Question 1

If the determinant of a 2×2 matrix A = [[a, b], [c, d]] is 7, then the determinant of adj(A) is:

Accepted Answer

We use the fundamental property relating a matrix, its adjugate, and determinant:
A · adj(A) = det(A) · I

Taking determinants of both sides:
det(A · adj(A)) = det(det(A) · I)

Using det(AB) = det(A) · det(B) on the left side:
det(A) · det(adj(A)) = det(det(A) · I)

For a 2×2 matrix, det(det(A) · I) = det(A)² · det(I) = det(A)² · 1 = det(A)²

Therefore:
7 · det(adj(A)) = 7² = 49
det(adj(A)) = 49/7 = 7

Alternatively, for an n×n matrix: det(adj(A)) = [det(A)]^(n-1)
For n = 2: det(adj(A)) = [det(A)]^(2-1) = det(A)^1 = 7.

Question 2

If A and B are 3×3 matrices with det(A) = 4 and det(B) = −2, then det(AB) is equal to:

Accepted Answer

We use the multiplicative property of determinants: det(AB) = det(A) · det(B).

Given:
det(A) = 4
det(B) = −2

Therefore:
det(AB) = det(A) · det(B) = 4 · (−2) = −8

Note: This property holds for any square matrices of the same order, and the order of multiplication does not matter for the determinant (though AB ≠ BA in general).

Question 3

If A is a 2×2 matrix with det(A) = 5, then det(A⁻¹) is equal to:

Accepted Answer

We use the property that for an invertible matrix A:
A · A⁻¹ = I (identity matrix)

Taking determinants on both sides:
det(A · A⁻¹) = det(I)
det(A) · det(A⁻¹) = 1 (using the multiplicative property and det(I) = 1)

Therefore:
det(A⁻¹) = 1/det(A) = 1/5

This formula holds for any invertible square matrix, regardless of its order.

Question 4

If A = [[2, 0], [0, 3]] and B = [[1, 1], [0, 1]], then det(AB) is equal to:

Accepted Answer

We use the property: det(AB) = det(A) · det(B)

Step 1: Compute det(A)
det(A) = (2)(3) − (0)(0) = 6

Step 2: Compute det(B)
det(B) = (1)(1) − (1)(0) = 1

Step 3: Apply the product rule
det(AB) = det(A) · det(B) = 6 · 1 = 6

Verification (optional): Compute AB directly:
AB = [[2·1 + 0·0, 2·1 + 0·1], [0·1 + 3·0, 0·1 + 3·1]] = [[2, 2], [0, 3]]
det(AB) = (2)(3) − (2)(0) = 6 ✓

Theorem: For square matrices A and B of the same order, det(AB) = det(A) · det(B).

Question 5

If A is a 2×2 matrix with det(A) = 3, then det(A⁻¹) is equal to:

Accepted Answer

We use the property: det(A⁻¹) = 1/det(A)

Given: det(A) = 3

Therefore:
det(A⁻¹) = 1/det(A) = 1/3

Justification: Since A · A⁻¹ = I (identity matrix),
det(A · A⁻¹) = det(I)
det(A) · det(A⁻¹) = 1  [by product rule]
det(A⁻¹) = 1/det(A)

Substituting det(A) = 3:
det(A⁻¹) = 1/3

Theorem: For an invertible matrix A, det(A⁻¹) = [det(A)]⁻¹ = 1/det(A).

Question 6

If A = [[1, 2, 3], [0, 4, 5], [0, 0, 6]] is an upper triangular matrix, then det(A) is:

Accepted Answer

For a triangular matrix (upper or lower), the determinant is the product of the diagonal elements.

Matrix A = [[1, 2, 3], [0, 4, 5], [0, 0, 6]]

Diagonal elements: 1, 4, 6

Therefore:
det(A) = 1 × 4 × 6 = 24

Verification using cofactor expansion along the first column:
det(A) = 1 · det([[4, 5], [0, 6]]) − 0 + 0
       = 1 · (4·6 − 5·0)
       = 1 · 24
       = 24 ✓

Theorem: For an upper triangular (or lower triangular) matrix, det(A) = product of diagonal elements.

