Find the rank of the matrix by minor method: − 1 3 4 − 7 3 − 4

Let A = − 1 3 4 − 7 3 − 4 Its order is 3 × 2 . So, rank of A is ρ ( A ) ≤ min { 3 , 2 } = 2 ⇒ ρ ( A ) ≤ 2 The highest order minor of A is 2 . A has several second order minor. Consider second order minor of A − 1 3 4 − 7 = 7 − 12 = - 5 ≠ 0 So, A has atleast one non-vanishing minor of order 2 . So, Rank of A , ρ ( A ) = 2 .

Find the rank of the matrix by minor method: − 1 3 4 − 7 3 − 4

Let A = − 1 3 4 − 7 3 − 4 Its order is 3 × 2 . So, rank of A is ρ ( A ) ≤ min { 3 , 2 } = 2 ⇒ ρ ( A ) ≤ 2 The highest order minor of A is 2 . A has several second order minor. Consider second order minor of A − 1 3 4 − 7 = 7 − 12 = - 5 ≠ 0 So, A has atleast one non-vanishing minor of order 2 . So, Rank of A , ρ ( A ) = 2 .

Find the rank of the matrix by minor method: 1 − 2 − 1 0 3 − 6 − 3 1

Let A = 1 − 2 − 1 0 3 − 6 − 3 1 Its order is 2 × 4 . So, rank of A , ρ ( A ) ≤ min ( 2 , 4 ) = 2 Hence, ρ ( A ) ≤ 2 The highest order minor of A is 2 . A has several minor of order 2 . Consider second order minor of A . − 1 0 − 3 1 = − 1 − 0 = − 1 ≠ 0 So, A has atleast one non-vanishing minor of order 2 . So, Rank of A , ρ ( A ) = 2 .

Find the rank of the matrix by minor method: 1 − 2 3 2 4 − 6 5 1 − 1

Let A = 1 − 2 3 2 4 − 6 5 1 − 1 Its order is 3 × 3 . So, rank of A , ρ ( A ) ≤ min { 3 , 3 } = 3 The highest order minor of A is 3 and A has only one third order minor. 1 − 2 3 2 4 − 6 5 1 − 1 = 1 ( − 4 + 6 ) + 2 ( − 2 + 30 ) + 3 ( 2 − 20 ) = 1 2 + 2 28 + 3 − 18 = 2 + 56 - 54 = 4 ≠ 0 So, A has a third order non-vanishing minor. Hence the rank of A , ρ ( A ) = 3 .

Find the rank of the matrix by minor method: 0 1 2 1 0 2 4 3 8 1 0 2

Let A = 0 1 2 1 0 2 4 3 8 1 0 2 Its order is 3 × 4 . So, rank of A , ρ ( A ) ≤ min { 3 , 4 } = 3 ⇒ ρ ( A ) ≤ 3 The highest order minor of A is 3 . A has several third order minor. Consider, the third order minor of A . 0 1 1 0 2 3 8 1 2 = 0 ( 4 − 3 ) − 1 ( 0 − 24 ) + 1 ( 0 − 16 ) = 0 + 24 − 16 = 8 ≠ 0 So, A has atleast one non-vanishing minor of order 3 . So, rank of A , ρ ( A ) = 3 .

Find the rank of the matrix by row reduction method: 1 1 1 3 2 − 1 3 4 5 − 1 7 11

Let A = 1 1 1 3 2 − 1 3 4 5 − 1 7 11 Apply, R 2 → R 2 − 2 R 1 and R 3 → R 3 − 5 R 1 = 1 1 1 3 0 - 3 1 - 2 0 - 6 2 - 4 Apply, R 3 → R 3 − 2 R 2 = 1 1 1 3 0 − 3 1 − 2 0 0 0 0 The last equivalent matrix is a row echelon matrix having two non-zero rows. Hence, rank of A is ρ ( A ) = 2 .

Find the adjoint of the matrix A = 1 2 3 - 5 and verify the result A ( adj A ) = ( adj A ) A = | A | · I .

