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

Question 1. Define a symmetric matrix. Show that the following matrices are symmetric.
(i) \( \left[\begin{array}{lll} a & h & g \\ h & b & f \\ g & f & c \end{array}\right] \)
(ii) \( \left[\begin{array}{ccc} 2+i & 1 & 3 \\ 1 & 2 & 3+2 i \\ 3 & 3+2 i & 4 \end{array}\right] \)
Answer: A matrix \( P \) is called a symmetric matrix if its transpose \( P' \) is equal to itself, meaning \( P' = P \). This happens when the elements \( a_{ij} \) are equal to \( a_{ji} \).
(i) Let \( P = \left[\begin{array}{lll} a & h & g \\ h & b & f \\ g & f & c \end{array}\right] \)
\( \implies P' = \left[\begin{array}{lll} a & h & g \\ h & b & f \\ g & f & c \end{array}\right]' = \left[\begin{array}{lll} a & h & g \\ h & b & f \\ g & f & c \end{array}\right] = P \) Thus, \( P \) is a symmetric matrix.
(ii) Let \( P = \left[\begin{array}{ccc} 2+i & 1 & 3 \\ 1 & 2 & 3+2 i \\ 3 & 3+2 i & 4 \end{array}\right] \)
\( \implies P' = \left[\begin{array}{ccc} 2+i & 1 & 3 \\ 1 & 2 & 3+2 i \\ 3 & 3+2 i & 4 \end{array}\right]' = \left[\begin{array}{ccc} 2+i & 1 & 3 \\ 1 & 2 & 3+2 i \\ 3 & 3+2 i & 4 \end{array}\right] = P \) Thus, \( P \) is a symmetric matrix.
In simple words: A symmetric matrix is like a mirror image across its main diagonal; if you swap rows and columns, the matrix stays the same. The numbers \( a_{ij} \) and \( a_{ji} \) are always equal.

🎯 Exam Tip: To show a matrix is symmetric, always calculate its transpose and then compare it to the original matrix. For full marks, explicitly state that \( P' = P \).

Question 2. Show that each of the following is a skew matrix.
(i) \( \left[\begin{array}{rrr} 0 & 5 & -4 \\ -5 & 0 & 7 \\ 4 & -7 & 0 \end{array}\right] \)
(ii) \( \left[\begin{array}{rrr} 0 & b & -c \\ -b & 0 & d \\ c & -d & 0 \end{array}\right] \)
(iii) \( \left[\begin{array}{rrr} 0 & 2 i & 3 \\ -2 i & 0 & 4 \\ -3 & -4 & 0 \end{array}\right] \)
Answer: A matrix \( Q \) is called a skew-symmetric matrix if its transpose \( Q' \) is equal to the negative of the original matrix, meaning \( Q' = -Q \). This also implies that all diagonal elements must be zero.
(i) Let \( Q = \left[\begin{array}{rrr} 0 & 5 & -4 \\ -5 & 0 & 7 \\ 4 & -7 & 0 \end{array}\right] \)
\( \implies Q' = \left[\begin{array}{rrr} 0 & -5 & 4 \\ 5 & 0 & -7 \\ -4 & 7 & 0 \end{array}\right] \) Now, taking the negative of \( Q \):
\( \implies -Q = -\left[\begin{array}{rrr} 0 & 5 & -4 \\ -5 & 0 & 7 \\ 4 & -7 & 0 \end{array}\right] = \left[\begin{array}{rrr} 0 & -5 & 4 \\ 5 & 0 & -7 \\ -4 & 7 & 0 \end{array}\right] \) Since \( Q' = -Q \), \( Q \) is a skew-symmetric matrix.
(ii) Let \( Q = \left[\begin{array}{rrr} 0 & b & -c \\ -b & 0 & d \\ c & -d & 0 \end{array}\right] \)
\( \implies Q' = \left[\begin{array}{rrr} 0 & -b & c \\ b & 0 & -d \\ -c & d & 0 \end{array}\right] \) Now, taking the negative of \( Q \):
\( \implies -Q = -\left[\begin{array}{rrr} 0 & b & -c \\ -b & 0 & d \\ c & -d & 0 \end{array}\right] = \left[\begin{array}{rrr} 0 & -b & c \\ b & 0 & -d \\ -c & d & 0 \end{array}\right] \) Since \( Q' = -Q \), \( Q \) is a skew-symmetric matrix.
(iii) Let \( Q = \left[\begin{array}{rrr} 0 & 2 i & 3 \\ -2 i & 0 & 4 \\ -3 & -4 & 0 \end{array}\right] \)
\( \implies Q' = \left[\begin{array}{rrr} 0 & -2 i & -3 \\ 2 i & 0 & -4 \\ 3 & 4 & 0 \end{array}\right] \) Now, taking the negative of \( Q \):
\( \implies -Q = -\left[\begin{array}{rrr} 0 & 2 i & 3 \\ -2 i & 0 & 4 \\ -3 & -4 & 0 \end{array}\right] = \left[\begin{array}{rrr} 0 & -2 i & -3 \\ 2 i & 0 & -4 \\ 3 & 4 & 0 \end{array}\right] \) Since \( Q' = -Q \), \( Q \) is a skew-symmetric matrix.
In simple words: A skew-symmetric matrix is one where its transpose is the exact opposite (negative) of the original matrix. The elements on the main diagonal will always be zero, and the elements across the diagonal will be negatives of each other.

🎯 Exam Tip: When proving a matrix is skew-symmetric, remember two key things: all diagonal elements must be zero, and \( a_{ij} = -a_{ji} \) for non-diagonal elements. Show that \( Q' = -Q \) clearly.

Question 3. Show that A + A' is a symmetric matrix if
(i) \( A = \left[\begin{array}{ll} 4 & 1 \\ 5 & 8 \end{array}\right] \)
(ii) \( \left[\begin{array}{ll} 2 & 4 \\ 5 & 6 \end{array}\right] \)
Answer: We need to show that the sum of a matrix A and its transpose A' (let's call it P) is symmetric, meaning \( P' = P \).
(i) Given \( A = \left[\begin{array}{ll} 4 & 1 \\ 5 & 8 \end{array}\right] \) First, find the transpose of A: \( A' = \left[\begin{array}{ll} 4 & 5 \\ 1 & 8 \end{array}\right] \) Now, let \( P = A + A' \):
\( P = \left[\begin{array}{ll} 4 & 1 \\ 5 & 8 \end{array}\right] + \left[\begin{array}{ll} 4 & 5 \\ 1 & 8 \end{array}\right] = \left[\begin{array}{ll} 4+4 & 1+5 \\ 5+1 & 8+8 \end{array}\right] = \left[\begin{array}{cc} 8 & 6 \\ 6 & 16 \end{array}\right] \) Next, find the transpose of P:
\( P' = \left[\begin{array}{cc} 8 & 6 \\ 6 & 16 \end{array}\right]' = \left[\begin{array}{cc} 8 & 6 \\ 6 & 16 \end{array}\right] \) Since \( P' = P \), the matrix \( A + A' \) is symmetric.
(ii) Given \( A = \left[\begin{array}{ll} 2 & 4 \\ 5 & 6 \end{array}\right] \) First, find the transpose of A: \( A' = \left[\begin{array}{ll} 2 & 5 \\ 4 & 6 \end{array}\right] \) Now, let \( P = A + A' \):
\( P = \left[\begin{array}{ll} 2 & 4 \\ 5 & 6 \end{array}\right] + \left[\begin{array}{ll} 2 & 5 \\ 4 & 6 \end{array}\right] = \left[\begin{array}{ll} 2+2 & 4+5 \\ 5+4 & 6+6 \end{array}\right] = \left[\begin{array}{cc} 4 & 9 \\ 9 & 12 \end{array}\right] \) Next, find the transpose of P:
\( P' = \left[\begin{array}{cc} 4 & 9 \\ 9 & 12 \end{array}\right]' = \left[\begin{array}{cc} 4 & 9 \\ 9 & 12 \end{array}\right] \) Since \( P' = P \), the matrix \( A + A' \) is symmetric.
In simple words: When you add a matrix to its own flipped version (transpose), the new matrix you get will always be symmetric. This is a general rule in matrix algebra.

