Question 1

Let A = [[3, 1], [1, 3]]. Which of the following statements about A is correct?

Accepted Answer

We need to check properties of A = [[3, 1], [1, 3]].

First, compute det(A):
det(A) = (3)(3) - (1)(1) = 9 - 1 = 8 ≠ 0

Since det(A) ≠ 0, the matrix A is invertible (non-singular).

Second, check if A is symmetric:
A^T = [[3, 1], [1, 3]] = A

Yes, A is symmetric because A = A^T.

Therefore, A is both invertible and symmetric.

Question 2

If A = [[1, 2], [3, 4]], then the transpose of A, denoted Aᵀ, is obtained by swapping rows and columns. Which of the following is Aᵀ?

Accepted Answer

The transpose of a matrix A is obtained by interchanging its rows and columns.

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

Row 1 of A is [1, 2] → becomes Column 1 of Aᵀ
Row 2 of A is [3, 4] → becomes Column 2 of Aᵀ

Therefore:
Aᵀ = [[1, 3], [2, 4]]

Verification: (Aᵀ)ᵢⱼ = Aⱼᵢ
(Aᵀ)₁₁ = A₁₁ = 1 ✓
(Aᵀ)₁₂ = A₂₁ = 3 ✓
(Aᵀ)₂₁ = A₁₂ = 2 ✓
(Aᵀ)₂₂ = A₂₂ = 4 ✓

Question 3

Let A = [[2, 1], [1, 2]]. The determinant of A, denoted det(A), is calculated as det(A) = ad − bc for a 2×2 matrix [[a, b], [c, d]]. What is det(A)?

Accepted Answer

For a 2×2 matrix A = [[a, b], [c, d]], the determinant is:
det(A) = ad − bc

Given A = [[2, 1], [1, 2]], we have:
a = 2, b = 1, c = 1, d = 2

det(A) = (2)(2) − (1)(1)
       = 4 − 1
       = 3

Therefore, det(A) = 3.

Question 4

If A = [[3, 1], [2, 1]], what is det(A)?

Accepted Answer

For a 2×2 matrix A = [[a, b], [c, d]], the determinant is det(A) = ad − bc.

Here, A = [[3, 1], [2, 1]], so:
- a = 3, b = 1
- c = 2, d = 1

det(A) = (3)(1) − (1)(2)
       = 3 − 2
       = 1

Therefore, det(A) = 1.

Question 5

Let A = [[2, 1], [3, 4]] and B = [[1, 0], [2, 1]]. Find the element in position (1,2) of the matrix product AB.

Accepted Answer

To find AB, we use matrix multiplication where the element at position (i,j) is the dot product of row i of A with column j of B.

A = [[2, 1], [3, 4]], B = [[1, 0], [2, 1]]

Element (1,2) of AB = (row 1 of A) · (column 2 of B)
= [2, 1] · [0, 1]
= 2(0) + 1(1)
= 0 + 1
= 1

Therefore, the element at position (1,2) of AB is 1.

Question 6

If A = [[1, 2], [3, 4]] and B = [[5, 6], [7, 8]], then (A + B) equals:

Accepted Answer

Matrix addition is performed element-wise: (A + B)ᵢⱼ = Aᵢⱼ + Bᵢⱼ.

A + B = [[1 + 5, 2 + 6], [3 + 7, 4 + 8]]
      = [[6, 8], [10, 12]]

Verification:
(A + B)₁₁ = 1 + 5 = 6 ✓
(A + B)₁₂ = 2 + 6 = 8 ✓
(A + B)₂₁ = 3 + 7 = 10 ✓
(A + B)₂₂ = 4 + 8 = 12 ✓

Question 7

If A = [[1, 2], [3, 4]], then the transpose of A, denoted A^T, is which of the following?

Accepted Answer

The transpose of a matrix is obtained by interchanging its rows and columns. If A = [[a, b], [c, d]], then A^T = [[a, c], [b, d]].

Given A = [[1, 2], [3, 4]]:
- Row 1 of A is [1, 2] → becomes Column 1 of A^T
- Row 2 of A is [3, 4] → becomes Column 2 of A^T

Therefore, A^T = [[1, 3], [2, 4]].

Question 8

Let A = [[2, 0], [0, 3]]. What is the determinant of A?

