{"@context":"https://schema.org/","@type":"Quiz","name":"

Determinant of $\\begin{bmatrix}4 & 5 & 4 \\\\2 & 1 & 2 \\\\ 3 & 7 & 3 \\end{bmatrix}$ is:

","hasPart":{"@type":"Question","typicalAgeRange":"12-22","educationalAlignment":[{"@type":"AlignmentObject","alignmentType":"educationalSubject","targetName":"NET"},{"@type":"AlignmentObject","alignmentType":"educationalLevel","targetName":"NET","educationalFramework":"Common Core","targetUrl":"https://classnotes.xyz/mcqs/determinant-of-begin-bmatrix-4-5-4-2-1-2-3-7-3-end-bmatrix-is"}],"about":{"@type":"Thing","name":"NET, NUST Entry Test, Mathematics, Matrices and Determinants, Quiz"},"educationalLevel":"intermediate","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","assesses":["NET","NUST Entry Test","Mathematics","Matrices and Determinants","Quiz"],"name":"

Determinant of $\\begin{bmatrix}4 & 5 & 4 \\\\2 & 1 & 2 \\\\ 3 & 7 & 3 \\end{bmatrix}$ is:

","comment":{"@type":"Comment","text":"

The determinant of a 3x3 matrix can be calculated using the formula:

det(A) = a11(a22a33 - a23a32) - a12(a21a33 - a23a31) + a13(a21a32 - a22a31)

where aij is the element in the i-th row and j-th column of the matrix.

For the matrix $\\begin{bmatrix}4 & 5 & 4 \\\\2 & 1 & 2 \\\\ 3 & 7 & 3 \\end{bmatrix}$, the determinant can be calculated as follows:

det(A) = 4(13- 22) - 5(23 - 23) + 4(22 - 13) = 4(1 - 4) + 5(0) + 4(4 - 3) = -16 + 16 = 0

So, the determinant of the matrix $\\begin{bmatrix}4 & 5 & 4 \\\\2 & 1 & 2 \\\\ 3 & 7 & 3 \\end{bmatrix}$ is 0.

Important Note: As the 1st column and the 3rd column are identical, the determinant will always be zero in such cases.

"},"encodingFormat":"text/markdown","text":"

Determinant of $\\begin{bmatrix}4 & 5 & 4 \\\\2 & 1 & 2 \\\\ 3 & 7 & 3 \\end{bmatrix}$ is:

","suggestedAnswer":[{"@type":"Answer","position":0,"encodingFormat":"text/html","text":"

"},{"@type":"Answer","position":1,"encodingFormat":"text/html","text":"

$-42$

"},{"@type":"Answer","position":2,"encodingFormat":"text/html","text":"

"}],"acceptedAnswer":{"@type":"Answer","position":3,"encodingFormat":"text/html","text":"

None of these

","comment":{"@type":"Comment","text":"

The determinant of a 3x3 matrix can be calculated using the formula:

det(A) = a11(a22a33 - a23a32) - a12(a21a33 - a23a31) + a13(a21a32 - a22a31)

where aij is the element in the i-th row and j-th column of the matrix.

For the matrix $\\begin{bmatrix}4 & 5 & 4 \\\\2 & 1 & 2 \\\\ 3 & 7 & 3 \\end{bmatrix}$, the determinant can be calculated as follows:

det(A) = 4(13- 22) - 5(23 - 23) + 4(22 - 13) = 4(1 - 4) + 5(0) + 4(4 - 3) = -16 + 16 = 0

So, the determinant of the matrix $\\begin{bmatrix}4 & 5 & 4 \\\\2 & 1 & 2 \\\\ 3 & 7 & 3 \\end{bmatrix}$ is 0.

Important Note: As the 1st column and the 3rd column are identical, the determinant will always be zero in such cases.

"},"answerExplanation":{"@type":"Comment","text":"

The determinant of a 3x3 matrix can be calculated using the formula:

det(A) = a11(a22a33 - a23a32) - a12(a21a33 - a23a31) + a13(a21a32 - a22a31)

where aij is the element in the i-th row and j-th column of the matrix.

For the matrix $\\begin{bmatrix}4 & 5 & 4 \\\\2 & 1 & 2 \\\\ 3 & 7 & 3 \\end{bmatrix}$, the determinant can be calculated as follows:

det(A) = 4(13- 22) - 5(23 - 23) + 4(22 - 13) = 4(1 - 4) + 5(0) + 4(4 - 3) = -16 + 16 = 0

So, the determinant of the matrix $\\begin{bmatrix}4 & 5 & 4 \\\\2 & 1 & 2 \\\\ 3 & 7 & 3 \\end{bmatrix}$ is 0.

Important Note: As the 1st column and the 3rd column are identical, the determinant will always be zero in such cases.

"}}},"url":"https://classnotes.xyz/mcqs/determinant-of-begin-bmatrix-4-5-4-2-1-2-3-7-3-end-bmatrix-is"}

Determinant of $\begin{bmatrix}4 & 5 & 4 \\2 & 1 & 2 \\ 3 & 7 & 3 \end{bmatrix}$ is: