Determinant Properties - Linear Algebra

1. Let $\vec{u}, \vec{v}, \vec{w}$ be vectors in $\mathbb{R}^3$. If the determinant of the matrix with columns $[\vec{u}, \vec{v}, \vec{w}]$ is 5, what is the determinant of the matrix with columns $[\vec{u}+\vec{v}, \vec{v}+\vec{w}, \vec{w}-\vec{u}]$?
   Answer: $0$ Let the columns of the new matrix be $\vec{c}_1' = \vec{u}+\vec{v}$, $\vec{c}_2' = \vec{v}+\vec{w}$, and $\vec{c}_3' = \vec{w}-\vec{u}$. The determinant of a matrix is zero if its columns are linearly dependent. Let's check for a linear combination of the columns that equals the zero vector. Consider the sum $\vec{c}_1' - \vec{c}_2' + \vec{c}_3'$: $(\vec{u}+\vec{v}) - (\vec{v}+\vec{w}) + (\vec{w}-\vec{u}) = \vec{u}+\vec{v}-\vec{v}-\vec{w}+\vec{w}-\vec{u} = \vec{0}$. Since a non-trivial linear combination of the columns equals the zero vector (the coefficients are 1, -1, and 1), the columns are linearly dependent. Therefore, the determinant of the matrix is 0.
2. A matrix $B$ is obtained from a $3 \times 3$ matrix $A$ by the single row operation $R_2 \to R_2 - 3R_1$. If $\det(B) = 12$, what is $\det(2A)$?
   Answer: $96$ This is a two-step problem. First, we find $\det(A)$, then we find $\det(2A)$. Step 1: The row operation $R_2 \to R_2 - 3R_1$ adds a multiple of one row to another. This type of operation does not change the value of the determinant. Therefore, $\det(A) = \det(B) = 12$. Step 2: We need to find $\det(2A)$. For an $n \times n$ matrix $A$ and a scalar $c$, the property is $\det(cA) = c^n \det(A)$. Here, $A$ is a $3 \times 3$ matrix, so $n=3$, and the scalar is $c=2$. $\det(2A) = 2^3 \det(A) = 8 \times 12 = 96$. Distractor A (24) is the result of the common error $2\det(A)$. Distractor C (32) results from first incorrectly concluding $\det(A) = 12/(-3)$ or $12/3$ and then correctly applying the scaling property.
3. A $3 \times 3$ matrix $B$ is formed from a matrix $A$ by first taking the transpose of $A$, and then swapping the first and third columns of the resulting matrix. If $\det(A) = k$, what is $\det(B)$?
   Answer: $-k$ There are two operations applied to find $B$ from $A$. 1. Taking the transpose: Let $C = A^T$. A fundamental property of determinants is that $\det(A^T) = \det(A)$. Therefore, $\det(C) = k$. 2. Swapping columns: Matrix $B$ is obtained by swapping columns 1 and 3 of $C$. Swapping any two columns (or rows) of a matrix multiplies its determinant by -1. Therefore, $\det(B) = -\det(C)$. Combining these steps, we get $\det(B) = -\det(C) = -\det(A^T) = -\det(A) = -k$.
4. Let $A$ be an $n \times n$ matrix with rows $\vec{r}_1, \vec{r}_2, \ldots, \vec{r}_n$. Let $B$ be a matrix where, for some $i \neq j$, the $i$-th row of $A$ has been replaced by the $j$-th row of $A$, while all other rows remain the same. Which of the following must be true?
   Answer: $\det(B) = 0$ In the new matrix $B$, the $i$-th row is now a copy of the $j$-th row, i.e., $\vec{b}_i = \vec{r}_j$. The $j$-th row of $B$ is also the $j$-th row of $A$, so $\vec{b}_j = \vec{r}_j$. This means matrix $B$ has two identical rows (the $i$-th and $j$-th rows). A matrix with two identical rows has linearly dependent rows, and its determinant is always 0. This can be shown by performing the row operation $R_i \to R_i - R_j$ on $B$, which results in a row of zeros at row $i$. This operation does not change the determinant, and the determinant of a matrix with a zero row is 0.
