) = -x - 4$ . Now, use the determinant property: $\det(AB) = \det(A) \det(B)$ $-24 = (4)(-x - 4)$ Divide both sides by 4: $-6 = -x - 4$ Add 4 to both sides: $-2 = -x$ Multiply by -1: $x = 2$ . Distractor A, $x=10$ , results from a sign error in $\det(B)$ , yielding $4(-x+4)=-24$ , so $-x+4=-6$ , and $-x=-10$ . Distractor B, $x=-1$ , results from a sign error in $\det(A)$ , yielding $8(-x-4)=-24$ , so $-x-4=-3$ , and $-x=1$ .
Let matrix $M = \begin{pmatrix} k-1 & 2 \\ 4 & k+1 \end{pmatrix}$ . For which set of values of $k$ is the matrix $M$ singular?
Answer: $\{3, -3\}$ A matrix is singular if its determinant is equal to zero. The determinant of a $2 \times 2$ matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ is $ad-bc$ . For matrix $M$ , the determinant is $\det(M) = (k-1)(k+1) - (2)(4)$ . Simplify the expression: $\det(M) = (k^2 - 1) - 8 = k^2 - 9$ . Set the determinant to zero to find the values of $k$ for which the matrix is singular: $k^2 - 9 = 0$ $(k-3)(k+3) = 0$ This gives two solutions: $k=3$ and $k=-3$ . Therefore, the set of values is $\{3, -3\}$ . Distractor B is incorrect; it results from the calculation error $(k-1)(k+1) + (2)(4) = k^2+7=0$ . Distractor C is incorrect; it results from incorrectly solving by setting the diagonal terms to zero individually. Distractor D is incorrect; it would be the answer if the quadratic equation had no real solutions.
What is the determinant of the matrix $B = \begin{pmatrix} 2 & -1 & 3 \\ 4 & -2 & 6 \\ 1 & 5 & -7 \end{pmatrix}$ ?
Answer: 0 This problem can be solved by observing the relationship between the rows of the matrix. Notice that Row 2 is a scalar multiple of Row 1: $(4, -2, 6) = 2 \times (2, -1, 3)$ . A fundamental property of determinants is that if one row (or column) of a matrix is a scalar multiple of another row (or column), the rows are linearly dependent, and the determinant of the matrix is 0. Without noticing this property, a full calculation would be: $\det(B) = 2((-2)(-7) - (6)(5)) - (-1)((4)(-7) - (6)(1)) + 3((4)(5) - (-2)(1))$ $= 2(14 - 30) + 1(-28 - 6) + 3(20 + 2)$ $= 2(-16) - 34 + 3(22)$ $= -32 - 34 + 66 = 0$ . Distractor A is the product of the diagonal elements. Distractor B is the sum of the diagonal elements (the trace). Distractor C is a plausible result of a sign error in the expansion.
Find the value of $x$ for which the determinant of matrix $C = \begin{pmatrix} 1 & 0 & -2 \\ 3 & x & 1 \\ 2 & -1 & 4 \end{pmatrix}$ is equal to 7.
Answer: 0 To find the value of $x$ , we first need to compute the determinant of matrix $C$ in terms of $x$ . It is efficient to use cofactor expansion along the first row, as it contains a zero. $\det(C) = 1 \cdot \det\begin{pmatrix} x & 1 \\ -1 & 4 \end{pmatrix} - 0 \cdot \det\begin{pmatrix} 3 & 1 \\ 2 & 4 \end{pmatrix} + (-2) \cdot \det\begin{pmatrix} 3 & x \\ 2 & -1 \end{pmatrix}$ Now compute the $2 \times 2$ determinants: $\det(C) = 1((x)(4) - (1)(-1)) - 0 + (-2)((3)(-1) - (x)(2))$ $= 1(4x + 1) - 2(-3 - 2x)$ $= 4x + 1 + 6 + 4x$ $= 8x + 7$ . We are given that the determinant is equal to 7. So, we set up the equation: $8x + 7 = 7$ $8x = 0$ $x = 0$ . Distractor A results from the calculation error $8x+5=7$ , which leads to $8x=2$ and $x=1/4$ . Distractors B and D result from other common algebraic errors when solving for $x$ .
Two students are asked to find $\det\begin{pmatrix} 5 & 8 \\ 3 & 2 \end{pmatrix}$ . Student A computes $(5)(2) - (8)(3) = 10 - 24 = -14$ . Student B computes $(5)(2) - (3)(8) = 10 - 24 = -14$ . Which statement is correct?