🎯 Exam Tip: Remember the property that \( (A+B)' = A' + B' \). For \( P = A + A' \), its transpose is \( P' = (A+A')' = A' + (A')' = A' + A = A + A' = P \), thus always symmetric.

Question 4. Show that A – A' is a skew-symmetric matrix if
(i) \( A = \left[\begin{array}{ll} 3 & 4 \\ 5 & 1 \end{array}\right] \)
(ii) \( A = \left[\begin{array}{ll} 3 & -4 \\ 1 & -1 \end{array}\right] \)
(iii) \( A = \left[\begin{array}{ll} 1 & 4 \\ 3 & 7 \end{array}\right] \)
Answer: We need to show that the difference between a matrix A and its transpose A' (let's call it P or Q) is skew-symmetric, meaning \( P' = -P \).
(i) Given \( A = \left[\begin{array}{ll} 3 & 4 \\ 5 & 1 \end{array}\right] \) First, find the transpose of A: \( A' = \left[\begin{array}{ll} 3 & 5 \\ 4 & 1 \end{array}\right] \) Now, let \( P = A - A' \):
\( P = \left[\begin{array}{ll} 3 & 4 \\ 5 & 1 \end{array}\right] - \left[\begin{array}{ll} 3 & 5 \\ 4 & 1 \end{array}\right] = \left[\begin{array}{cc} 3-3 & 4-5 \\ 5-4 & 1-1 \end{array}\right] = \left[\begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array}\right] \) Next, find the transpose of P:
\( P' = \left[\begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array}\right]' = \left[\begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array}\right] \) And find the negative of P:
\( -P = -\left[\begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array}\right] = \left[\begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array}\right] \) Since \( P' = -P \), the matrix \( A - A' \) is skew-symmetric.
(ii) Given \( A = \left[\begin{array}{ll} 3 & -4 \\ 1 & -1 \end{array}\right] \) First, find the transpose of A: \( A' = \left[\begin{array}{rr} 3 & 1 \\ -4 & -1 \end{array}\right] \) Now, let \( Q = A - A' \):
\( Q = \left[\begin{array}{rr} 3 & -4 \\ 1 & -1 \end{array}\right] - \left[\begin{array}{rr} 3 & 1 \\ -4 & -1 \end{array}\right] = \left[\begin{array}{cc} 3-3 & -4-1 \\ 1-(-4) & -1-(-1) \end{array}\right] = \left[\begin{array}{cc} 0 & -5 \\ 5 & 0 \end{array}\right] \) Next, find the transpose of Q:
\( Q' = \left[\begin{array}{cc} 0 & -5 \\ 5 & 0 \end{array}\right]' = \left[\begin{array}{cc} 0 & 5 \\ -5 & 0 \end{array}\right] \) And find the negative of Q:
\( -Q = -\left[\begin{array}{cc} 0 & -5 \\ 5 & 0 \end{array}\right] = \left[\begin{array}{cc} 0 & 5 \\ -5 & 0 \end{array}\right] \) Since \( Q' = -Q \), the matrix \( A - A' \) is skew-symmetric.
(iii) Given \( A = \left[\begin{array}{ll} 1 & 4 \\ 3 & 7 \end{array}\right] \) First, find the transpose of A: \( A' = \left[\begin{array}{ll} 1 & 3 \\ 4 & 7 \end{array}\right] \) Now, let \( Q = A - A' \):
\( Q = \left[\begin{array}{ll} 1 & 4 \\ 3 & 7 \end{array}\right] - \left[\begin{array}{ll} 1 & 3 \\ 4 & 7 \end{array}\right] = \left[\begin{array}{cc} 1-1 & 4-3 \\ 3-4 & 7-7 \end{array}\right] = \left[\begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array}\right] \) Next, find the transpose of Q:
\( Q' = \left[\begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array}\right]' = \left[\begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array}\right] \) And find the negative of Q:
\( -Q = -\left[\begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array}\right] = \left[\begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array}\right] \) Since \( Q' = -Q \), the matrix \( A - A' \) is skew-symmetric.
In simple words: When you subtract a matrix from its own flipped version (transpose), the resulting matrix will always be skew-symmetric. This means its transpose will be its negative.