Accepted Answer

For a 2×2 matrix A = [[a, b], [c, d]], the determinant is calculated as:
det(A) = ad - bc

Given A = [[2, 0], [0, 3]]:
det(A) = (2)(3) - (0)(0)
= 6 - 0
= 6

Therefore, det(A) = 6.

Question 9

For a 2×2 matrix A = [[a, b], [c, d]], the adjugate (or adjoint) matrix adj(A) is defined as [[d, −b], [−c, a]]. If A = [[3, 2], [1, 4]], what is adj(A)?

Accepted Answer

For a 2×2 matrix A = [[a, b], [c, d]], the adjugate matrix is:
adj(A) = [[d, −b], [−c, a]]

Given A = [[3, 2], [1, 4]], we identify:
a = 3, b = 2, c = 1, d = 4

Therefore:
adj(A) = [[4, −2], [−1, 3]]

Verification of the formula:
- Top-left: d = 4 ✓
- Top-right: −b = −2 ✓
- Bottom-left: −c = −1 ✓
- Bottom-right: a = 3 ✓

Question 10

If A = [[1, 2], [2, 1]], find A².

Accepted Answer

To find A², we compute A × A using matrix multiplication.

A = [[1, 2], [2, 1]]

A² = A × A = [[1, 2], [2, 1]] × [[1, 2], [2, 1]]

Element (1,1): (1)(1) + (2)(2) = 1 + 4 = 5
Element (1,2): (1)(2) + (2)(1) = 2 + 2 = 4
Element (2,1): (2)(1) + (1)(2) = 2 + 2 = 4
Element (2,2): (2)(2) + (1)(1) = 4 + 1 = 5

Therefore, A² = [[5, 4], [4, 5]].

Question 11

If A = [[1, 2], [2, 1]], then A + A^T equals which of the following?

Accepted Answer

First, find A^T by transposing A:
A = [[1, 2], [2, 1]]
A^T = [[1, 2], [2, 1]]

Note: A is a symmetric matrix, so A = A^T.

Now compute A + A^T:
A + A^T = [[1, 2], [2, 1]] + [[1, 2], [2, 1]]
= [[1+1, 2+2], [2+2, 1+1]]
= [[2, 4], [4, 2]]

Therefore, A + A^T = [[2, 4], [4, 2]].

Question 12

If A = [[1, 2], [3, 4]], what is the transpose of A, denoted Aᵀ?

Accepted Answer

The transpose of a matrix is obtained by interchanging its rows and columns. Each element aᵢⱼ becomes aⱼᵢ.

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

Row 1 of A is [1, 2] → becomes column 1 of Aᵀ
Row 2 of A is [3, 4] → becomes column 2 of Aᵀ

Alternatively:
- Element (1,1): a₁₁ = 1 → (Aᵀ)₁₁ = 1
- Element (1,2): a₁₂ = 2 → (Aᵀ)₂₁ = 2
- Element (2,1): a₂₁ = 3 → (Aᵀ)₁₂ = 3
- Element (2,2): a₂₂ = 4 → (Aᵀ)₂₂ = 4

Therefore, Aᵀ = [[1, 3], [2, 4]].

Question 13

If A and B are 2×2 matrices such that AB = BA = I, then which statement is necessarily false?

Accepted Answer

Step 1: Given AB = BA = I, this means B is the inverse of A, i.e., B = A⁻¹.

Step 2: From AB = I, we know A is invertible and B = A⁻¹.

Step 3: From BA = I, we confirm the same relationship (since matrix multiplication is not commutative in general, both conditions together guarantee B = A⁻¹).

Step 4: Analyze each statement:
(a) A is invertible: TRUE (since A⁻¹ = B exists)
(b) det(A) · det(B) = 1: TRUE (since det(AB) = det(I) ⟹ det(A)·det(B) = 1)
(c) A + B = I: NOT NECESSARILY TRUE. Counterexample: A = [[2, 0], [0, 2]], B = [[1/2, 0], [0, 1/2]]. Then AB = BA = I, but A + B = [[5/2, 0], [0, 5/2]] ≠ I.
(d) (AB)ᵀ = I: TRUE (since (AB)ᵀ = Bᵀ·Aᵀ, and if AB = I, then (AB)ᵀ = Iᵀ = I)

Therefore, statement (c) is necessarily false.