Answer: Both students used correct methods and arrived at the correct answer When you encounter a 2×2 determinant, you're applying a fundamental formula, but there's an important property about matrix operations that this question highlights: the determinant formula has some flexibility in how you can think about it. For a 2×2 matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ , the determinant is $ad - bc$ . Student A computed $(5)(2) - (8)(3)$ , which follows this standard formula exactly: multiply the main diagonal elements and subtract the product of the off-diagonal elements. Student B computed $(5)(2) - (3)(8)$ , which rearranges the subtraction term but gives the same result since multiplication is commutative: $bc = cb$ . Both approaches are mathematically valid because $(8)(3) = (3)(8) = 24$ . Option A is wrong because both students used valid mathematical reasoning—there were no errors. Option B incorrectly suggests Student B made an error, but rearranging factors in multiplication doesn't constitute an error. Option C makes the opposite mistake, claiming Student A was wrong when they used the textbook formula perfectly. Option D is correct because both students applied mathematically sound methods and reached the right answer of -14. Study tip: Remember that while the determinant formula $ad - bc$ is typically written in a specific order, the commutative property of multiplication means you can rearrange factors within each term without changing the result. Focus on getting the sign pattern right (positive for main diagonal, negative for off-diagonal) rather than worrying about the exact order of multiplication.
What is the determinant of the matrix $R = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$ ?
Answer: 1 The determinant of a $2 \times 2$ matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ is $ad-bc$ . For the given matrix $R$ , we have $a = \cos\theta$ , $b = -\sin\theta$ , $c = \sin\theta$ , and $d = \cos\theta$ . Calculate the determinant: $\det(R) = (\cos\theta)(\cos\theta) - (-\sin\theta)(\sin\theta)$ $= \cos^2\theta + \sin^2\theta$ This is the fundamental Pythagorean identity in trigonometry, which states that $\cos^2\theta + \sin^2\theta = 1$ for any angle $\theta$ . Thus, the determinant is 1. Distractor D, $\cos(2\theta)$ , is the value of the similar-looking expression $\cos^2\theta - \sin^2\theta$ . Distractor B or C would result from sign errors or incorrect application of trigonometric identities.
The determinant of a $3 \times 3$ matrix $A$ is found to be 0. Which of the following statements must be true about the matrix $A$ ?
Answer: The column vectors of the matrix are coplanar. A determinant of zero for a $3 \times 3$ matrix indicates several equivalent conditions. The matrix is singular (not invertible), its rows are linearly dependent, and its columns are linearly dependent. Geometrically, the absolute value of the determinant of a $3 \times 3$ matrix represents the volume of the parallelepiped formed by its three column (or row) vectors. If the determinant is zero, this volume is zero. This happens if and only if the three vectors lie in the same plane (i.e., they are coplanar). Let's analyze the distractors: - A: A matrix with a row of zeros has a determinant of 0, but a matrix can have a determinant of 0 without a row of zeros (e.g., if one row is a multiple of another). So, this is a sufficient but not a necessary condition. - B: A matrix with two identical columns has a determinant of 0, but this is also a sufficient, not a necessary condition. - C: The column vectors being coplanar is the geometric definition of linear dependence in 3D, which is equivalent to the determinant being zero. This statement must be true. - D: The trace (the sum of the main diagonal elements) has no required relationship with the determinant being zero. For example, the identity matrix has a trace of 3 and a determinant of 1, while $\begin{pmatrix} 1 & 1 & 1 \\ 2 & 2 & 2 \\ 3 & 4 & 5 \end{pmatrix}$ has a determinant of 0 but a trace of 8.
Let $A$ be a $3 \times 3$ matrix with $\det(A) = 10$ . A new matrix $B$ is formed by performing the following sequence of row operations on $A$ : 1. Swap Row 1 and Row 3. ( $R_1 \leftrightarrow R_3$ ) 2. Multiply Row 2 by a factor of 4. ( $R_2 \to 4R_2$ ) 3. Add -2 times Row 2 to Row 1. ( $R_1 \to R_1 - 2R_2$ ) What is the value of $\det(B)$ ?
Answer: -40 We need to track how each elementary row operation affects the determinant. Starting with $\det(A) = 10$ . 1. Swap Row 1 and Row 3: Swapping any two rows of a matrix multiplies its determinant by -1. After this step, the new determinant is $-1 \times \det(A) = -1 \times 10 = -10$ . 2. Multiply Row 2 by a factor of 4: Multiplying a single row of a matrix by a scalar $k$ multiplies the determinant by $k$ . After this step, the new determinant is $4 \times (-10) = -40$ . 3. Add -2 times Row 2 to Row 1: Adding a multiple of one row to another row does not change the determinant. After this step, the determinant remains $-40$ . Therefore, $\det(B) = -40$ . Distractor A results from forgetting the sign change due to the row swap. Distractor C results from incorrectly believing the third operation multiplies the determinant by -2. Distractor D results from ignoring the effect of the second operation.