🎯 Exam Tip: The property \( (A-B)' = A' - B' \) is useful here. For \( Q = A - A' \), its transpose is \( Q' = (A-A')' = A' - (A')' = A' - A = -(A - A') = -Q \), thus always skew-symmetric.

Question 5. If \( A = \left[\begin{array}{cc} 4 & x+2 \\ 2 x-3 & x+1 \end{array}\right] \) is a symmetric matrix. Find the value of x.
Answer: A matrix A is symmetric if its transpose \( A' \) is equal to A itself, meaning \( A' = A \). This means that for a symmetric matrix, the elements \( a_{ij} \) must be equal to \( a_{ji} \).
Given \( A = \left[\begin{array}{cc} 4 & x+2 \\ 2 x-3 & x+1 \end{array}\right] \) First, find the transpose of A: \( A' = \left[\begin{array}{cr} 4 & 2 x-3 \\ x+2 & x+1 \end{array}\right] \) Since A is a symmetric matrix, we must have \( A' = A \).
\( \implies \left[\begin{array}{cr} 4 & 2 x-3 \\ x+2 & x+1 \end{array}\right] = \left[\begin{array}{cc} 4 & x+2 \\ 2 x-3 & x+1 \end{array}\right] \) For these matrices to be equal, their corresponding elements must be equal. We can set the off-diagonal elements equal to each other:
\( \implies 2x - 3 = x + 2 \) Subtract x from both sides: \( x - 3 = 2 \) Add 3 to both sides: \( x = 5 \) Alternatively, using the other off-diagonal elements:
\( \implies x + 2 = 2x - 3 \) Subtract x from both sides: \( 2 = x - 3 \) Add 3 to both sides: \( x = 5 \) Thus, the value of \( x \) is 5.
In simple words: For a matrix to be symmetric, the numbers opposite each other across the main diagonal must be the same. We use this rule to set the expressions with 'x' equal to each other and solve for 'x'.

🎯 Exam Tip: Always remember that for a symmetric matrix \( A = [a_{ij}] \), we have \( a_{ij} = a_{ji} \). This means elements like \( a_{12} \) and \( a_{21} \) must be equal. This is the key to solving for unknown variables.

