10th Class Mathematics Unit 3 Matrices and Determinants Numericals with Solutions

3.3.1.Which of the following matrices are conformable for addition and subtraction? A=[[3,4],[-2,1]], B=[[3],[4]], C=[5,2], D=[[1,7],[2,5]], E=[2], F=[7,11], G=[[a],[b]], H=[3], M=[[l],[m]]

Orders A:2 × 2, B:2 × 1, C:1 × 2, D:2 × 2, E:1 × 1, F:1 × 2, G:2 × 1, H:1 × 1, M:2 × 1

Result

Conformable pairs (same order) (A,D), (B,G,M), (E,H), (C,F)

3.3.2.If X=[[1,-1],[-2,2]], Y=[[1,2],[3,4]] and Z=[[3,0],[0,-2]], then find: (i) X+Y (ii) Y+7Z (iii) 4X-Z (iv) X+2Y+3Z (v) X-4Y+Z (vi) Z-Z

Result

(i) X+Y=begin{bmatrix}2&11&6end{bmatrix}

(ii) Y+7Z=begin{bmatrix}22&23&-10end{bmatrix}

(iii) 4X-Z=begin{bmatrix}1&-4-8&10end{bmatrix}

(iv) X+2Y+3Z=begin{bmatrix}12&34&4end{bmatrix}

(v) X-4Y+Z=begin{bmatrix}0&-9-14&-16end{bmatrix}

(vi) Z-Z=begin{bmatrix}0&00&0end{bmatrix}

3.3.3.Find the additive inverse of the following matrices: (i) P=[[5],[-7]] (ii) Q=[9,-3] (iii) R=[[2,-1],[-2,3]] (iv) S=[[1,0],[0,1]]

Result

Additive inverses -P=begin{bmatrix}-57end{bmatrix}, -Q=begin{bmatrix}-9&3end{bmatrix}, -R=begin{bmatrix}-2&12&-3end{bmatrix}, -S=begin{bmatrix}-1&00&-1end{bmatrix}

3.3.4.If A=[[2,3],[-3,2]], B=[[3,4],[5,6]] and C=[[1,-2],[0,5]], then verify the following: (i) A+B=B+A (ii) (A+B)+C=A+(B+C) (iii) (2A+B)+C=2A+(B+C) (iv) 3(A+B)=3A+3B

(i) A+B=begin{bmatrix}5&72&8end{bmatrix}=B+A

Result

(i) Verified A+B=B+A

(ii) (A+B)+C=begin{bmatrix}6&52&13end{bmatrix}=A+(B+C)

(ii) Verified (A+B)+C=A+(B+C)

(iii) (2A+B)+C=begin{bmatrix}8&8-1&15end{bmatrix}=2A+(B+C)

(iii) Verified (2A+B)+C=2A+(B+C)

(iv) 3(A+B)=begin{bmatrix}15&216&24end{bmatrix}=3A+3B

(iv) Verified 3(A+B)=3A+3B

3.3.5.If A=[[5,6],[7,-2]] and B=[[-5,-6],[-7,2]], then show that B is additive inverse of A and A is additive inverse of B.

A+B=begin{bmatrix}0&00&0end{bmatrix},quad B+A=begin{bmatrix}0&00&0end{bmatrix}

Result

Conclusion B is additive inverse of A and A is additive inverse of B

3.3.6.If A=[[6,-2],[0,3]] and B=[[8,-1],[3,0]], then verify that (i) (A+B)^t = A^t+B^t (ii) (A-B)^t = A^t-B^t

(i) A+B=begin{bmatrix}14&-33&3end{bmatrix} ⇒ (A+B)^t=begin{bmatrix}14&3-3&3end{bmatrix}; A^t+B^t=begin{bmatrix}14&3-3&3end{bmatrix}

Result

(i) Verified (A+B)^t=A^t+B^t

(ii) A-B=begin{bmatrix}-2&-1-3&3end{bmatrix} ⇒ (A-B)^t=begin{bmatrix}-2&-3-1&3end{bmatrix}; A^t-B^t=begin{bmatrix}-2&-3-1&3end{bmatrix}

(ii) Verified (A-B)^t=A^t-B^t

Unit 3: Matrices and Determinants — Exercise 3 3