Invertible Matrices

The number of 3\times3 matrices whose entries are either 0 or 1 is calculated as follows: each entry has 2 choices (either 0 or 1). Therefore, the total number of such matrices is 2 9 = 512. Next, we need to compute how many of these matrices have a sum of all entries that is a prime number. The sum of the entries is a number between 0 and 9 (inclusive) since the matrix can have 0 ones (all zero matrix) to 9 ones (all one matrix). We evaluate which of these sums are prime numbers. The prime numbers between 0 and 9 are: 2, 3, 5, 7. We compute the number of matrices for each prime sum: Sum = 2: Choose 2 positions from 9 to be 1, i.e., combinations of 9 taken 2 = C(9,2)=36. Sum = 3: Choose 3 positions from 9 to be 1, i.e., combinations of 9 taken 3 = C(9,3)=84. Sum = 5: Choose 5 positions from 9 to be 1, i.e., combinations of 9 taken 5 = C(9,5)=126. Sum = 7: Choose 7 positions from 9 to be 1, i.e., combinations of 9 taken 7 = C(9,7)=36. Adding these gives: 36 + 84 + 126 + 36 = 282 This confirms that the computed number of such matrices is 282, which is within the expected range (282,282). Therefore, the number of 3\times3 matrices with entries 0 or 1, and a prime sum is 282 .

To determine which matrix cannot be obtained from the given matrix $%5Cbegin%7Bbmatrix%7D%20-1%20%26%202%20%5C%5C%201%20%26%20-1%20%5Cend%7Bbmatrix%7D$ by a single elementary row operation, let's first recall the three types of elementary row operations:

Swapping two rows.
Multiplying a row by a non-zero scalar.
Adding or subtracting a multiple of one row to another row.

Let's examine the options one by one:

$%5Cbegin%7Bbmatrix%7D%200%20%26%201%20%5C%5C%201%20%26%20-1%20%5Cend%7Bbmatrix%7D$

To achieve this, we can add the first row to the second row:

R_1%20%2B%20R_2%20%5Crightarrow%20R_1

%5Cbegin%7Bbmatrix%7D%20-1%20%26%202%20%5C%5C%201%20%26%20-1%20%5Cend%7Bbmatrix%7D%20%5Crightarrow%20%5Cbegin%7Bbmatrix%7D%200%20%26%201%20%5C%5C%201%20%26%20-1%20%5Cend%7Bbmatrix%7D

$%5Cbegin%7Bbmatrix%7D%201%20%26%20-1%20%5C%5C%20-1%20%26%202%20%5Cend%7Bbmatrix%7D$

To obtain this matrix, multiply the first row by $-1$ and the second row by $-1$ :

-R_1%20%5Crightarrow%20R_1

and

-R_2%20%5Crightarrow%20R_2

%5Cbegin%7Bbmatrix%7D%20-1%20%26%202%20%5C%5C%201%20%26%20-1%20%5Cend%7Bbmatrix%7D%20%5Crightarrow%20%5Cbegin%7Bbmatrix%7D%201%20%26%20-1%20%5C%5C%20-1%20%26%202%20%5Cend%7Bbmatrix%7D

$%5Cbegin%7Bbmatrix%7D%20-1%20%26%202%20%5C%5C%20-2%20%26%207%20%5Cend%7Bbmatrix%7D$

To achieve this transformation, it requires adding some scalar multiple of the first row to the second row; let's verify:

%5Cbegin%7Bbmatrix%7D%20-1%20%26%202%20%5C%5C%201%20%26%20-1%20%5Cend%7Bbmatrix%7D%20%5Crightarrow%20%5Cbegin%7Bbmatrix%7D%20-1%20%26%202%20%5C%5C%20(-2)(-1)%20%2B%201%20%26%20(-2)(2)%20%2B%20(-1)%20%5Cend%7Bbmatrix%7D

%5CRightarrow%20(-2)%20%5Ctimes%20R_1%20%2B%20R_2%20%5Crightarrow%20R_2

By calculating: $-2%20%5Ctimes%20(-1)%20%2B%201%20%3D%202%20%2B%201%20%3D%203$ and $-2%20%5Ctimes%202%20%2B%20(-1)%20%3D%20-4%20-%201%20%3D%20-5$ . Thus incorrect, it cannot be obtained with a single operation.

$%5Cbegin%7Bbmatrix%7D%20-1%20%26%202%20%5C%5C%20-1%20%26%203%20%5Cend%7Bbmatrix%7D$

Achieve this by subtracting the first row from the second row:

R_2%20-%20R_1%20%5Crightarrow%20R_2

%5Cbegin%7Bbmatrix%7D%20-1%20%26%202%20%5C%5C%201%20%26%20-1%20%5Cend%7Bbmatrix%7D%20%5Crightarrow%20%5Cbegin%7Bbmatrix%7D%20-1%20%26%202%20%5C%5C%20-1%20%26%203%20%5Cend%7Bbmatrix%7D

Therefore, the matrix that cannot be obtained from the original matrix $%5Cbegin%7Bbmatrix%7D%20-1%20%26%202%20%5C%5C%201%20%26%20-1%20%5Cend%7Bbmatrix%7D$ by a single elementary row operation is $%5Cbegin%7Bbmatrix%7D%20-1%20%26%202%20%5C%5C%20-2%20%26%207%20%5Cend%7Bbmatrix%7D$ .

High-Yield Trend

Chapter Questions
2 MCQs

Official Solution

Official Solution

Invertible Matrices

High-Yield Trend

Chapter Questions 2 MCQs

Official Solution

Official Solution

Chapter Questions
2 MCQs