Question 7

If A is a square matrix of order 2 and det(A) = 3, then det(A · adj(A)) is equal to:

Accepted Answer

We use the fundamental property: A · adj(A) = det(A) · I, where I is the identity matrix.

Taking determinants on both sides:
det(A · adj(A)) = det(det(A) · I)

For a 2×2 matrix, det(det(A) · I) = (det(A))² · det(I) = (det(A))² · 1 = (det(A))²

Since det(A) = 3, we have:
det(A · adj(A)) = 3² = 9

Alternatively, using det(AB) = det(A) · det(B):
det(A · adj(A)) = det(A) · det(adj(A)) = 3 · 3¹ = 9 (since for 2×2, det(adj(A)) = (det(A))^(n-1) = 3¹ = 3)

Question 8

If the determinant of a 3×3 matrix A is 6, and B is obtained by swapping two rows of A, then det(B) is:

Accepted Answer

We use the row operation property of determinants: When two rows (or columns) of a matrix are interchanged, the determinant changes sign.

Given: det(A) = 6
B is obtained by swapping two rows of A.

By the row swap property:
det(B) = −det(A) = −6

This property holds regardless of which two rows are swapped and applies to any square matrix.

Question 9

If A is a 3×3 matrix such that det(A) = 5, then det(2A) is equal to:

Accepted Answer

We use the property that for an n×n matrix A and scalar k, det(kA) = k^n · det(A).

Here, n = 3 (since A is 3×3), k = 2, and det(A) = 5.

Therefore:
det(2A) = 2³ · det(A)
        = 8 · 5
        = 40

The key theorem is: When all entries of an n×n matrix are multiplied by a scalar k, the determinant is multiplied by k^n, not k.

Question 10

Let A = [[1, 2], [3, 4]]. If B is the matrix obtained by interchanging the rows of A, then det(B) is:

Accepted Answer

Matrix A = [[1, 2], [3, 4]]

First, compute det(A):
det(A) = (1)(4) − (2)(3) = 4 − 6 = −2

When we interchange two rows of a matrix, the determinant changes sign.

After interchanging rows, B = [[3, 4], [1, 2]]

det(B) = (3)(2) − (4)(1) = 6 − 4 = 2

Alternatively, by the row interchange property:
det(B) = −det(A) = −(−2) = 2

Theorem used: If two rows (or columns) of a matrix are interchanged, the determinant changes sign.

Question 11

If A = [[2, 1], [1, 2]] and B = [[1, 0], [0, 1]], then det(A − B) is equal to:

Accepted Answer

First, compute A − B:
A − B = [[2, 1], [1, 2]] − [[1, 0], [0, 1]]
     = [[2−1, 1−0], [1−0, 2−1]]
     = [[1, 1], [1, 1]]

Next, compute det(A − B):
det(A − B) = det([[1, 1], [1, 1]])
           = (1)(1) − (1)(1)
           = 1 − 1
           = 0

Note: B is the identity matrix I, so A − B = A − I. The determinant is 0 because the two rows (and columns) of (A − B) are identical, making them linearly dependent.

Question 12

If A is a 3×3 matrix and det(A) = 0, which of the following statements is necessarily true?

Accepted Answer

When det(A) = 0, the matrix A is singular (non-invertible). This has several equivalent consequences:

1. The rows of A are linearly dependent.
2. The columns of A are linearly dependent.
3. The matrix A does not have an inverse (A⁻¹ does not exist).
4. The system Ax = 0 has non-trivial solutions.
5. The rank of A is less than 3.

All of these are logically equivalent statements. The question asks which is "necessarily true," meaning it must follow directly from det(A) = 0.

The most fundamental consequence is that A is singular and therefore A⁻¹ does not exist. This is the defining property of a singular matrix.

Question 13

Let A = [[1, 2], [3, 4]]. If B is a matrix such that det(B) = 2, then det(A²B) is equal to:

Accepted Answer

First, calculate det(A):
det(A) = (1)(4) − (2)(3) = 4 − 6 = −2

Next, find det(A²):
Using the property det(A²) = det(A · A) = det(A) · det(A) = [det(A)]²:
det(A²) = (−2)² = 4

Now find det(A²B):
Using det(AB) = det(A) · det(B):
det(A²B) = det(A²) · det(B) = 4 · 2 = 8

The key steps are: (1) compute det(A) correctly including sign, (2) square it to get det(A²), (3) multiply by det(B).