Question 6. Express the following as the sum of a symmetric and skew symmetric matrix.
(i) \( \left[\begin{array}{ll} 7 & 4 \\ 5 & 3 \end{array}\right] \)
(ii) \( \left[\begin{array}{ll} 3 & 2 \\ 4 & 5 \end{array}\right] \)
(iii) \( \left[\begin{array}{rrr} 2 & 4 & -6 \\ 7 & 3 & 5 \\ 1 & -2 & 4 \end{array}\right] \)
Answer: Any square matrix A can be expressed as the sum of a symmetric matrix P and a skew-symmetric matrix Q, where \( P = \frac{1}{2}(A + A') \) and \( Q = \frac{1}{2}(A - A') \).
(i) Let \( A = \left[\begin{array}{ll} 7 & 4 \\ 5 & 3 \end{array}\right] \) First, find the transpose of A: \( A' = \left[\begin{array}{ll} 7 & 5 \\ 4 & 3 \end{array}\right] \) Now, find the symmetric part \( P \):
\( P = \frac{1}{2}(A + A') = \frac{1}{2}\left( \left[\begin{array}{ll} 7 & 4 \\ 5 & 3 \end{array}\right] + \left[\begin{array}{ll} 7 & 5 \\ 4 & 3 \end{array}\right] \right) = \frac{1}{2}\left[\begin{array}{ll} 7+7 & 4+5 \\ 5+4 & 3+3 \end{array}\right] = \frac{1}{2}\left[\begin{array}{cc} 14 & 9 \\ 9 & 6 \end{array}\right] = \left[\begin{array}{cc} 7 & 9/2 \\ 9/2 & 3 \end{array}\right] \) To verify, \( P' = \left[\begin{array}{cc} 7 & 9/2 \\ 9/2 & 3 \end{array}\right]' = \left[\begin{array}{cc} 7 & 9/2 \\ 9/2 & 3 \end{array}\right] = P \). So, P is symmetric. Next, find the skew-symmetric part \( Q \):
\( Q = \frac{1}{2}(A - A') = \frac{1}{2}\left( \left[\begin{array}{ll} 7 & 4 \\ 5 & 3 \end{array}\right] - \left[\begin{array}{ll} 7 & 5 \\ 4 & 3 \end{array}\right] \right) = \frac{1}{2}\left[\begin{array}{ll} 7-7 & 4-5 \\ 5-4 & 3-3 \end{array}\right] = \frac{1}{2}\left[\begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array}\right] = \left[\begin{array}{cc} 0 & -1/2 \\ 1/2 & 0 \end{array}\right] \) To verify, \( Q' = \left[\begin{array}{cc} 0 & -1/2 \\ 1/2 & 0 \end{array}\right]' = \left[\begin{array}{cc} 0 & 1/2 \\ -1/2 & 0 \end{array}\right] = -\left[\begin{array}{cc} 0 & -1/2 \\ 1/2 & 0 \end{array}\right] = -Q \). So, Q is skew-symmetric. Finally, \( A = P + Q = \left[\begin{array}{cc} 7 & 9/2 \\ 9/2 & 3 \end{array}\right] + \left[\begin{array}{cc} 0 & -1/2 \\ 1/2 & 0 \end{array}\right] = \left[\begin{array}{cc} 7+0 & 9/2-1/2 \\ 9/2+1/2 & 3+0 \end{array}\right] = \left[\begin{array}{cc} 7 & 8/2 \\ 10/2 & 3 \end{array}\right] = \left[\begin{array}{ll} 7 & 4 \\ 5 & 3 \end{array}\right] \). This matches the original A.
(ii) Let \( A = \left[\begin{array}{ll} 3 & 2 \\ 4 & 5 \end{array}\right] \) First, find the transpose of A: \( A' = \left[\begin{array}{ll} 3 & 4 \\ 2 & 5 \end{array}\right] \) Now, find the symmetric part \( P \):
\( P = \frac{1}{2}(A + A') = \frac{1}{2}\left( \left[\begin{array}{ll} 3 & 2 \\ 4 & 5 \end{array}\right] + \left[\begin{array}{ll} 3 & 4 \\ 2 & 5 \end{array}\right] \right) = \frac{1}{2}\left[\begin{array}{ll} 3+3 & 2+4 \\ 4+2 & 5+5 \end{array}\right] = \frac{1}{2}\left[\begin{array}{cc} 6 & 6 \\ 6 & 10 \end{array}\right] = \left[\begin{array}{cc} 3 & 3 \\ 3 & 5 \end{array}\right] \) To verify, \( P' = \left[\begin{array}{cc} 3 & 3 \\ 3 & 5 \end{array}\right]' = \left[\begin{array}{cc} 3 & 3 \\ 3 & 5 \end{array}\right] = P \). So, P is symmetric. Next, find the skew-symmetric part \( Q \):