5. Let $A$ be a $4 \times 4$ matrix with $\det(A) = -2$. What is the value of $\det(3A)$?
   Answer: $-162$ For an $n \times n$ matrix $A$ and a scalar $c$, the property of determinants states that $\det(cA) = c^n \det(A)$. In this case, $A$ is a $4 \times 4$ matrix, so $n=4$. The scalar is $c=3$, and $\det(A) = -2$. Applying the formula: $\det(3A) = 3^4 \det(A) = 81 \times (-2) = -162$. Distractor A, -6, results from the common mistake of calculating $3 \det(A)$. Distractor B, -54, results from incorrectly using $n-1$ for the exponent: $3^{4-1} \det(A)$. Distractor D, -128, results from confusing the scalar and the dimension: $4^3 \det(A)$.
6. If $\det\begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} = 7$, what is the value of $\det\begin{pmatrix} a+2d & b+2e & c+2f \\ d & e & f \\ 3g & 3h & 3i \end{pmatrix}$?
   Answer: $21$ The correct answer is B. Using linearity of determinants in the first row: $\det\begin{pmatrix} a+2d & b+2e & c+2f \\ d & e & f \\ 3g & 3h & 3i \end{pmatrix} = \det\begin{pmatrix} a & b & c \\ d & e & f \\ 3g & 3h & 3i \end{pmatrix} + 2\det\begin{pmatrix} d & e & f \\ d & e & f \\ 3g & 3h & 3i \end{pmatrix}$. The second determinant equals 0 because rows 1 and 2 are identical. For the first determinant, factoring out 3 from the third row gives $3\det\begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} = 3 \times 7 = 21$. Choice A incorrectly multiplies by 6 instead of 3. Choice C incorrectly multiplies by 2. Choice D ignores the row scaling entirely.
7. Let $A$ be a $4 \times 4$ matrix. If swapping rows 1 and 3 of $A$ gives matrix $B$, and then multiplying row 2 of $B$ by $\frac{1}{2}$ gives matrix $C$, which relationship correctly describes $\det(C)$ in terms of $\det(A)$?
   Answer: $\det(C) = -\frac{1}{2}\det(A)$ The correct answer is A. Row operations affect determinants systematically: swapping two rows multiplies the determinant by $-1$, and multiplying a row by a scalar $k$ multiplies the determinant by $k$. Starting with $\det(A)$, swapping rows 1 and 3 gives $\det(B) = -\det(A)$. Then multiplying row 2 by $\frac{1}{2}$ gives $\det(C) = \frac{1}{2} \cdot \det(B) = \frac{1}{2} \cdot (-\det(A)) = -\frac{1}{2}\det(A)$. Choice B forgets the sign change from swapping. Choices C and D incorrectly use the reciprocal of the scaling factor.
8. What is the determinant of the matrix $M = \begin{pmatrix} 1 & 2 & 3 & 4 \\ 0 & 5 & 6 & 7 \\ 0 & 0 & 8 & 9 \\ 2 & 4 & 6 & 9 \end{pmatrix}$?
   Answer: $40$ We can simplify the matrix using a row operation that does not change the determinant. Notice that the fourth row is very similar to twice the first row. Let's perform the row operation $R_4 \to R_4 - 2R_1$. The new fourth row is: $(2, 4, 6, 9) - 2(1, 2, 3, 4) = (2-2, 4-4, 6-6, 9-8) = (0, 0, 0, 1)$. Since this operation does not change the determinant, we have: $\det(M) = \det \begin{pmatrix} 1 & 2 & 3 & 4 \\ 0 & 5 & 6 & 7 \\ 0 & 0 & 8 & 9 \\ 0 & 0 & 0 & 1 \end{pmatrix}$. This new matrix is upper triangular. The determinant of a triangular matrix is the product of its diagonal entries. $\det(M) = 1 \times 5 \times 8 \times 1 = 40$. Distractor A (0) is a trap for those who incorrectly assume $R_4$ is exactly $2R_1$. Distractor C (360) is the product of the diagonal entries of the original matrix, which is not triangular.
9. If $\det\begin{pmatrix} 2u & 2v & 2w \\ x & y & z \\ 3u+x & 3v+y & 3w+z \end{pmatrix} = 48$, what is $\det\begin{pmatrix} u & v & w \\ x & y & z \\ u & v & w \end{pmatrix}$?
   Answer: $0$ When you encounter determinant problems involving matrix transformations, focus on how row operations and linear combinations affect determinant values. Let's work with the given matrix systematically. First, notice that the third row is $3u+x, 3v+y, 3w+z$, which can be written as $3(u,v,w) + (x,y,z)$. Using the linearity property of determinants, we can split this: $\det\begin{pmatrix} 2u & 2v & 2w \\ x & y & z \\ 3u+x & 3v+y & 3w+z \end{pmatrix} = \det\begin{pmatrix} 2u & 2v & 2w \\ x & y & z \\ 3u & 3v & 3w \end{pmatrix} + \det\begin{pmatrix} 2u & 2v & 2w \\ x & y & z \\ x & y & z \end{pmatrix}$ The second determinant equals zero because it has two identical rows (the second and third rows are the same). So we have: $48 = \det\begin{pmatrix} 2u & 2v & 2w \\ x & y & z \\ 3u & 3v & 3w \end{pmatrix}$ Now factor out constants: factor 2 from the first row and 3 from the third row: $48 = 2 \cdot 3 \cdot \det\begin{pmatrix} u & v & w \\ x & y & z \\ u & v & w \end{pmatrix} = 6 \cdot \det\begin{pmatrix} u & v & w \\ x & y & z \\ u & v & w \end{pmatrix}$ But the target matrix has identical first and third rows, making its determinant zero. Choice A (24) assumes you divided 48 by 2 only. Choice B (8) assumes you divided by 6 but ignored the repeated rows. Choice C (12) assumes you divided by 4 somehow. Key insight: Any matrix with two identical rows has determinant zero, regardless of other calculations. Always check for this pattern first.
10. Let $A$ be a $3 \times 3$ matrix with $\det(A) = 5$. If matrix $B$ is obtained from $A$ by first multiplying the second row by $-2$, then adding 3 times the first row to the third row, what is $\det(B)$?
    Answer: $-10$ The correct answer is A. Row operations affect determinants as follows: (1) Multiplying a row by a scalar $k$ multiplies the determinant by $k$, and (2) Adding a multiple of one row to another row does not change the determinant. First, multiplying the second row by $-2$ changes the determinant from $5$ to $5 \times (-2) = -10$. Then, adding 3 times the first row to the third row leaves the determinant unchanged at $-10$. Choice B incorrectly uses the absolute value. Choice C incorrectly multiplies by both $-2$ and $3$. Choice D combines both errors from B and C.
11. Let $D(\vec{r}_1, \vec{r}_2, \vec{r}_3)$ denote the determinant of a $3 \times 3$ matrix with rows $\vec{r}_1, \vec{r}_2, \vec{r}_3$. Which of the following expressions is a correct statement of the linearity property of determinants?
    Answer: $D(\vec{r}_1, c\vec{r}_2 + \vec{s}_2, \vec{r}_3) = c D(\vec{r}_1, \vec{r}_2, \vec{r}_3) + D(\vec{r}_1, \vec{s}_2, \vec{r}_3)$ When you encounter questions about determinant properties, focus on understanding that determinants are multilinear functions. This means they are linear in each row (or column) individually, but not necessarily linear across all rows simultaneously. The linearity property states that if you have a linear combination in any single row, the determinant distributes over that combination while keeping other rows fixed. Option D perfectly demonstrates this: $D(\vec{r}_1, c\vec{r}_2 + \vec{s}_2, \vec{r}_3) = c D(\vec{r}_1, \vec{r}_2, \vec{r}_3) + D(\vec{r}_1, \vec{s}_2, \vec{r}_3)$. The determinant is linear in the second row while the first and third rows remain unchanged. Option A is incorrect because scaling all rows by factor $c$ scales the determinant by $c^3$, not $c$. The correct relationship would be $D(c\vec{r}_1, c\vec{r}_2, c\vec{r}_3) = c^3 D(\vec{r}_1, \vec{r}_2, \vec{r}_3)$. Option B fails because determinants aren't linear across multiple rows simultaneously. Adding the same vector to all rows doesn't follow any standard determinant property, and $D(\vec{s}, \vec{s}, \vec{s}) = 0$ since identical rows make the determinant zero. Option C makes no mathematical sense—it's not even a proper equation since it claims something equals itself minus something else (unless that "something else" is zero, which isn't generally true). Study tip: Remember that determinants are linear in each individual row or column, but not across multiple rows/columns at once. Practice identifying which single row or column is being modified in linearity problems.
12. Let $A$ be a $3 \times 3$ matrix with columns $\vec{a}_1, \vec{a}_2, \vec{a}_3$. If $\det(A) = 3$, and $\vec{v}$ is a vector in $\mathbb{R}^3$ such that the determinant of the matrix with columns $[\vec{v}, \vec{a}_2, \vec{a}_3]$ is 4, what is the determinant of the matrix with columns $[\vec{a}_1 + 2\vec{v}, \vec{a}_2, \vec{a}_3]$?
    Answer: $11$ The determinant is a linear function of each of its columns. Using the linearity property on the first column, we can write: $\det([\vec{a}_1 + 2\vec{v}, \vec{a}_2, \vec{a}_3]) = \det([\vec{a}_1, \vec{a}_2, \vec{a}_3]) + \det([2\vec{v}, \vec{a}_2, \vec{a}_3])$. Further, we can factor out the scalar from the first column of the second term: $\det([2\vec{v}, \vec{a}_2, \vec{a}_3]) = 2 \det([\vec{v}, \vec{a}_2, \vec{a}_3])$. We are given that $\det([\vec{a}_1, \vec{a}_2, \vec{a}_3]) = \det(A) = 3$ and $\det([\vec{v}, \vec{a}_2, \vec{a}_3]) = 4$. Substituting these values, we get: $\det([\vec{a}_1 + 2\vec{v}, \vec{a}_2, \vec{a}_3]) = 3 + 2(4) = 3 + 8 = 11$.
13. Let $A$ be a $3 \times 3$ matrix with columns $\vec{c}_1, \vec{c}_2, \vec{c}_3$, and let $\det(A)=10$. What is the determinant of the matrix $B$ with columns $[\vec{c}_1 + \vec{c}_2, \vec{c}_2 + \vec{c}_3, \vec{c}_3 + \vec{c}_1]$?
    Answer: $20$ The columns of $B$ are linear combinations of the columns of $A$. We can represent this transformation using matrix multiplication. Let $B = [\vec{b}_1, \vec{b}_2, \vec{b}_3]$. Then: $\vec{b}_1 = 1\vec{c}_1 + 1\vec{c}_2 + 0\vec{c}_3$ $\vec{b}_2 = 0\vec{c}_1 + 1\vec{c}_2 + 1\vec{c}_3$ $\vec{b}_3 = 1\vec{c}_1 + 0\vec{c}_2 + 1\vec{c}_3$ This means $B = A K$, where $K = \begin{pmatrix} 1 & 0 & 1 \\ 1 & 1 & 0 \\ 0 & 1 & 1 \end{pmatrix}$. Using the property $\det(AK) = \det(A)\det(K)$, we first find $\det(K)$: $\det(K) = 1(1\cdot1 - 0\cdot1) - 0(1\cdot1 - 0\cdot0) + 1(1\cdot1 - 1\cdot0) = 1(1) - 0 + 1(1) = 2$. Now, we can find $\det(B)$: $\det(B) = \det(A) \det(K) = 10 \times 2 = 20$.
14. Consider matrices $A$ and $B$ where $B$ is obtained from $A$ by the sequence of operations: (1) multiply row 1 by $k$, (2) swap rows 2 and 3, (3) add 5 times row 1 to row 2. If $\det(A) = 6$ and $\det(B) = -18$, what is the value of $k$?
    Answer: $3$ The correct answer is C. Track how each operation affects the determinant: (1) Multiplying row 1 by $k$ changes the determinant from $6$ to $6k$. (2) Swapping rows 2 and 3 changes the determinant from $6k$ to $-6k$. (3) Adding 5 times row 1 to row 2 leaves the determinant unchanged at $-6k$. Since $\det(B) = -18$, we have $-6k = -18$, which gives $k = 3$. Choice A gives the wrong magnitude. Choice B has the wrong sign and magnitude. Choice D has the correct magnitude but wrong sign, resulting from forgetting that the row swap introduces a negative sign.
15. A student claims that for any $3 \times 3$ matrix $A$, if $B$ is obtained by adding the first row to the second row and then subtracting the third row from the first row, then $\det(B) = \det(A)$. Which statement best evaluates this claim?
