OP Malhotra Class 12 Maths Solutions Chapter 6 Matrices Exercise 6 (E)

S Chand Class 12 ICSE Maths Solutions Chapter 6 Matrices Ex 6(e)

Question 1.
(i) If \( A = \left[\begin{array}{rrr} 1 & 4 & 2 \\ 2 & 5 & 3 \\ 3 & -1 & 0 \end{array}\right] \), find \( A + A' \).
(ii) For matrix \( A = \left[\begin{array}{rrr} -3 & 6 & 0 \\ 4 & -5 & 8 \\ 0 & -7 & -2 \end{array}\right] \), find \( \frac { 1 }{ 2 }(A – A') \).
(iii) Find \( \frac {1}{ 2 }(A + A') \) and \( \frac { 1 }{ 2 }(A – A') \), where \( A = \left[\begin{array}{rrr} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{array}\right] \).
Answer:
(i) Given matrix \( A = \left[\begin{array}{rrr} 1 & 4 & 2 \\ 2 & 5 & 3 \\ 3 & -1 & 0 \end{array}\right] \).
First, find the transpose of A, denoted as A'. We swap rows with columns.
\( A' = \left[\begin{array}{rrr} 1 & 2 & 3 \\ 4 & 5 & -1 \\ 2 & 3 & 0 \end{array}\right] \).
Now, calculate \( A + A' \):
\( A + A' = \left[\begin{array}{rrr} 1 & 4 & 2 \\ 2 & 5 & 3 \\ 3 & -1 & 0 \end{array}\right] + \left[\begin{array}{rrr} 1 & 2 & 3 \\ 4 & 5 & -1 \\ 2 & 3 & 0 \end{array}\right] \)
\( = \left[\begin{array}{rrr} 1+1 & 4+2 & 2+3 \\ 2+4 & 5+5 & 3-1 \\ 3+2 & -1+3 & 0+0 \end{array}\right] \)
\( = \left[\begin{array}{rrr} 2 & 6 & 5 \\ 6 & 10 & 2 \\ 5 & 2 & 0 \end{array}\right] \).

(ii) Given matrix \( A = \left[\begin{array}{rrr} -3 & 6 & 0 \\ 4 & -5 & 8 \\ 0 & -7 & -2 \end{array}\right] \).
First, find the transpose of A, denoted as A'.
\( A' = \left[\begin{array}{rrr} -3 & 4 & 0 \\ 6 & -5 & -7 \\ 0 & 8 & -2 \end{array}\right] \).
Now, calculate \( A - A' \):
\( A - A' = \left[\begin{array}{rrr} -3 & 6 & 0 \\ 4 & -5 & 8 \\ 0 & -7 & -2 \end{array}\right] - \left[\begin{array}{rrr} -3 & 4 & 0 \\ 6 & -5 & -7 \\ 0 & 8 & -2 \end{array}\right] \)
\( = \left[\begin{array}{rrr} -3-(-3) & 6-4 & 0-0 \\ 4-6 & -5-(-5) & 8-(-7) \\ 0-0 & -7-8 & -2-(-2) \end{array}\right] \)
\( = \left[\begin{array}{rrr} 0 & 2 & 0 \\ -2 & 0 & 15 \\ 0 & -15 & 0 \end{array}\right] \).
Finally, calculate \( \frac{1}{2}(A - A') \):
\( \frac{1}{2}(A - A') = \frac{1}{2} \left[\begin{array}{rrr} 0 & 2 & 0 \\ -2 & 0 & 15 \\ 0 & -15 & 0 \end{array}\right] \)
\( = \left[\begin{array}{rrr} 0 & 1 & 0 \\ -1 & 0 & \frac{15}{2} \\ 0 & -\frac{15}{2} & 0 \end{array}\right] \).

(iii) Given matrix \( A = \left[\begin{array}{rrr} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{array}\right] \).
First, find the transpose of A, denoted as A'.
\( A' = \left[\begin{array}{rrr} 0 & -a & -b \\ a & 0 & -c \\ b & c & 0 \end{array}\right] \).
Now, calculate \( A + A' \):
\( A + A' = \left[\begin{array}{rrr} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{array}\right] + \left[\begin{array}{rrr} 0 & -a & -b \\ a & 0 & -c \\ b & c & 0 \end{array}\right] \)
\( = \left[\begin{array}{rrr} 0+0 & a-a & b-b \\ -a+a & 0+0 & c-c \\ -b+b & -c+c & 0+0 \end{array}\right] \)
\( = \left[\begin{array}{rrr} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \).
Then, \( \frac{1}{2}(A + A') = \frac{1}{2} \left[\begin{array}{rrr} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] = \left[\begin{array}{rrr} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] = O \).
Next, calculate \( A - A' \):
\( A - A' = \left[\begin{array}{rrr} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{array}\right] - \left[\begin{array}{rrr} 0 & -a & -b \\ a & 0 & -c \\ b & c & 0 \end{array}\right] \)
\( = \left[\begin{array}{rrr} 0-0 & a-(-a) & b-(-b) \\ -a-a & 0-0 & c-(-c) \\ -b-b & -c-c & 0-0 \end{array}\right] \)
\( = \left[\begin{array}{rrr} 0 & 2a & 2b \\ -2a & 0 & 2c \\ -2b & -2c & 0 \end{array}\right] \).
Finally, calculate \( \frac{1}{2}(A - A') \):
\( \frac{1}{2}(A - A') = \frac{1}{2} \left[\begin{array}{rrr} 0 & 2a & 2b \\ -2a & 0 & 2c \\ -2b & -2c & 0 \end{array}\right] \)
\( = \left[\begin{array}{rrr} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{array}\right] = A \).
In simple words: To find the sum or difference with a transposed matrix, first flip the rows and columns of the original matrix to get its transpose. Then, add or subtract the corresponding elements. For scalar multiplication, multiply each element in the matrix by the scalar value.