\( Q = \frac{1}{2}(A - A') = \frac{1}{2}\left( \left[\begin{array}{ll} 3 & 2 \\ 4 & 5 \end{array}\right] - \left[\begin{array}{ll} 3 & 4 \\ 2 & 5 \end{array}\right] \right) = \frac{1}{2}\left[\begin{array}{ll} 3-3 & 2-4 \\ 4-2 & 5-5 \end{array}\right] = \frac{1}{2}\left[\begin{array}{cc} 0 & -2 \\ 2 & 0 \end{array}\right] = \left[\begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array}\right] \) To verify, \( Q' = \left[\begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array}\right]' = \left[\begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array}\right] = -\left[\begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array}\right] = -Q \). So, Q is skew-symmetric. Finally, \( A = P + Q = \left[\begin{array}{cc} 3 & 3 \\ 3 & 5 \end{array}\right] + \left[\begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array}\right] = \left[\begin{array}{cc} 3+0 & 3-1 \\ 3+1 & 5+0 \end{array}\right] = \left[\begin{array}{cc} 3 & 2 \\ 4 & 5 \end{array}\right] \). This matches the original A.
(iii) Let \( A = \left[\begin{array}{rrr} 2 & 4 & -6 \\ 7 & 3 & 5 \\ 1 & -2 & 4 \end{array}\right] \) First, find the transpose of A: \( A' = \left[\begin{array}{rrr} 2 & 7 & 1 \\ 4 & 3 & -2 \\ -6 & 5 & 4 \end{array}\right] \) Now, find the symmetric part \( P \):
\( P = \frac{1}{2}(A + A') = \frac{1}{2}\left( \left[\begin{array}{rrr} 2 & 4 & -6 \\ 7 & 3 & 5 \\ 1 & -2 & 4 \end{array}\right] + \left[\begin{array}{rrr} 2 & 7 & 1 \\ 4 & 3 & -2 \\ -6 & 5 & 4 \end{array}\right] \right) \)
\( P = \frac{1}{2}\left[\begin{array}{rrr} 2+2 & 4+7 & -6+1 \\ 7+4 & 3+3 & 5+(-2) \\ 1+(-6) & -2+5 & 4+4 \end{array}\right] = \frac{1}{2}\left[\begin{array}{rrr} 4 & 11 & -5 \\ 11 & 6 & 3 \\ -5 & 3 & 8 \end{array}\right] = \left[\begin{array}{rrr} 2 & 11/2 & -5/2 \\ 11/2 & 3 & 3/2 \\ -5/2 & 3/2 & 4 \end{array}\right] \) To verify, \( P' = P \), so P is symmetric. Next, find the skew-symmetric part \( Q \):
\( Q = \frac{1}{2}(A - A') = \frac{1}{2}\left( \left[\begin{array}{rrr} 2 & 4 & -6 \\ 7 & 3 & 5 \\ 1 & -2 & 4 \end{array}\right] - \left[\begin{array}{rrr} 2 & 7 & 1 \\ 4 & 3 & -2 \\ -6 & 5 & 4 \end{array}\right] \right) \)
\( Q = \frac{1}{2}\left[\begin{array}{rrr} 2-2 & 4-7 & -6-1 \\ 7-4 & 3-3 & 5-(-2) \\ 1-(-6) & -2-5 & 4-4 \end{array}\right] = \frac{1}{2}\left[\begin{array}{rrr} 0 & -3 & -7 \\ 3 & 0 & 7 \\ 7 & -7 & 0 \end{array}\right] = \left[\begin{array}{rrr} 0 & -3/2 & -7/2 \\ 3/2 & 0 & 7/2 \\ 7/2 & -7/2 & 0 \end{array}\right] \) To verify, \( Q' = -Q \), so Q is skew-symmetric. Finally, \( A = P + Q = \left[\begin{array}{rrr} 2 & 11/2 & -5/2 \\ 11/2 & 3 & 3/2 \\ -5/2 & 3/2 & 4 \end{array}\right] + \left[\begin{array}{rrr} 0 & -3/2 & -7/2 \\ 3/2 & 0 & 7/2 \\ 7/2 & -7/2 & 0 \end{array}\right] \)
\( A = \left[\begin{array}{rrr} 2+0 & 11/2-3/2 & -5/2-7/2 \\ 11/2+3/2 & 3+0 & 3/2+7/2 \\ -5/2+7/2 & 3/2-7/2 & 4+0 \end{array}\right] = \left[\begin{array}{rrr} 2 & 8/2 & -12/2 \\ 14/2 & 3 & 10/2 \\ 2/2 & -4/2 & 4 \end{array}\right] = \left[\begin{array}{rrr} 2 & 4 & -6 \\ 7 & 3 & 5 \\ 1 & -2 & 4 \end{array}\right] \). This matches the original A.
In simple words: Any square matrix can be neatly split into two parts: one part that is symmetric (stays the same when flipped) and another part that is skew-symmetric (becomes its negative when flipped). You find these parts by adding or subtracting the matrix and its transpose, then dividing by two.