    Answer: The claim is correct because both operations preserve the determinant value When you encounter questions about row operations and determinants, focus on which specific operations preserve the determinant value versus those that change it. Let's analyze the two operations step by step. First, adding the first row to the second row is an elementary row operation of type "add a multiple of one row to another row." This operation always preserves the determinant value. Second, subtracting the third row from the first row is also an "add a multiple of one row to another row" operation (adding $-1$ times the third row to the first row). This also preserves the determinant value. Since both operations preserve the determinant independently, performing them in sequence means $\det(B) = \det(A)$, making the claim correct. Looking at the wrong answers: Choice A incorrectly suggests that invertibility matters here. Elementary row operations that preserve determinants work the same way regardless of whether the matrix is invertible or singular. Choice B contains a fundamental misconception—subtracting a row (which is adding a multiple of one row to another) does not change the determinant's sign. You might be thinking of row swapping, which does flip the sign. Choice C incorrectly suggests that the order of these determinant-preserving operations matters. Since both operations preserve the determinant value, their order is irrelevant to the final result. Remember this key pattern: Adding any multiple of one row to a different row always preserves the determinant, while swapping rows changes the sign, and scaling a row by a constant multiplies the determinant by that constant.
16. If $A$ is a $4 \times 4$ matrix and $\det(A) = 0$, which of the following statements must be true?
    Answer: One row of $A$ can be expressed as a linear combination of the other rows. When you encounter a determinant problem, you're dealing with one of the most fundamental concepts in linear algebra that reveals crucial information about a matrix's properties and the linear independence of its rows and columns. If $\det(A) = 0$, the matrix is called singular or non-invertible. This immediately tells you that the rows (and columns) of the matrix are linearly dependent, meaning at least one row can be written as a linear combination of the others. This is exactly what option D states, making it correct. Let's examine why the other options fail. Option A claims the columns are linearly independent, but this is false – when $\det(A) = 0$, the columns are actually linearly dependent. Option B suggests the homogeneous equation $A\vec{x} = \vec{0}$ has only the trivial solution, but this is wrong too. A zero determinant guarantees that this equation has infinitely many non-trivial solutions, forming the null space of A. Option C makes an extreme claim that every element is zero. While the zero matrix does have determinant zero, many non-zero matrices also have zero determinants – for example, any matrix with two identical rows. The key insight is that $\det(A) = 0$ is equivalent to linear dependence among rows (or columns). Think of the determinant as a "linear independence detector" – it's zero precisely when the vectors that form the matrix's rows or columns don't span the full space. Remember: Zero determinant always means linear dependence, which opens the door to non-trivial solutions and eliminates invertibility.
17. Let $A = \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{pmatrix}$. Let $B$ be the matrix obtained from $A$ by the sequence of row operations $R_2 \to R_2 - 4R_1$ and then $R_3 \to R_3 - 7R_1$. What is $\det(B)$?
    Answer: $0$ The row operations performed are of the type $R_i \to R_i - cR_j$. These operations do not change the value of the determinant. Therefore, $\det(B) = \det(A)$. We can calculate $\det(A)$ directly: $\det(A) = 1(5\cdot9 - 6\cdot8) - 2(4\cdot9 - 6\cdot7) + 3(4\cdot8 - 5\cdot7)$ $\det(A) = 1(45 - 48) - 2(36 - 42) + 3(32 - 35)$ $\det(A) = 1(-3) - 2(-6) + 3(-3) = -3 + 12 - 9 = 0$. Alternatively, after performing the row operations on $A$, we get: $B = \begin{pmatrix} 1 & 2 & 3 \\ 0 & -3 & -6 \\ 0 & -6 & -12 \end{pmatrix}$. In matrix $B$, the third row is a multiple of the second row ($R_3 = 2R_2$). If one row of a matrix is a multiple of another row, the rows are linearly dependent, and the determinant is 0.
18. Let $A$ be a $3 \times 3$ matrix with $\det(A) = 5$. A new matrix $B$ is created by performing the following sequence of elementary row operations on $A$: 1. Swap Row 1 and Row 3 ($R_1 \leftrightarrow R_3$). 2. Multiply Row 2 by 4 ($R_2 \to 4R_2$). 3. Add -2 times Row 1 to Row 3 ($R_3 \to R_3 - 2R_1$). What is the value of $\det(B)$?
    Answer: $-20$ Let's track the effect of each row operation on the determinant. Starting with $\det(A) = 5$. 1. The operation $R_1 \leftrightarrow R_3$ swaps two rows, which multiplies the determinant by $-1$. The new determinant is $-1 \times 5 = -5$. 2. The operation $R_2 \to 4R_2$ scales a single row by 4, which multiplies the current determinant by 4. The new determinant is $4 \times (-5) = -20$. 3. The operation $R_3 \to R_3 - 2R_1$ adds a multiple of one row to another. This operation does not change the value of the determinant. The determinant remains $-20$. Therefore, $\det(B) = -20$.

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