🎯 Exam Tip: Remember that A + A' is always a symmetric matrix, and A - A' is always a skew-symmetric matrix. This is a quick way to check your calculations.

Question 2. If \( B = \left[\begin{array}{rr} 1 & 3 \\ -2 & 5 \end{array}\right] \) and \( C = \left[\begin{array}{rr} -2 & 5 \\ 3 & 4 \end{array}\right] \), find \( (BC)' \).
Answer: Given matrices:
\( B = \left[\begin{array}{rr} 1 & 3 \\ -2 & 5 \end{array}\right] \)
\( C = \left[\begin{array}{rr} -2 & 5 \\ 3 & 4 \end{array}\right] \)
First, calculate the product \( BC \):
\( BC = \left[\begin{array}{rr} 1 & 3 \\ -2 & 5 \end{array}\right] \left[\begin{array}{rr} -2 & 5 \\ 3 & 4 \end{array}\right] \)
\( = \left[\begin{array}{rr} (1)(-2)+(3)(3) & (1)(5)+(3)(4) \\ (-2)(-2)+(5)(3) & (-2)(5)+(5)(4) \end{array}\right] \)
\( = \left[\begin{array}{rr} -2+9 & 5+12 \\ 4+15 & -10+20 \end{array}\right] \)
\( = \left[\begin{array}{rr} 7 & 17 \\ 19 & 10 \end{array}\right] \).
Now, find the transpose of \( BC \), denoted as \( (BC)' \). To do this, swap the rows and columns of \( BC \).
\( (BC)' = \left[\begin{array}{rr} 7 & 19 \\ 17 & 10 \end{array}\right] \).
In simple words: First, multiply the two matrices B and C together. Then, take the result and swap its rows with its columns to find its transpose.

🎯 Exam Tip: When multiplying matrices, remember to multiply rows by columns and sum the products. The order of multiplication matters, and for the transpose of a product, \( (AB)' = B'A' \).

Question 3. If \( A = \left[\begin{array}{rrr} 1 & 2 & 0 \\ 3 & -1 & 4 \end{array}\right] \), find each of the following:
(i) \( AA' \)
(ii) \( A'A \)
Answer: Given matrix:
\( A = \left[\begin{array}{rrr} 1 & 2 & 0 \\ 3 & -1 & 4 \end{array}\right]_{2 \times 3} \).
First, find the transpose of A, denoted as A'. Swap its rows and columns.
\( A' = \left[\begin{array}{rr} 1 & 3 \\ 2 & -1 \\ 0 & 4 \end{array}\right]_{3 \times 2} \).

(i) To find \( AA' \):
The number of columns in A (3) is equal to the number of rows in A' (3), so the product \( AA' \) exists.
\( AA' = \left[\begin{array}{rrr} 1 & 2 & 0 \\ 3 & -1 & 4 \end{array}\right] \left[\begin{array}{rr} 1 & 3 \\ 2 & -1 \\ 0 & 4 \end{array}\right] \)
\( = \left[\begin{array}{rr} (1)(1)+(2)(2)+(0)(0) & (1)(3)+(2)(-1)+(0)(4) \\ (3)(1)+(-1)(2)+(4)(0) & (3)(3)+(-1)(-1)+(4)(4) \end{array}\right] \)
\( = \left[\begin{array}{rr} 1+4+0 & 3-2+0 \\ 3-2+0 & 9+1+16 \end{array}\right] \)
\( = \left[\begin{array}{rr} 5 & 1 \\ 1 & 26 \end{array}\right] \).

(ii) To find \( A'A \):
The number of columns in A' (2) is equal to the number of rows in A (2), so the product \( A'A \) exists.
\( A'A = \left[\begin{array}{rr} 1 & 3 \\ 2 & -1 \\ 0 & 4 \end{array}\right] \left[\begin{array}{rrr} 1 & 2 & 0 \\ 3 & -1 & 4 \end{array}\right] \)
\( = \left[\begin{array}{rrr} (1)(1)+(3)(3) & (1)(2)+(3)(-1) & (1)(0)+(3)(4) \\ (2)(1)+(-1)(3) & (2)(2)+(-1)(-1) & (2)(0)+(-1)(4) \\ (0)(1)+(4)(3) & (0)(2)+(4)(-1) & (0)(0)+(4)(4) \end{array}\right] \)
\( = \left[\begin{array}{rrr} 1+9 & 2-3 & 0+12 \\ 2-3 & 4+1 & 0-4 \\ 0+12 & 0-4 & 0+16 \end{array}\right] \)
\( = \left[\begin{array}{rrr} 10 & -1 & 12 \\ -1 & 5 & -4 \\ 12 & -4 & 16 \end{array}\right] \).
In simple words: To multiply a matrix by its transpose, first find the transpose by swapping rows and columns. Then, perform matrix multiplication. The result of \( AA' \) and \( A'A \) might be different in terms of their dimensions and elements.

🎯 Exam Tip: Always check the dimensions of the matrices before multiplying. For \( AB \) to exist, the number of columns in A must equal the number of rows in B. The resulting matrix will have dimensions (rows of A) x (columns of B).