🎯 Exam Tip: Remember the formulas \( P = \frac{1}{2}(A + A') \) for the symmetric part and \( Q = \frac{1}{2}(A - A') \) for the skew-symmetric part. Always verify that P is symmetric (\( P'=P \)) and Q is skew-symmetric (\( Q'=-Q \)).

Question 7. Find the symmetric and skew symmetric parts of the matrix \( A = \left[\begin{array}{III} 1 & 2 & 4 \\ 6 & 8 & 1 \\ 3 & 5 & 7 \end{array}\right] \)
Answer: Any square matrix A can be written as the sum of a symmetric matrix P and a skew-symmetric matrix Q. The formulas for these parts are \( P = \frac{1}{2}(A + A') \) and \( Q = \frac{1}{2}(A - A') \).
Given \( A = \left[\begin{array}{lll} 1 & 2 & 4 \\ 6 & 8 & 1 \\ 3 & 5 & 7 \end{array}\right] \) First, find the transpose of A by swapping rows and columns:
\( A' = \left[\begin{array}{lll} 1 & 6 & 3 \\ 2 & 8 & 5 \\ 4 & 1 & 7 \end{array}\right] \) Now, calculate the symmetric part, P:
\( P = \frac{1}{2}(A + A') = \frac{1}{2}\left( \left[\begin{array}{lll} 1 & 2 & 4 \\ 6 & 8 & 1 \\ 3 & 5 & 7 \end{array}\right] + \left[\begin{array}{lll} 1 & 6 & 3 \\ 2 & 8 & 5 \\ 4 & 1 & 7 \end{array}\right] \right) \)
\( P = \frac{1}{2}\left[\begin{array}{lll} 1+1 & 2+6 & 4+3 \\ 6+2 & 8+8 & 1+5 \\ 3+4 & 5+1 & 7+7 \end{array}\right] = \frac{1}{2}\left[\begin{array}{ccc} 2 & 8 & 7 \\ 8 & 16 & 6 \\ 7 & 6 & 14 \end{array}\right] \)
\( P = \left[\begin{array}{ccc} 1 & 4 & 7/2 \\ 4 & 8 & 3 \\ 7/2 & 3 & 7 \end{array}\right] \) To verify, \( P' = \left[\begin{array}{ccc} 1 & 4 & 7/2 \\ 4 & 8 & 3 \\ 7/2 & 3 & 7 \end{array}\right]' = \left[\begin{array}{ccc} 1 & 4 & 7/2 \\ 4 & 8 & 3 \\ 7/2 & 3 & 7 \end{array}\right] = P \). So P is the symmetric part.
Next, calculate the skew-symmetric part, Q:
\( Q = \frac{1}{2}(A - A') = \frac{1}{2}\left( \left[\begin{array}{lll} 1 & 2 & 4 \\ 6 & 8 & 1 \\ 3 & 5 & 7 \end{array}\right] - \left[\begin{array}{lll} 1 & 6 & 3 \\ 2 & 8 & 5 \\ 4 & 1 & 7 \end{array}\right] \right) \)
\( Q = \frac{1}{2}\left[\begin{array}{lll} 1-1 & 2-6 & 4-3 \\ 6-2 & 8-8 & 1-5 \\ 3-4 & 5-1 & 7-7 \end{array}\right] = \frac{1}{2}\left[\begin{array}{rrr} 0 & -4 & 1 \\ 4 & 0 & -4 \\ -1 & 4 & 0 \end{array}\right] \)
\( Q = \left[\begin{array}{rrr} 0 & -2 & 1/2 \\ 2 & 0 & -2 \\ -1/2 & 2 & 0 \end{array}\right] \) To verify, \( Q' = \left[\begin{array}{rrr} 0 & -2 & 1/2 \\ 2 & 0 & -2 \\ -1/2 & 2 & 0 \end{array}\right]' = \left[\begin{array}{rrr} 0 & 2 & -1/2 \\ -2 & 0 & 2 \\ 1/2 & -2 & 0 \end{array}\right] = -\left[\begin{array}{rrr} 0 & -2 & 1/2 \\ 2 & 0 & -2 \\ -1/2 & 2 & 0 \end{array}\right] = -Q \). So Q is the skew-symmetric part.
In simple words: We took the original matrix A, found its flipped version A', then calculated two new matrices. One is P, which is half of (A + A'), and it's always symmetric. The other is Q, which is half of (A - A'), and it's always skew-symmetric. When you add P and Q together, you get the original matrix A back.