Question 14

Let A = [[a, b], [c, d]] be a 2×2 matrix with det(A) = 5. If B is obtained from A by swapping its rows, then det(B) is:

Accepted Answer

Step 1: Recall the elementary row operation property: swapping two rows of a matrix multiplies the determinant by −1.

Step 2: Matrix A = [[a, b], [c, d]] has det(A) = ad − bc = 5.

Step 3: When we swap rows, B = [[c, d], [a, b]].

Step 4: Calculate det(B):
det(B) = c·b − d·a = −(ad − bc) = −det(A) = −5

Alternatively, using the row swap property directly:
det(B) = −det(A) = −5

Step 5: Verify with an example: A = [[1, 2], [3, 4]], det(A) = 1·4 − 2·3 = −2.
B = [[3, 4], [1, 2]], det(B) = 3·2 − 4·1 = 2 = −(−2) ✓

Question 15

Let A = [[2, 1], [1, 2]] and B = [[1, -1], [-1, 1]]. If C = A² - 2AB + B², then the trace of C is:

Accepted Answer

We need to compute A², AB, B², then form C = A² - 2AB + B².

Step 1: Compute A².
A² = [[2, 1], [1, 2]] × [[2, 1], [1, 2]]
   = [[2·2 + 1·1, 2·1 + 1·2], [1·2 + 2·1, 1·1 + 2·2]]
   = [[5, 4], [4, 5]]

Step 2: Compute AB.
AB = [[2, 1], [1, 2]] × [[1, -1], [-1, 1]]
   = [[2·1 + 1·(-1), 2·(-1) + 1·1], [1·1 + 2·(-1), 1·(-1) + 2·1]]
   = [[1, -1], [-1, 1]]

Step 3: Compute B².
B² = [[1, -1], [-1, 1]] × [[1, -1], [-1, 1]]
   = [[1·1 + (-1)·(-1), 1·(-1) + (-1)·1], [(-1)·1 + 1·(-1), (-1)·(-1) + 1·1]]
   = [[2, -2], [-2, 2]]

Step 4: Compute C = A² - 2AB + B².
C = [[5, 4], [4, 5]] - 2[[1, -1], [-1, 1]] + [[2, -2], [-2, 2]]
  = [[5, 4], [4, 5]] - [[2, -2], [-2, 2]] + [[2, -2], [-2, 2]]
  = [[5 - 2 + 2, 4 - (-2) + (-2)], [4 - (-2) + (-2), 5 - 2 + 2]]
  = [[5, 4], [4, 5]]

Step 5: Compute trace(C).
trace(C) = 5 + 5 = 10

Note: The expression C = A² - 2AB + B² is NOT equal to (A - B)² in general because matrix multiplication is non-commutative. However, the numerical result here is 10.

Question 16

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

Accepted Answer

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

Given: det(A) = 5, and A is 3×3, so n = 3.

We need det(2A).
By the scalar multiplication property:
det(2A) = 2³ · det(A) = 8 · 5 = 40

Explanation of the property: When we multiply a matrix by a scalar k, each row is multiplied by k. Since the determinant is multilinear in rows, multiplying one row by k multiplies the determinant by k. Multiplying all n rows by k multiplies the determinant by k^n.

Question 17

Let A = [[1, 2], [3, 4]] and B = [[5, 6], [7, 8]]. If (AB)ᵀ = BᵀAᵀ, then which of the following is true?

Accepted Answer

We verify the transpose property: (AB)ᵀ = BᵀAᵀ.

Step 1: Compute AB.
AB = [[1, 2], [3, 4]] × [[5, 6], [7, 8]]
   = [[1·5 + 2·7, 1·6 + 2·8], [3·5 + 4·7, 3·6 + 4·8]]
   = [[19, 22], [43, 50]]

Step 2: Compute (AB)ᵀ.
(AB)ᵀ = [[19, 43], [22, 50]]

Step 3: Compute Aᵀ and Bᵀ.
Aᵀ = [[1, 3], [2, 4]]
Bᵀ = [[5, 7], [6, 8]]

Step 4: Compute BᵀAᵀ.
BᵀAᵀ = [[5, 7], [6, 8]] × [[1, 3], [2, 4]]
     = [[5·1 + 7·2, 5·3 + 7·4], [6·1 + 8·2, 6·3 + 8·4]]
     = [[19, 43], [22, 50]]

Step 5: Verify.
(AB)ᵀ = [[19, 43], [22, 50]] = BᵀAᵀ ✓

This is a fundamental property of matrix transpose: (AB)ᵀ = BᵀAᵀ (note the reversal of order). This holds for ANY matrices A and B (of compatible dimensions), not just these specific ones.

