Inverse matrix questions
Slides do conjunto (5)
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- 1a. Find the value of x in the singular matrix [x -3][1 2]
- b. Find the values of x in the singular matrix [1 x][5x x]
- 2a. Find the inverse of the matrix A which is [2 4][-1 3]
- b. Verify that AA-1 = I
- 3. A = [1 1][3 -1] , B = [-1 1][2 3] , Find matrix X when:
- a. AX=B
- b. XB=[1 -1][3 2]
- 4. A=[3 -1][2 1] , BA=[-1 1][3 5] , Find matrix B
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- 1a. Find the value of x in the singular matrix [x -3][1 2]
- b. Find the values of x in the singular matrix [1 x][5x x]
- a). ad-bc=0 in a singular matrix, so we sub in values. (x)(2)-(-3)(1) = 0 , 2x+3=0 , 2x=3 , x=3/2
- b). Again, ad-bc=0 as we have a singular matrix, so (1)(x) - (x)(5x)=0 , x-5x^2=0 , factorise to get x(5x)=0, so x=0 or x=1/5
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- 2a. Find the inverse of the matrix A which is [2 4][-1 3]
- b. Verify that AA-1 = I
- a). Use the formula : A-1=1/ad-bc(d -b)(-c a), so A-1=1/(2)(3)-(4)(-1) x [3 -4][1 2] , A-1=[3/10 -4/10][1/10 2/10]
- b). [2 4][-1 3] x [3/10 -4/10][1/10 2/10] = [1 0][1 0]
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- 3. A = [1 1][3 -1] , B = [-1 1][2 3] , Find matrix X when:
- a. AX=B
- b. XB=[1 -1][3 2]
- a). AX=B , AA-1X=BA-1 , IX = BA-1 , X=BA-1. So A-1=1/(1)(-1)-(1)(3) x [-1 -1][-3 1] , A-1 = [1/4 1/4][3/4 -1/4]. Now we do B x A-1, so [-1 1][2 3] multiplied by [1/4 1/4][3/4 -1/4] = [1/4 1][-5/4 0] , which can also we written as 1/4[1 4][-5 0] , so X = 1/4[1 4][-5 0]
- b). XB=[1 -1][3 2] , XBB-1=[1 -1][3 2]B-1 , XI=[1 -1][3 2]B-1, X=[1 -1][3 2]B-1 . So B-1 = 1/(-1)(3)-(1)(2) x [3 -1][-2 -1] , B-1 = [-3/5 1/5][2/5 1/5]. Now we multiply our original matrix by this, so [1 -1][3 2] x [-3/5 1/5][2/5 1/5] = [-1 0][-1 1] , so X = [-1 0][-1 1]
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- 4. A=[3 -1][2 1] , BA=[-1 1][3 5] , Find matrix B
- Not yet discovered how to do.
Informações
- Descrição
- AS level Further Maths (Matrices) Slides sobre Inverse matrix questions, criado por Doc Boff em 03-10-2018.
- Criado por
- Doc Boff · out 2018
- Atualizado
- 4 out 2018
- Visualizações
- 2
- Avaliação