11th Class Mathematics Exercise 4.1 Solutions

2.If A = [0, -1, 2; 3, 2, 1; -1, 0, 4], B = [2, 1, -1; 1, 2, 4; -1, 2, 1] and C = [1, 0, -2; -1, 5, 0; 3, 4, -1], then find (i) A-B (ii) B-C (iii) (A-B)-C (iv) A-(B-C)

Given

Matrix A A = begin{bmatrix} 0 & -1 & 2 3 & 2 & 1 -1 & 0 & 4 end{bmatrix}

Matrix B B = begin{bmatrix} 2 & 1 & -1 1 & 2 & 4 -1 & 2 & 1 end{bmatrix}

Matrix C C = begin{bmatrix} 1 & 0 & -2 -1 & 5 & 0 3 & 4 & -1 end{bmatrix}

(i) A-B element-wise A - B = begin{bmatrix} 0-2 & -1-1 & 2-(-1) 3-1 & 2-2 & 1-4 -1-(-1) & 0-2 & 4-1 end{bmatrix}

Result

(i) A-B A - B = begin{bmatrix} -2 & -2 & 3 2 & 0 & -3 0 & -2 & 3 end{bmatrix}

(ii) B-C element-wise B - C = begin{bmatrix} 2-1 & 1-0 & -1-(-2) 1-(-1) & 2-5 & 4-0 -1-3 & 2-4 & 1-(-1) end{bmatrix}

(ii) B-C B - C = begin{bmatrix} 1 & 1 & 1 2 & -3 & 4 -4 & -2 & 2 end{bmatrix}

(iii) (A-B)-C (A-B) - C = begin{bmatrix} -3 & -2 & 5 3 & -5 & -3 -3 & -6 & 4 end{bmatrix}

(iv) A-(B-C) A - (B-C) = begin{bmatrix} -1 & -2 & 1 1 & 5 & -3 3 & 2 & 2 end{bmatrix}

3.If A = [i, 2i; 1, -i], B = [-i, 1; 2i, 1] and C = [2i, 1; -i, i], then show that (i) (AB)C = A(BC) (ii) A(B+C) = AB+AC

Given

Matrix A (complex) A = begin{bmatrix} i & 2i 1 & -i end{bmatrix}

Matrix B (complex) B = begin{bmatrix} -i & 1 2i & 1 end{bmatrix}

Matrix C (complex) C = begin{bmatrix} 2i & 1 -i & i end{bmatrix}

(i) Compute AB AB = begin{bmatrix} i(-i)+2i(2i) & i(1)+2i(1) 1(-i)+(-i)(2i) & 1(1)+(-i)(1) end{bmatrix} = begin{bmatrix} -3 & 3i 2-i & 1-i end{bmatrix}

(i) Compute (AB)C (AB)C = begin{bmatrix} -3 & 3i 2-i & 1-i end{bmatrix}begin{bmatrix} 2i & 1 -i & i end{bmatrix} = begin{bmatrix} 3-6i & -6 1+3i & 3 end{bmatrix}

(i) Compute BC BC = begin{bmatrix} -i & 1 2i & 1 end{bmatrix}begin{bmatrix} 2i & 1 -i & i end{bmatrix} = begin{bmatrix} 2-i & 0 -4-i & 3i end{bmatrix}

(i) Compute A(BC) A(BC) = begin{bmatrix} i & 2i 1 & -i end{bmatrix}begin{bmatrix} 2-i & 0 -4-i & 3i end{bmatrix} = begin{bmatrix} 3-6i & -6 1+3i & 3 end{bmatrix}

Result

(i) Associativity verified (AB)C = A(BC) = begin{bmatrix} 3-6i & -6 1+3i & 3 end{bmatrix}

(ii) Compute B+C B + C = begin{bmatrix} -i+2i & 1+1 2i-i & 1+i end{bmatrix} = begin{bmatrix} i & 2 i & 1+i end{bmatrix}

(ii) Compute A(B+C) A(B+C) = begin{bmatrix} i & 2i 1 & -i end{bmatrix}begin{bmatrix} i & 2 i & 1+i end{bmatrix} = begin{bmatrix} -3 & -2+4i 1+i & 3-i end{bmatrix}

(ii) Compute AC AC = begin{bmatrix} i & 2i 1 & -i end{bmatrix}begin{bmatrix} 2i & 1 -i & i end{bmatrix} = begin{bmatrix} 0 & -2+i -1+2i & 2 end{bmatrix}