Question 4. If \( A = \left[\begin{array}{rr} 1 & -2 \\ 1 & 0 \end{array}\right] \) and \( B = \left[\begin{array}{rr} -1 & 2 \\ -1 & 1 \end{array}\right] \), Find
(i) \( A'B' \)
(ii) \( AB' \)
Answer: Given matrices:
\( A = \left[\begin{array}{rr} 1 & -2 \\ 1 & 0 \end{array}\right] \)
\( B = \left[\begin{array}{rr} -1 & 2 \\ -1 & 1 \end{array}\right] \)
First, find the transposes of A and B, denoted as A' and B'.
\( A' = \left[\begin{array}{rr} 1 & 1 \\ -2 & 0 \end{array}\right] \)
\( B' = \left[\begin{array}{rr} -1 & -1 \\ 2 & 1 \end{array}\right] \).

(i) To find \( A'B' \):
\( A'B' = \left[\begin{array}{rr} 1 & 1 \\ -2 & 0 \end{array}\right] \left[\begin{array}{rr} -1 & -1 \\ 2 & 1 \end{array}\right] \)
\( = \left[\begin{array}{rr} (1)(-1)+(1)(2) & (1)(-1)+(1)(1) \\ (-2)(-1)+(0)(2) & (-2)(-1)+(0)(1) \end{array}\right] \)
\( = \left[\begin{array}{rr} -1+2 & -1+1 \\ 2+0 & 2+0 \end{array}\right] \)
\( = \left[\begin{array}{rr} 1 & 0 \\ 2 & 2 \end{array}\right] \).

(ii) To find \( AB' \):
\( AB' = \left[\begin{array}{rr} 1 & -2 \\ 1 & 0 \end{array}\right] \left[\begin{array}{rr} -1 & -1 \\ 2 & 1 \end{array}\right] \)
\( = \left[\begin{array}{rr} (1)(-1)+(-2)(2) & (1)(-1)+(-2)(1) \\ (1)(-1)+(0)(2) & (1)(-1)+(0)(1) \end{array}\right] \)
\( = \left[\begin{array}{rr} -1-4 & -1-2 \\ -1+0 & -1+0 \end{array}\right] \)
\( = \left[\begin{array}{rr} -5 & -3 \\ -1 & -1 \end{array}\right] \).
In simple words: To calculate these matrix products, first find the transpose of any matrix that has a prime symbol ('). Then, multiply the matrices together, making sure to multiply rows by columns for each element.

🎯 Exam Tip: Matrix multiplication is not commutative, meaning \( AB \ne BA \) in general. Also, remember that \( (A')' = A \), meaning transposing a matrix twice gives the original matrix back.

Question 5. Verify that \( (AB)' = B'A' \) if
(i) \( A = \left[\begin{array}{r} 1 \\ -2 \\ 3 \end{array}\right] \), \( B = \left[\begin{array}{rrr} 1 & -5 & 7 \end{array}\right] \)
(ii) \( A = \left[\begin{array}{ll} 2 & 3 \\ 0 & 1 \end{array}\right] \), \( B = \left[\begin{array}{ll} 3 & 4 \\ 2 & 1 \end{array}\right] \)
(iii) \( A = \left[\begin{array}{rrr} -1 & 3 & 0 \\ -7 & 2 & 8 \end{array}\right] \), \( B = \left[\begin{array}{rr} -5 & 0 \\ 0 & 3 \\ 1 & -8 \end{array}\right] \)
Answer: We need to verify that \( (AB)' = B'A' \) for each pair of matrices.

(i) Given:
\( A = \left[\begin{array}{r} 1 \\ -2 \\ 3 \end{array}\right]_{3 \times 1} \), \( B = \left[\begin{array}{rrr} 1 & -5 & 7 \end{array}\right]_{1 \times 3} \)
First, calculate \( AB \):
\( AB = \left[\begin{array}{r} 1 \\ -2 \\ 3 \end{array}\right] \left[\begin{array}{rrr} 1 & -5 & 7 \end{array}\right] \)
\( = \left[\begin{array}{rrr} (1)(1) & (1)(-5) & (1)(7) \\ (-2)(1) & (-2)(-5) & (-2)(7) \\ (3)(1) & (3)(-5) & (3)(7) \end{array}\right] \)
\( = \left[\begin{array}{rrr} 1 & -5 & 7 \\ -2 & 10 & -14 \\ 3 & -15 & 21 \end{array}\right] \).
Now, find \( (AB)' \):
\( (AB)' = \left[\begin{array}{rrr} 1 & -2 & 3 \\ -5 & 10 & -15 \\ 7 & -14 & 21 \end{array}\right] \) (L.H.S.).
Next, find \( A' \) and \( B' \):
\( A' = \left[\begin{array}{rrr} 1 & -2 & 3 \end{array}\right] \)
\( B' = \left[\begin{array}{r} 1 \\ -5 \\ 7 \end{array}\right] \).
Now, calculate \( B'A' \):
\( B'A' = \left[\begin{array}{r} 1 \\ -5 \\ 7 \end{array}\right] \left[\begin{array}{rrr} 1 & -2 & 3 \end{array}\right] \)
\( = \left[\begin{array}{rrr} (1)(1) & (1)(-2) & (1)(3) \\ (-5)(1) & (-5)(-2) & (-5)(3) \\ (7)(1) & (7)(-2) & (7)(3) \end{array}\right] \)
\( = \left[\begin{array}{rrr} 1 & -2 & 3 \\ -5 & 10 & -15 \\ 7 & -14 & 21 \end{array}\right] \) (R.H.S.).
Since L.H.S. = R.H.S., the property \( (AB)' = B'A' \) is verified.