Given that, the matrix is A = 1 2 3 - 5 Adjoint of a 2 × 2 matrix is obtained by interchanging diagonal elements and adding negative sign in front of other elements. ∴ Adj A = - 5 - 2 - 3 1 Now, A ( adj A ) = 1 2 3 - 5 - 5 - 2 - 3 1 = - 5 - 6 - 2 + 2 - 15 + 15 - 6 - 5 = - 11 0 0 - 11 Similarly, ( adj A ) A = - 5 - 2 - 3 1 1 2 3 - 5 = - 5 - 6 - 10 + 10 - 3 + 3 - 6 - 5 = - 11 0 0 - 11 Also, A · I = - 11 1 0 0 1 = - 11 0 0 - 11 Hence, A ( adj A ) = ( adj A ) A = | A | · I

MEDIUM

12th Tamil Nadu Board

IMPORTANT

Earn 100

Find the rank of the matrix by minor method: $[\begin{array}{rr} - 1 3 \\ 4 - 7 \\ 3 - 4 \end{array}]$

Important Questions on Applications of Matrices and Determinants

MEDIUM

12th Tamil Nadu Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 1 - Applications of Matrices and Determinants>Exercise>Q 20

Find the rank of the matrix by minor method:

[\begin{array}{rrrr} 1 - 2 - 1 0 \\ 3 - 6 - 3 1 \end{array}]

MEDIUM

12th Tamil Nadu Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 1 - Applications of Matrices and Determinants>Exercise>Q 21

Find the rank of the matrix by minor method:

[\begin{array}{ccc} 1 & - 2 & 3 \\ 2 & 4 & - 6 \\ 5 & 1 & - 1 \end{array}]

MEDIUM

12th Tamil Nadu Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 1 - Applications of Matrices and Determinants>Exercise>Q 22

Find the rank of the matrix by minor method:

[\begin{array}{llll} 0 1 2 1 \\ 0 2 4 3 \\ 8 1 0 2 \end{array}]

HARD

12th Tamil Nadu Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 1 - Applications of Matrices and Determinants>Exercise>Q 23

Find the rank of the matrix by row reduction method:

[\begin{array}{cccc} 1 & 1 & 1 & 3 \\ 2 & - 1 & 3 & 4 \\ 5 & - 1 & 7 & 11 \end{array}]

HARD

12th Tamil Nadu Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 1 - Applications of Matrices and Determinants>Exercise>Q 24

Solve the following equations by Gauss-Jordan method

3 x + 4 y + 5 z = 18, 2 x - y + 8 z = 13

and

5 x - 2 y + 7 z = 20

HARD

12th Tamil Nadu Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 1 - Applications of Matrices and Determinants>Exercise>Q 25

Solve the system of equation by using Gauss-Jordan method. Also determine whether the system has a unique solution or infinite number of solutions or no solution and find the solutions if exists.

$5 x - 6 y + 4 z = 15, 7 x + 4 y - 3 z = 19, 2 x + y + 6 z = 46$

HARD

12th Tamil Nadu Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 1 - Applications of Matrices and Determinants>Exercise>Q 26

Solve the system of equation by Gauss-Jordan method. Also determine whether the system has a unique solution or infinite number of solutions or no solution and find the solutions if exists.

$x + y + z = 1, 2 x + 2 y + 3 z = 6, x + 4 y + 9 z = 3$

HARD

12th Tamil Nadu Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 1 - Applications of Matrices and Determinants>Exercise>Q 27

Solve the system of equation by using Gauss-Jordan method. Also determine whether the system has a unique solution or infinite number of solutions or no solution and find the solution if exists.

$x - y + 3 z = 5, 4 x + 2 y - z = 0, - x + 3 y + z = 5$

Find the rank of the matrix by minor method: −1 34 −73 −4

Important Questions on Applications of Matrices and Determinants

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Find the rank of the matrix by minor method: $[\begin{array}{rr} - 1 3 \\ 4 - 7 \\ 3 - 4 \end{array}]$