(ii) Compute AB+AC AB + AC = begin{bmatrix} -3 & 3i 2-i & 1-i end{bmatrix} + begin{bmatrix} 0 & -2+i -1+2i & 2 end{bmatrix} = begin{bmatrix} -3 & -2+4i 1+i & 3-i end{bmatrix}

(ii) Distributivity verified A(B+C) = AB + AC = begin{bmatrix} -3 & -2+4i 1+i & 3-i end{bmatrix}

5.If A = [-1, 2, 3; 1, 0, 2; -3, 5, 3], then find A+A', A-A', AA', A'A and (A')'

Given

Matrix A A = begin{bmatrix} -1 & 2 & 3 1 & 0 & 2 -3 & 5 & 3 end{bmatrix}

Transpose A' A' = begin{bmatrix} -1 & 1 & -3 2 & 0 & 5 3 & 2 & 3 end{bmatrix}

A + A' A + A' = begin{bmatrix} -2 & 3 & 0 3 & 0 & 7 0 & 7 & 6 end{bmatrix}

A - A' A - A' = begin{bmatrix} 0 & 1 & 6 -1 & 0 & -3 -6 & 3 & 0 end{bmatrix}

AA' matrix multiplication AA' = begin{bmatrix} (-1)(-1)+(2)(2)+(3)(3) & (-1)(1)+(2)(0)+(3)(2) & (-1)(-3)+(2)(5)+(3)(3) (1)(-1)+(0)(2)+(2)(3) & (1)(1)+(0)(0)+(2)(2) & (1)(-3)+(0)(5)+(2)(3) (-3)(-1)+(5)(2)+(3)(3) & (-3)(1)+(5)(0)+(3)(2) & (-3)(-3)+(5)(2)+(3)(3) end{bmatrix}

AA' simplified AA' = begin{bmatrix} 14 & 5 & 22 5 & 5 & 3 22 & 3 & 43 end{bmatrix}

A'A matrix multiplication A'A = begin{bmatrix} 11 & -17 & -10 -17 & 29 & 21 -10 & 21 & 22 end{bmatrix}

(A')' transpose of transpose (A')' = begin{bmatrix} -1 & 2 & 3 1 & 0 & 2 -3 & 5 & 3 end{bmatrix} = A

Result

A+A' A + A' = begin{bmatrix} -2 & 3 & 0 3 & 0 & 7 0 & 7 & 6 end{bmatrix}

A-A' A - A' = begin{bmatrix} 0 & 1 & 6 -1 & 0 & -3 -6 & 3 & 0 end{bmatrix}

AA' AA' = begin{bmatrix} 14 & 5 & 22 5 & 5 & 3 22 & 3 & 43 end{bmatrix}

A'A A'A = begin{bmatrix} 11 & -17 & -10 -17 & 29 & 21 -10 & 21 & 22 end{bmatrix}

(A')' (A')' = begin{bmatrix} -1 & 2 & 3 1 & 0 & 2 -3 & 5 & 3 end{bmatrix}

6.Solve the matrix equation A² - 5A + 4I - X = 0 if A = [2, 0, 1; 2, 1, 3; 1, -1, 0]

Given

Matrix A A = begin{bmatrix} 2 & 0 & 1 2 & 1 & 3 1 & -1 & 0 end{bmatrix}

Compute A² A² = AA = begin{bmatrix} 2 & 0 & 1 2 & 1 & 3 1 & -1 & 0 end{bmatrix}begin{bmatrix} 2 & 0 & 1 2 & 1 & 3 1 & -1 & 0 end{bmatrix}

A² calculated A² = begin{bmatrix} 5 & -1 & 2 9 & -2 & 5 0 & -1 & -2 end{bmatrix}

Compute 5A 5A = begin{bmatrix} 10 & 0 & 5 10 & 5 & 15 5 & -5 & 0 end{bmatrix}

Compute 4I 4I = begin{bmatrix} 4 & 0 & 0 0 & 4 & 0 0 & 0 & 4 end{bmatrix}

Rearrange equation A² - 5A + 4I - X = 0 ⇒ X = A² - 5A + 4I

Compute X = A² - 5A + 4I X = begin{bmatrix} 5 & -1 & 2 9 & -2 & 5 0 & -1 & -2 end{bmatrix} - begin{bmatrix} 10 & 0 & 5 10 & 5 & 15 5 & -5 & 0 end{bmatrix} + begin{bmatrix} 4 & 0 & 0 0 & 4 & 0 0 & 0 & 4 end{bmatrix}

Result

Solution for X X = begin{bmatrix} -1 & -1 & -3 -1 & -3 & -10 -5 & 4 & 2 end{bmatrix}

Unit 4: Matrices and Determinants — Exercise 4 1