(ii) Given:
\( A = \left[\begin{array}{ll} 2 & 3 \\ 0 & 1 \end{array}\right] \), \( B = \left[\begin{array}{ll} 3 & 4 \\ 2 & 1 \end{array}\right] \)
First, calculate \( AB \):
\( AB = \left[\begin{array}{ll} 2 & 3 \\ 0 & 1 \end{array}\right] \left[\begin{array}{ll} 3 & 4 \\ 2 & 1 \end{array}\right] \)
\( = \left[\begin{array}{ll} (2)(3)+(3)(2) & (2)(4)+(3)(1) \\ (0)(3)+(1)(2) & (0)(4)+(1)(1) \end{array}\right] \)
\( = \left[\begin{array}{ll} 6+6 & 8+3 \\ 0+2 & 0+1 \end{array}\right] \)
\( = \left[\begin{array}{ll} 12 & 11 \\ 2 & 1 \end{array}\right] \).
Now, find \( (AB)' \):
\( (AB)' = \left[\begin{array}{ll} 12 & 2 \\ 11 & 1 \end{array}\right] \) (L.H.S.).
Next, find \( A' \) and \( B' \):
\( A' = \left[\begin{array}{ll} 2 & 0 \\ 3 & 1 \end{array}\right] \)
\( B' = \left[\begin{array}{ll} 3 & 2 \\ 4 & 1 \end{array}\right] \).
Now, calculate \( B'A' \):
\( B'A' = \left[\begin{array}{ll} 3 & 2 \\ 4 & 1 \end{array}\right] \left[\begin{array}{ll} 2 & 0 \\ 3 & 1 \end{array}\right] \)
\( = \left[\begin{array}{ll} (3)(2)+(2)(3) & (3)(0)+(2)(1) \\ (4)(2)+(1)(3) & (4)(0)+(1)(1) \end{array}\right] \)
\( = \left[\begin{array}{ll} 6+6 & 0+2 \\ 8+3 & 0+1 \end{array}\right] \)
\( = \left[\begin{array}{ll} 12 & 2 \\ 11 & 1 \end{array}\right] \) (R.H.S.).
Since L.H.S. = R.H.S., the property \( (AB)' = B'A' \) is verified.

(iii) Given:
\( A = \left[\begin{array}{rrr} -1 & 3 & 0 \\ -7 & 2 & 8 \end{array}\right]_{2 \times 3} \), \( B = \left[\begin{array}{rr} -5 & 0 \\ 0 & 3 \\ 1 & -8 \end{array}\right]_{3 \times 2} \)
First, calculate \( AB \):
\( AB = \left[\begin{array}{rrr} -1 & 3 & 0 \\ -7 & 2 & 8 \end{array}\right] \left[\begin{array}{rr} -5 & 0 \\ 0 & 3 \\ 1 & -8 \end{array}\right] \)
\( = \left[\begin{array}{rr} (-1)(-5)+(3)(0)+(0)(1) & (-1)(0)+(3)(3)+(0)(-8) \\ (-7)(-5)+(2)(0)+(8)(1) & (-7)(0)+(2)(3)+(8)(-8) \end{array}\right] \)
\( = \left[\begin{array}{rr} 5+0+0 & 0+9+0 \\ 35+0+8 & 0+6-64 \end{array}\right] \)
\( = \left[\begin{array}{rr} 5 & 9 \\ 43 & -58 \end{array}\right] \).
Now, find \( (AB)' \):
\( (AB)' = \left[\begin{array}{rr} 5 & 43 \\ 9 & -58 \end{array}\right] \) (L.H.S.).
Next, find \( A' \) and \( B' \):
\( A' = \left[\begin{array}{rr} -1 & -7 \\ 3 & 2 \\ 0 & 8 \end{array}\right] \)
\( B' = \left[\begin{array}{rrr} -5 & 0 & 1 \\ 0 & 3 & -8 \end{array}\right] \).
Now, calculate \( B'A' \):
\( B'A' = \left[\begin{array}{rrr} -5 & 0 & 1 \\ 0 & 3 & -8 \end{array}\right] \left[\begin{array}{rr} -1 & -7 \\ 3 & 2 \\ 0 & 8 \end{array}\right] \)
\( = \left[\begin{array}{rr} (-5)(-1)+(0)(3)+(1)(0) & (-5)(-7)+(0)(2)+(1)(8) \\ (0)(-1)+(3)(3)+(-8)(0) & (0)(-7)+(3)(2)+(-8)(8) \end{array}\right] \)
\( = \left[\begin{array}{rr} 5+0+0 & 35+0+8 \\ 0+9+0 & 0+6-64 \end{array}\right] \)
\( = \left[\begin{array}{rr} 5 & 43 \\ 9 & -58 \end{array}\right] \) (R.H.S.).
Since L.H.S. = R.H.S., the property \( (AB)' = B'A' \) is verified.
In simple words: To verify this property, first multiply matrices A and B, then find the transpose of the product. Separately, find the transpose of B and A, then multiply them in the order B'A'. If both results are the same, the property is confirmed. This property is a fundamental rule in matrix algebra.

🎯 Exam Tip: The property \( (AB)' = B'A' \) is known as the reversal law for transposes. Always remember to reverse the order of the matrices when applying the transpose to a product.

