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Worksheet featuring practice problems on matrices and simultaneous equations, covering topics like determinants, matrix inverses, and solving linear systems.

Practice questions on matrices and simultaneous equations, including evaluating determinants, finding inverses, and solving systems of equations using matrix methods.

Practice questions on matrices and simultaneous equations, including evaluating determinants, finding inverses, and solving systems of equations using matrix methods.

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Show Answer Key & Explanations Step-by-step solution for: Matrices and Simultaneous Equations - CSEC Math Tutor
1. a. det(A) = (3)(-3) - (2)(2) = -9 - 4 = -13
b. A⁻¹ = (1/-13) * [ -3 -2 ]
[ -2 3 ] = [ 3/13 2/13 ]
[ 2/13 -3/13 ]
c. [x] = A⁻¹ * [1] = [ 3/13 2/13 ] * [1] = [ (3/13)(1) + (2/13)(5) ] = [ 13/13 ] = [1]
[y] [5] [ 2/13 -3/13 ] [5] [ (2/13)(1) + (-3/13)(5) ] [-13/13] [-1]

2. a. [1 -1][x] = [-5]
[3 2][y] [-5]
b. Let B = [1 -1]. det(B) = (1)(2) - (-1)(3) = 2 + 3 = 5.
[3 2]
B⁻¹ = (1/5) * [ 2 1] = [ 2/5 1/5]
[-3 1] [-3/5 1/5]
c. [x] = B⁻¹ * [-5] = [ 2/5 1/5] * [-5] = [ (2/5)(-5) + (1/5)(-5) ] = [ -10/5 - 5/5 ] = [ -15/5 ] = [-3]
[y] [-5] [-3/5 1/5] [-5] [ (-3/5)(-5) + (1/5)(-5) ] [ 15/5 - 5/5 ] [ 10/5 ] [ 2]

3. a. Let C = [1 4]. det(C) = (1)(1) - (4)(2) = 1 - 8 = -7.
[2 1]
C⁻¹ = (1/-7) * [ 1 -4] = [ -1/7 4/7]
[-2 1] [ 2/7 -1/7]
b. From C * [a b] = [4 13], we have [a b] = C⁻¹ * [4 13].
[c d] [8 5] [c d] [8 5]
[a b] = [ -1/7 4/7] * [4 13] = [ (-1/7)(4) + (4/7)(8) (-1/7)(13) + (4/7)(5) ] = [ (-4+32)/7 (-13+20)/7 ] = [28/7 7/7] = [4 1]
[c d] [ 2/7 -1/7] [8 5] [ (2/7)(4) + (-1/7)(8) (2/7)(13) + (-1/7)(5) ] [ (8-8)/7 (26-5)/7 ] [ 0/7 21/7] [0 3]
So, a=4, b=1, c=0, d=3.

4. a. P = [2 1]. det(P) = (2)(-3) - (1)(5) = -6 - 5 = -11.
[5 -3]
P⁻¹ = (1/-11) * [ -3 -1] = [ 3/11 1/11]
[ -5 2] [ 5/11 -2/11]
b. The equation x[2] + y[1] = [7] is equivalent to P*[x] = [7].
[5] [-3] [1] [y] [1]
So, [x] = P⁻¹ * [7] = [ 3/11 1/11] * [7] = [ (3/11)(7) + (1/11)(1) ] = [ (21+1)/11 ] = [22/11] = [2]
[y] [1] [ 5/11 -2/11] [1] [ (5/11)(7) + (-2/11)(1) ] [ (35-2)/11 ] [33/11] [3]

5. a. M = [2 5]. det(M) = (2)(15) - (5)(7) = 30 - 35 = -5 ≠ 0, so M is non-singular.
[7 15]
b. M⁻¹ = (1/-5) * [ 15 -5] = [ -3 1]
[ -7 2] [ 7/5 -2/5]
c. M * M⁻¹ = I = [1 0]
[0 1]
d. Pre-multiplying both sides of M*[x] = [-3] by M⁻¹ gives:
[y] [17]
M⁻¹ * M * [x] = M⁻¹ * [-3]
[y] [17]
I * [x] = M⁻¹ * [-3]
[y] [17]
e. [x] = M⁻¹ * [-3] = [ -3 1] * [-3] = [ (-3)(-3) + (1)(17) ] = [ 9 + 17 ] = [26]
[y] [17] [ 7/5 -2/5] [17] [ (7/5)(-3) + (-2/5)(17) ] [ (-21 - 34)/5 ] [-55/5] = [-11]
So, x = 26, y = -11.
Parent Tip: Review the logic above to help your child master the concept of solving systems using matrices worksheet.
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