If $\det\begin{pmatrix} a & b \\ c & d \end{pmatrix} = 7$ and $\det\begin{pmatrix} 2a & 2b \\ 3c & 3d \end{pmatrix} = k \cdot 7$ , what is the value of $k$ ?
Answer: $k = 6$ We can factor out scalars from rows: $\det\begin{pmatrix} 2a & 2b \\ 3c & 3d \end{pmatrix} = \det\begin{pmatrix} 2 & 0 \\ 0 & 3 \end{pmatrix} \cdot \det\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ . Actually, this is incorrect. The proper approach uses the fact that multiplying a row by a scalar multiplies the determinant by that scalar. So $\det\begin{pmatrix} 2a & 2b \\ 3c & 3d \end{pmatrix} = 2 \cdot 3 \cdot \det\begin{pmatrix} a & b \\ c & d \end{pmatrix} = 6 \cdot 7 = 42$ . Therefore $k = 6$ . Choice A assumes $k = 2 \cdot 2 = 4$ . Choice B assumes $k = 2 + 3 = 5$ . Choice D assumes $k = 2 \cdot 2 \cdot 3 = 12$ .
Evaluate the determinant of the matrix $M = \begin{pmatrix} x & y & z \\ x+1 & y+1 & z+1 \\ x+2 & y+2 & z+2 \end{pmatrix}$ .
Answer: 0 The determinant can be evaluated using row operations, which simplify the matrix without changing the determinant's value (for operations of the type $R_i \to R_i + kR_j$ ). Let's perform the following row operations: 1. Replace Row 2 with Row 2 - Row 1 ( $R_2 \to R_2 - R_1$ ). 2. Replace Row 3 with Row 3 - Row 1 ( $R_3 \to R_3 - R_1$ ). This gives a new matrix $M'$ with the same determinant: $R_2 - R_1 = ((x+1)-x, (y+1)-y, (z+1)-z) = (1, 1, 1)$ $R_3 - R_1 = ((x+2)-x, (y+2)-y, (z+2)-z) = (2, 2, 2)$ So, $M' = \begin{pmatrix} x & y & z \\ 1 & 1 & 1 \\ 2 & 2 & 2 \end{pmatrix}$ . In matrix $M'$ , Row 3 is a scalar multiple of Row 2 (specifically, $R_3 = 2R_2$ ). A matrix with one row being a multiple of another has linearly dependent rows, and its determinant is 0. Therefore, $\det(M) = \det(M') = 0$ . The other choices represent plausible but incorrect guesses if one does not see the structural property of the matrix.
What is the determinant of the matrix $A = \begin{pmatrix} 5 & 8 & -4 \\ 0 & -2 & 9 \\ 0 & 0 & 3 \end{pmatrix}$ ?
Answer: -30 The given matrix $A$ is an upper triangular matrix because all the entries below the main diagonal are zero. A property of triangular matrices (both upper and lower) is that their determinant is the product of the elements on the main diagonal. So, $\det(A) = (5) \times (-2) \times (3)$ . $\det(A) = -10 \times 3 = -30$ . Alternatively, one could use cofactor expansion along the first column: $\det(A) = 5 \cdot \det\begin{pmatrix} -2 & 9 \\ 0 & 3 \end{pmatrix} - 0 \cdot (...) + 0 \cdot (...) = 5((-2)(3) - (9)(0)) = 5(-6) = -30$ . Distractor A is the sum of the diagonal elements (the trace). Distractor B is the correct magnitude but with an incorrect sign. Distractor D could result from a calculation error or incorrectly assuming the presence of zeros makes the determinant zero.
For which value of $c$ does the system of linear equations below lack a unique solution? $\begin{cases} (c-2)x + 5y = 10 \\ 3x + cy = 15 \end{cases}$
Answer: 5 A system of linear equations of the form $A\mathbf{x} = \mathbf{b}$ has a unique solution if and only if the determinant of the coefficient matrix $A$ is non-zero. Therefore, the system will lack a unique solution (i.e., have no solution or infinitely many solutions) if the determinant of the coefficient matrix is zero. The coefficient matrix for this system is $A = \begin{pmatrix} c-2 & 5 \\ 3 & c \end{pmatrix}$ . We need to find the value of $c$ for which $\det(A) = 0$ . $\det(A) = (c-2)(c) - (5)(3)$ $= c^2 - 2c - 15$ Set the determinant to zero and solve for $c$ : $c^2 - 2c - 15 = 0$ Factor the quadratic equation: $(c-5)(c+3) = 0$ The values of $c$ for which the system does not have a unique solution are $c=5$ and $c=-3$ . Of the choices provided, $c=5$ is listed. Distractor C, $c=2$ , is a value that makes an entry in the matrix zero but does not make the determinant zero. Distractors A and B result from sign errors when factoring the quadratic equation.

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