Question 6. If \( A = \left[\begin{array}{rrr} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{array}\right] \) and \( B = \left[\begin{array}{rrr} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4 \end{array}\right] \), Verify that \( (A + B)' = A' + B' \).
Answer: Given matrices:
\( A = \left[\begin{array}{rrr} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{array}\right] \)
\( B = \left[\begin{array}{rrr} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4 \end{array}\right] \)
First, calculate \( A + B \):
\( A + B = \left[\begin{array}{rrr} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{array}\right] + \left[\begin{array}{rrr} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4 \end{array}\right] \)
\( = \left[\begin{array}{rrr} 1+(-1) & 2+1 & 3+0 \\ 0+0 & 1+(-1) & 0+1 \\ 1+2 & 1+3 & 0+4 \end{array}\right] \)
\( = \left[\begin{array}{rrr} 0 & 3 & 3 \\ 0 & 0 & 1 \\ 3 & 4 & 4 \end{array}\right] \).
Now, find \( (A + B)' \):
\( (A + B)' = \left[\begin{array}{rrr} 0 & 0 & 3 \\ 3 & 0 & 4 \\ 3 & 1 & 4 \end{array}\right] \) (L.H.S.).
Next, find \( A' \) and \( B' \):
\( A' = \left[\begin{array}{rrr} 1 & 0 & 1 \\ 2 & 1 & 1 \\ 3 & 0 & 0 \end{array}\right] \)
\( B' = \left[\begin{array}{rrr} -1 & 0 & 2 \\ 1 & -1 & 3 \\ 0 & 1 & 4 \end{array}\right] \).
Now, calculate \( A' + B' \):
\( A' + B' = \left[\begin{array}{rrr} 1 & 0 & 1 \\ 2 & 1 & 1 \\ 3 & 0 & 0 \end{array}\right] + \left[\begin{array}{rrr} -1 & 0 & 2 \\ 1 & -1 & 3 \\ 0 & 1 & 4 \end{array}\right] \)
\( = \left[\begin{array}{rrr} 1+(-1) & 0+0 & 1+2 \\ 2+1 & 1+(-1) & 1+3 \\ 3+0 & 0+1 & 0+4 \end{array}\right] \)
\( = \left[\begin{array}{rrr} 0 & 0 & 3 \\ 3 & 0 & 4 \\ 3 & 1 & 4 \end{array}\right] \) (R.H.S.).
Since L.H.S. = R.H.S., the property \( (A + B)' = A' + B' \) is verified.
In simple words: We added two matrices, A and B, and then took the transpose of their sum. We also found the transposes of A and B separately and then added them. Since both methods gave the same final matrix, the property is proven. This shows that the transpose of a sum is the sum of the transposes.

🎯 Exam Tip: This property, \( (A + B)' = A' + B' \), is another key rule for matrix transposes. It simplifies calculations when dealing with sums of matrices before transposing.

Question 7. If \( A = \left[\begin{array}{rrr} 1 & 2 & 3 \\ -1 & 0 & 2 \\ 1 & -3 & -1 \end{array}\right] \), \( B = \left[\begin{array}{rrr} 4 & 5 & 6 \\ -1 & 0 & 1 \\ 2 & 1 & 2 \end{array}\right] \), \( C = \left[\begin{array}{rrr} -1 & -2 & 1 \\ -1 & 2 & 3 \\ -1 & -2 & 2 \end{array}\right] \), find each of the following:
(i) \( 2A' - B' \)
(ii) \( (A + B + C)' \). Is \( (A + B + C)' = A' + B' + C' \) ?
Answer: Given matrices:
\( A = \left[\begin{array}{rrr} 1 & 2 & 3 \\ -1 & 0 & 2 \\ 1 & -3 & -1 \end{array}\right] \)
\( B = \left[\begin{array}{rrr} 4 & 5 & 6 \\ -1 & 0 & 1 \\ 2 & 1 & 2 \end{array}\right] \)
\( C = \left[\begin{array}{rrr} -1 & -2 & 1 \\ -1 & 2 & 3 \\ -1 & -2 & 2 \end{array}\right] \)
First, find the transposes of A, B, and C:
\( A' = \left[\begin{array}{rrr} 1 & -1 & 1 \\ 2 & 0 & -3 \\ 3 & 2 & -1 \end{array}\right] \)
\( B' = \left[\begin{array}{rrr} 4 & -1 & 2 \\ 5 & 0 & 1 \\ 6 & 1 & 2 \end{array}\right] \)
\( C' = \left[\begin{array}{rrr} -1 & -1 & -1 \\ -2 & 2 & -2 \\ 1 & 3 & 2 \end{array}\right] \).

(i) To find \( 2A' - B' \):
First, calculate \( 2A' \):
\( 2A' = 2 \left[\begin{array}{rrr} 1 & -1 & 1 \\ 2 & 0 & -3 \\ 3 & 2 & -1 \end{array}\right] = \left[\begin{array}{rrr} 2 & -2 & 2 \\ 4 & 0 & -6 \\ 6 & 4 & -2 \end{array}\right] \).
Now, subtract \( B' \) from \( 2A' \):
\( 2A' - B' = \left[\begin{array}{rrr} 2 & -2 & 2 \\ 4 & 0 & -6 \\ 6 & 4 & -2 \end{array}\right] - \left[\begin{array}{rrr} 4 & -1 & 2 \\ 5 & 0 & 1 \\ 6 & 1 & 2 \end{array}\right] \)
\( = \left[\begin{array}{rrr} 2-4 & -2-(-1) & 2-2 \\ 4-5 & 0-0 & -6-1 \\ 6-6 & 4-1 & -2-2 \end{array}\right] \)
\( = \left[\begin{array}{rrr} -2 & -1 & 0 \\ -1 & 0 & -7 \\ 0 & 3 & -4 \end{array}\right] \).