Question 14

Let A = [[2, 1], [3, 4]] and B = [[1, 0], [0, 1]]. If det(A − λB) = 0, then the sum of all possible values of λ is:

Accepted Answer

We compute A − λB = [[2−λ, 1], [3, 4−λ]]. The characteristic equation is det(A − λB) = 0. Expanding: (2−λ)(4−λ) − 3 = 0 ⟹ 8 − 2λ − 4λ + λ² − 3 = 0 ⟹ λ² − 6λ + 5 = 0. By Vieta's formulas, the sum of roots = 6. Alternatively, the sum of eigenvalues equals the trace of A: tr(A) = 2 + 4 = 6.

Question 15

Let A be a 3×3 matrix such that det(A) = 5. If B is the matrix obtained by interchanging the first and second rows of A, and C is the matrix obtained by multiplying the third row of A by 3, then det(B) + det(C) equals:

Accepted Answer

We use the properties of determinants:

1. When two rows of a matrix are interchanged, the determinant changes sign.
   Therefore: det(B) = −det(A) = −5

2. When one row of a matrix is multiplied by a scalar k, the determinant is multiplied by k.
   Therefore: det(C) = 3·det(A) = 3(5) = 15

3. Thus: det(B) + det(C) = −5 + 15 = 10

Question 16

If A is a 3×3 matrix with det(A) = 4, then det(2A) equals:

Accepted Answer

We use the property that when a matrix of order n is multiplied by a scalar k, the determinant is multiplied by k^n.

For a 3×3 matrix A multiplied by scalar 2:
det(2A) = 2³·det(A) = 8·det(A) = 8(4) = 32

The key insight is that the scalar multiplies EVERY element of the matrix, affecting each of the 3 rows independently.

Question 17

Let A = [1, 2; 3, 4] (a 2×2 matrix). If B is obtained by adding 2 times the first row of A to the second row, then det(B) − det(A) equals:

Accepted Answer

We use the property that adding a multiple of one row to another row does NOT change the determinant.

Step 1: Calculate det(A).
A = [1, 2; 3, 4]
det(A) = (1)(4) − (2)(3) = 4 − 6 = −2

Step 2: Find B by adding 2 times row 1 to row 2.
Row 1 of A: [1, 2]
Row 2 of A: [3, 4]
New Row 2: [3, 4] + 2[1, 2] = [3 + 2, 4 + 4] = [5, 8]
B = [1, 2; 5, 8]

Step 3: Calculate det(B).
det(B) = (1)(8) − (2)(5) = 8 − 10 = −2

Step 4: Compute det(B) − det(A).
det(B) − det(A) = −2 − (−2) = 0

Question 18

Let A be a 3×3 matrix such that det(A) = 6. If A^T denotes the transpose of A, then det(A^T) + det(A) equals:

Accepted Answer

We use the property that the determinant of a matrix equals the determinant of its transpose.

Property: det(A^T) = det(A)

Given: det(A) = 6
Therefore: det(A^T) = 6

Thus: det(A^T) + det(A) = 6 + 6 = 12

This property holds because the determinant can be computed by expanding along any row or column, and transposing converts rows to columns (and vice versa), but the value remains unchanged.

Question 19

Let A and B be 3×3 matrices with det(A) = 3 and det(B) = −2. If C = AB (matrix product), then det(C) equals:

Accepted Answer

We use the multiplicative property of determinants: det(AB) = det(A)·det(B).

Given:
det(A) = 3
det(B) = −2
C = AB

By the multiplicative property:
det(C) = det(AB) = det(A)·det(B) = 3·(−2) = −6

Question 20

Let A be a 3×3 matrix such that det(A) = 5. If B is the matrix obtained by interchanging the first and third rows of A, then det(B) is:

Accepted Answer

By the property of determinants, when two rows (or columns) of a matrix are interchanged, the determinant changes sign. Since we interchange rows 1 and 3 of matrix A to obtain B, we have: det(B) = −det(A) = −5.

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