Question 18

If A = [[a, b], [c, d]] is a 2×2 matrix with det(A) = 10, then det(adj(A)) equals:

Accepted Answer

We use the fundamental property relating a matrix, its adjugate, and its determinant.

For an n×n matrix A:
A · adj(A) = det(A) · I

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

Using det(XY) = det(X) · det(Y) on the left:
det(A) · det(adj(A)) = det(det(A) · I)

For the right side, det(kI) = k^n for an n×n matrix:
det(A) · det(adj(A)) = (det(A))^n

Dividing both sides by det(A) (assuming det(A) ≠ 0):
det(adj(A)) = (det(A))^(n-1)

For n = 2:
det(adj(A)) = (det(A))^(2-1) = det(A)^1 = det(A) = 10

Alternatively, for a 2×2 matrix A = [[a, b], [c, d]]:
adj(A) = [[d, -b], [-c, a]]
det(adj(A)) = d·a - (-b)·(-c) = ad - bc = det(A) = 10

Question 19

Let A = [[2, 0], [0, 3]] and B = [[1, 1], [0, 1]]. If X is a 2×2 matrix such that AX = B, then the sum of all entries of X is:

Accepted Answer

We need to solve AX = B for X, then sum all entries of X.

Step 1: Find A⁻¹.
A = [[2, 0], [0, 3]] is diagonal.
det(A) = 2 · 3 = 6 ≠ 0, so A is invertible.
A⁻¹ = [[1/2, 0], [0, 1/3]]

Step 2: Solve for X.
AX = B
X = A⁻¹B

Step 3: Compute X = A⁻¹B.
X = [[1/2, 0], [0, 1/3]] × [[1, 1], [0, 1]]
  = [[1/2·1 + 0·0, 1/2·1 + 0·1], [0·1 + 1/3·0, 0·1 + 1/3·1]]
  = [[1/2, 1/2], [0, 1/3]]

Step 4: Verify AX = B.
AX = [[2, 0], [0, 3]] × [[1/2, 1/2], [0, 1/3]]
   = [[2·1/2 + 0·0, 2·1/2 + 0·1/3], [0·1/2 + 3·0, 0·1/2 + 3·1/3]]
   = [[1, 1], [0, 1]] = B ✓

Step 5: Sum all entries of X.
Sum = 1/2 + 1/2 + 0 + 1/3 = 1 + 1/3 = 4/3

Question 20

Let A = [[1, 2], [3, 4]] and B = [[5, 6], [7, 8]]. If (AB)^T = B^T A^T, which of the following is true?

Accepted Answer

We verify the fundamental property: (AB)^T = B^T A^T (note the reversal of order).

Step 1: Compute AB.
AB = [[1, 2], [3, 4]] × [[5, 6], [7, 8]]
= [[1(5) + 2(7), 1(6) + 2(8)], [3(5) + 4(7), 3(6) + 4(8)]]
= [[5 + 14, 6 + 16], [15 + 28, 18 + 32]]
= [[19, 22], [43, 50]]

Step 2: Compute (AB)^T.
(AB)^T = [[19, 43], [22, 50]]

Step 3: Compute B^T and A^T.
B^T = [[5, 7], [6, 8]]
A^T = [[1, 3], [2, 4]]

Step 4: Compute B^T A^T.
B^T A^T = [[5, 7], [6, 8]] × [[1, 3], [2, 4]]
= [[5(1) + 7(2), 5(3) + 7(4)], [6(1) + 8(2), 6(3) + 8(4)]]
= [[5 + 14, 15 + 28], [6 + 16, 18 + 32]]
= [[19, 43], [22, 50]]

Step 5: Compare.
(AB)^T = [[19, 43], [22, 50]] = B^T A^T ✓

The statement is TRUE. This is a fundamental property of transposes: the transpose of a product reverses the order of multiplication.

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