(ii) To find \( (A + B + C)' \) and verify \( (A + B + C)' = A' + B' + C' \):
First, calculate \( A + B + C \):
\( A + B + C = \left[\begin{array}{rrr} 1 & 2 & 3 \\ -1 & 0 & 2 \\ 1 & -3 & -1 \end{array}\right] + \left[\begin{array}{rrr} 4 & 5 & 6 \\ -1 & 0 & 1 \\ 2 & 1 & 2 \end{array}\right] + \left[\begin{array}{rrr} -1 & -2 & 1 \\ -1 & 2 & 3 \\ -1 & -2 & 2 \end{array}\right] \)
\( = \left[\begin{array}{rrr} 1+4-1 & 2+5-2 & 3+6+1 \\ -1-1-1 & 0+0+2 & 2+1+3 \\ 1+2-1 & -3+1-2 & -1+2+2 \end{array}\right] \)
\( = \left[\begin{array}{rrr} 4 & 5 & 10 \\ -3 & 2 & 6 \\ 2 & -4 & 3 \end{array}\right] \).
Now, find \( (A + B + C)' \):
\( (A + B + C)' = \left[\begin{array}{rrr} 4 & -3 & 2 \\ 5 & 2 & -4 \\ 10 & 6 & 3 \end{array}\right] \) (L.H.S.).
Next, calculate \( A' + B' + C' \):
\( A' + B' + C' = \left[\begin{array}{rrr} 1 & -1 & 1 \\ 2 & 0 & -3 \\ 3 & 2 & -1 \end{array}\right] + \left[\begin{array}{rrr} 4 & -1 & 2 \\ 5 & 0 & 1 \\ 6 & 1 & 2 \end{array}\right] + \left[\begin{array}{rrr} -1 & -1 & -1 \\ -2 & 2 & -2 \\ 1 & 3 & 2 \end{array}\right] \)
\( = \left[\begin{array}{rrr} 1+4+(-1) & -1+(-1)+(-1) & 1+2+(-1) \\ 2+5+(-2) & 0+0+2 & -3+1+(-2) \\ 3+6+1 & 2+1+3 & -1+2+2 \end{array}\right] \)
\( = \left[\begin{array}{rrr} 4 & -3 & 2 \\ 5 & 2 & -4 \\ 10 & 6 & 3 \end{array}\right] \) (R.H.S.).
Since L.H.S. = R.H.S., the property \( (A + B + C)' = A' + B' + C' \) is verified.
In simple words: For part (i), we multiplied matrix A' by 2 and then subtracted matrix B' from it. For part (ii), we added matrices A, B, and C, then took the transpose of their sum. We also separately took the transpose of each matrix and then added them up. Both calculations gave the same result, confirming the property.

🎯 Exam Tip: Remember that scalar multiplication distributes over addition (e.g., \( k(A+B) = kA + kB \)) and that the transpose of a sum is the sum of the transposes. These properties are very useful for simplifying matrix expressions.

Question 8. If \( A = \begin{bmatrix} 2 & 5 & 7 & 9 \end{bmatrix} \) and \( B = \begin{bmatrix} 3 \\ 0 \\ 2 \\ 4 \end{bmatrix} \) then prove that \( (A + B') = (A' + B)' \).
Answer:
Given matrix \( A = \begin{bmatrix} 2 & 5 & 7 & 9 \end{bmatrix} \) and \( B = \begin{bmatrix} 3 \\ 0 \\ 2 \\ 4 \end{bmatrix} \).
First, find the transpose of \( B \), which is \( B' \):
\( B' = \begin{bmatrix} 3 & 0 & 2 & 4 \end{bmatrix} \)
Now, calculate the Left Hand Side (L.H.S.) \( A + B' \):
\( A + B' = \begin{bmatrix} 2 & 5 & 7 & 9 \end{bmatrix} + \begin{bmatrix} 3 & 0 & 2 & 4 \end{bmatrix} \)
\( = \begin{bmatrix} 2+3 & 5+0 & 7+2 & 9+4 \end{bmatrix} \)
\( = \begin{bmatrix} 5 & 5 & 9 & 13 \end{bmatrix} \)
Next, find the transpose of \( A \), which is \( A' \):
\( A' = \begin{bmatrix} 2 \\ 5 \\ 7 \\ 9 \end{bmatrix} \)
Now, calculate \( A' + B \):
\( A' + B = \begin{bmatrix} 2 \\ 5 \\ 7 \\ 9 \end{bmatrix} + \begin{bmatrix} 3 \\ 0 \\ 2 \\ 4 \end{bmatrix} \)
\( = \begin{bmatrix} 2+3 \\ 5+0 \\ 7+2 \\ 9+4 \end{bmatrix} \)
\( = \begin{bmatrix} 5 \\ 5 \\ 9 \\ 13 \end{bmatrix} \)
Finally, calculate the Right Hand Side (R.H.S.) \( (A' + B)' \):
\( (A' + B)' = \begin{bmatrix} 5 \\ 5 \\ 9 \\ 13 \end{bmatrix}' \)
\( = \begin{bmatrix} 5 & 5 & 9 & 13 \end{bmatrix} \)
Since \( A + B' = \begin{bmatrix} 5 & 5 & 9 & 13 \end{bmatrix} \) and \( (A' + B)' = \begin{bmatrix} 5 & 5 & 9 & 13 \end{bmatrix} \), we have \( (A + B') = (A' + B)' \). This property is often called the transpose of a sum, and it holds true for compatible matrices. It shows how the transpose operation distributes over matrix addition.
In simple words: First, we change matrix B into its sideways version (transpose B'). Then, we add matrix A and transpose B'. Next, we change matrix A into its sideways version (transpose A'). We add transpose A' and matrix B. Finally, we change this result sideways (transpose (A' + B)'). Both final results are the same, proving the given statement.

🎯 Exam Tip: Remember that for matrix addition or subtraction, the matrices must have the same dimensions. The transpose operation simply swaps rows and columns.

Question 9. Find x and y if the matrix \( A = \frac{1}{3}\begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & -2 \\ x & 2 & y \end{bmatrix} \) may satisfy the condition \( AA' = A'A = I_3 \).
Answer:
Given matrix \( A = \frac{1}{3}\begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & -2 \\ x & 2 & y \end{bmatrix} \).
First, find the transpose of A, \( A' \):
\( A' = \frac{1}{3}\begin{bmatrix} 1 & 2 & x \\ 2 & 1 & 2 \\ 2 & -2 & y \end{bmatrix} \)
Now, calculate \( AA' \):
\( AA' = \left( \frac{1}{3}\begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & -2 \\ x & 2 & y \end{bmatrix} \right) \left( \frac{1}{3}\begin{bmatrix} 1 & 2 & x \\ 2 & 1 & 2 \\ 2 & -2 & y \end{bmatrix} \right) \)
\( = \frac{1}{9}\begin{bmatrix} 1(1)+2(2)+2(2) & 1(2)+2(1)+2(-2) & 1(x)+2(2)+2(y) \\ 2(1)+1(2)+(-2)(2) & 2(2)+1(1)+(-2)(-2) & 2(x)+1(2)+(-2)(y) \\ x(1)+2(2)+y(2) & x(2)+2(1)+y(-2) & x(x)+2(2)+y(y) \end{bmatrix} \)
\( = \frac{1}{9}\begin{bmatrix} 1+4+4 & 2+2-4 & x+4+2y \\ 2+2-4 & 4+1+4 & 2x+2-2y \\ x+4+2y & 2x+2-2y & x^2+4+y^2 \end{bmatrix} \)
\( = \frac{1}{9}\begin{bmatrix} 9 & 0 & x+4+2y \\ 0 & 9 & 2x+2-2y \\ x+4+2y & 2x+2-2y & x^2+4+y^2 \end{bmatrix} \)

Next, calculate \( A'A \):
\( A'A = \left( \frac{1}{3}\begin{bmatrix} 1 & 2 & x \\ 2 & 1 & 2 \\ 2 & -2 & y \end{bmatrix} \right) \left( \frac{1}{3}\begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & -2 \\ x & 2 & y \end{bmatrix} \right) \)
\( = \frac{1}{9}\begin{bmatrix} 1(1)+2(2)+x(x) & 1(2)+2(1)+x(2) & 1(2)+2(-2)+x(y) \\ 2(1)+1(2)+2(x) & 2(2)+1(1)+2(2) & 2(2)+1(-2)+2(y) \\ 2(1)+(-2)(2)+y(x) & 2(2)+(-2)(1)+y(2) & 2(2)+(-2)(-2)+y(y) \end{bmatrix} \)
\( = \frac{1}{9}\begin{bmatrix} 1+4+x^2 & 2+2+2x & 2-4+xy \\ 2+2+2x & 4+1+4 & 4-2+2y \\ 2-4+xy & 4-2+2y & 4+4+y^2 \end{bmatrix} \)
\( = \frac{1}{9}\begin{bmatrix} 5+x^2 & 4+2x & -2+xy \\ 4+2x & 9 & 2+2y \\ -2+xy & 2+2y & 8+y^2 \end{bmatrix} \)

We are given the condition \( AA' = A'A = I_3 \), where \( I_3 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \).
So, \( \frac{1}{9}\begin{bmatrix} 9 & 0 & x+4+2y \\ 0 & 9 & 2x+2-2y \\ x+4+2y & 2x+2-2y & x^2+4+y^2 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \)
This means: \( \begin{bmatrix} 9 & 0 & x+4+2y \\ 0 & 9 & 2x+2-2y \\ x+4+2y & 2x+2-2y & x^2+4+y^2 \end{bmatrix} = \begin{bmatrix} 9 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 9 \end{bmatrix} \)

Comparing elements, we get a system of equations:
1) \( x+4+2y = 0 \implies x+2y = -4 \) (Equation 1)
2) \( 2x+2-2y = 0 \implies 2x-2y = -2 \implies x-y = -1 \) (Equation 2)

Let's solve Equation 1 and Equation 2:
Add (1) and (2):
\( (x+2y) + (x-y) = -4 + (-1) \)
\( 2x + y = -5 \) (This is an intermediate step, not helpful for direct solution by adding)

A simpler way to solve is to add Eq 1 and Eq 2 term by term:
\( (x+2y) + (x-y) = -4 + (-1) \)
\( 2x + y = -5 \) (This is not correct, I will re-do the addition to find x and y).

Equation 1: \( x+2y = -4 \)
Equation 2: \( x-y = -1 \)
Subtract Equation 2 from Equation 1:
\( (x+2y) - (x-y) = -4 - (-1) \)
\( x+2y-x+y = -4+1 \)
\( 3y = -3 \)
\( \implies y = -1 \)

Substitute \( y = -1 \) into Equation 2:
\( x - (-1) = -1 \)
\( x+1 = -1 \)
\( \implies x = -2 \)

Now, we check these values with the corresponding elements from \( A'A \):
Comparing \( A'A \) with \( I_3 \):
\( \frac{1}{9}\begin{bmatrix} 5+x^2 & 4+2x & -2+xy \\ 4+2x & 9 & 2+2y \\ -2+xy & 2+2y & 8+y^2 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \)
This gives: \( \begin{bmatrix} 5+x^2 & 4+2x & -2+xy \\ 4+2x & 9 & 2+2y \\ -2+xy & 2+2y & 8+y^2 \end{bmatrix} = \begin{bmatrix} 9 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 9 \end{bmatrix} \)
Check with \( x = -2 \) and \( y = -1 \):
\( 5+x^2 = 5+(-2)^2 = 5+4 = 9 \) (Correct)
\( 4+2x = 4+2(-2) = 4-4 = 0 \) (Correct)
\( -2+xy = -2+(-2)(-1) = -2+2 = 0 \) (Correct)
\( 2+2y = 2+2(-1) = 2-2 = 0 \) (Correct)
\( 8+y^2 = 8+(-1)^2 = 8+1 = 9 \) (Correct)
The values \( x = -2 \) and \( y = -1 \) satisfy all conditions for \( AA' = A'A = I_3 \). This means the matrix A is an orthogonal matrix, a type of matrix where its inverse is equal to its transpose.
In simple words: We are given a matrix A and told that when we multiply it by its sideways version (A-transpose), we get the identity matrix. We did the multiplication for A multiplied by A-transpose and A-transpose multiplied by A. We then compared these results to the identity matrix to find matching parts. This helped us set up equations to solve for x and y, which turned out to be -2 and -1.

🎯 Exam Tip: When verifying conditions like \( AA' = I \), remember to equate corresponding elements of the resulting matrix with the identity matrix to form a system of equations. Be careful with calculations and signs, as a small error can lead to incorrect values for x and y.

Question 10. If \( A = \begin{bmatrix} 2 & 3 \\ 5 & -7 \end{bmatrix} \), then verify that \( (A^2)' = (A')^2 \).
Answer:
Given matrix \( A = \begin{bmatrix} 2 & 3 \\ 5 & -7 \end{bmatrix} \).
First, let's find \( A^2 \):
\( A^2 = A \cdot A = \begin{bmatrix} 2 & 3 \\ 5 & -7 \end{bmatrix} \begin{bmatrix} 2 & 3 \\ 5 & -7 \end{bmatrix} \)
\( = \begin{bmatrix} 2(2)+3(5) & 2(3)+3(-7) \\ 5(2)+(-7)(5) & 5(3)+(-7)(-7) \end{bmatrix} \)
\( = \begin{bmatrix} 4+15 & 6-21 \\ 10-35 & 15+49 \end{bmatrix} \)
\( = \begin{bmatrix} 19 & -15 \\ -25 & 64 \end{bmatrix} \)
Now, find \( (A^2)' \), which is the transpose of \( A^2 \):
\( (A^2)' = \begin{bmatrix} 19 & -15 \\ -25 & 64 \end{bmatrix}' \)
\( = \begin{bmatrix} 19 & -25 \\ -15 & 64 \end{bmatrix} \) (Equation 1)

Next, let's find \( A' \), the transpose of \( A \):
\( A' = \begin{bmatrix} 2 & 3 \\ 5 & -7 \end{bmatrix}' \)
\( = \begin{bmatrix} 2 & 5 \\ 3 & -7 \end{bmatrix} \)
Now, find \( (A')^2 \):
\( (A')^2 = A' \cdot A' = \begin{bmatrix} 2 & 5 \\ 3 & -7 \end{bmatrix} \begin{bmatrix} 2 & 5 \\ 3 & -7 \end{bmatrix} \)
\( = \begin{bmatrix} 2(2)+5(3) & 2(5)+5(-7) \\ 3(2)+(-7)(3) & 3(5)+(-7)(-7) \end{bmatrix} \)
\( = \begin{bmatrix} 4+15 & 10-35 \\ 6-21 & 15+49 \end{bmatrix} \)
\( = \begin{bmatrix} 19 & -25 \\ -15 & 64 \end{bmatrix} \) (Equation 2)

From (1) and (2), we can see that \( (A^2)' = (A')^2 \). This property is a useful rule for matrix transposes and powers, simplifying calculations when working with these operations.
In simple words: First, we square matrix A, meaning we multiply A by itself, then we flip its rows and columns to get the transpose. This is our first answer. Next, we flip matrix A first (get A-transpose), and then we multiply that flipped matrix by itself. This is our second answer. Both answers are identical, which means the statement is true.

🎯 Exam Tip: Remember the order of operations: \( (A^2)' \) means square first, then transpose; while \( (A')^2 \) means transpose first, then square. For square matrices, these two operations often lead to the same result.

Question 11. If \( A = \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix} \), then verify that \( A'A = I \).
Answer:
Given matrix \( A = \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix} \).
First, find the transpose of A, \( A' \):
\( A' = \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix} \)
Now, calculate \( A'A \):
\( A'A = \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix} \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix} \)
\( = \begin{bmatrix} (\cos \alpha)(\cos \alpha) + (-\sin \alpha)(-\sin \alpha) & (\cos \alpha)(\sin \alpha) + (-\sin \alpha)(\cos \alpha) \\ (\sin \alpha)(\cos \alpha) + (\cos \alpha)(-\sin \alpha) & (\sin \alpha)(\sin \alpha) + (\cos \alpha)(\cos \alpha) \end{bmatrix} \)
\( = \begin{bmatrix} \cos^2 \alpha + \sin^2 \alpha & \cos \alpha \sin \alpha - \sin \alpha \cos \alpha \\ \sin \alpha \cos \alpha - \cos \alpha \sin \alpha & \sin^2 \alpha + \cos^2 \alpha \end{bmatrix} \)
Using the trigonometric identity \( \cos^2 \alpha + \sin^2 \alpha = 1 \):
\( = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)
We know that \( I \) is the identity matrix, \( I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \).
Thus, \( A'A = I \). This shows that matrix A is an orthogonal matrix, meaning its inverse is equal to its transpose. This type of matrix is often used in rotations in geometry.
In simple words: We have a special matrix A that uses sine and cosine values. First, we flip A to get its transpose (A'). Then, we multiply A' by A. When we do the math, using a basic rule from trigonometry, all the numbers outside the main diagonal become zero, and the numbers on the diagonal become one. This gives us the identity matrix, which is like the number 1 for matrices.

🎯 Exam Tip: Questions involving trigonometric functions in matrices often require the use of fundamental trigonometric identities like \( \sin^2 \theta + \cos^2 \theta = 1 \) and \( \sin \theta \cos \theta - \cos \theta \sin \theta = 0 \). Keep these in mind to simplify your matrix products.