🎯 Exam Tip: Clearly label P as the symmetric part and Q as the skew-symmetric part. Always check that \( P+Q \) equals the original matrix A as a final verification step.

Question 8. Let \( A = \begin{bmatrix} 3 & 2 & 7 \\ 1 & 4 & 3 \\ -2 & 5 & 8 \end{bmatrix} \), find X and Y such that X + Y = A and X is a symmetric and Y a skew symmetric matrix.
Answer:We know that any square matrix A can be written as the sum of a symmetric matrix X and a skew-symmetric matrix Y. Here, \( X = \frac{1}{2}(A + A') \) and \( Y = \frac{1}{2}(A - A') \). Given matrix \( A = \begin{bmatrix} 3 & 2 & 7 \\ 1 & 4 & 3 \\ -2 & 5 & 8 \end{bmatrix} \). First, find its transpose, \( A' \): \( A' = \begin{bmatrix} 3 & 1 & -2 \\ 2 & 4 & 5 \\ 7 & 3 & 8 \end{bmatrix} \) Now, calculate X, the symmetric part: \( A + A' = \begin{bmatrix} 3 & 2 & 7 \\ 1 & 4 & 3 \\ -2 & 5 & 8 \end{bmatrix} + \begin{bmatrix} 3 & 1 & -2 \\ 2 & 4 & 5 \\ 7 & 3 & 8 \end{bmatrix} \)
\( \implies A + A' = \begin{bmatrix} 3+3 & 2+1 & 7-2 \\ 1+2 & 4+4 & 3+5 \\ -2+7 & 5+3 & 8+8 \end{bmatrix} \)
\( \implies A + A' = \begin{bmatrix} 6 & 3 & 5 \\ 3 & 8 & 8 \\ 5 & 8 & 16 \end{bmatrix} \)
\( \implies X = \frac{1}{2}(A + A') = \frac{1}{2} \begin{bmatrix} 6 & 3 & 5 \\ 3 & 8 & 8 \\ 5 & 8 & 16 \end{bmatrix} \)
\( \implies X = \begin{bmatrix} 3 & 3/2 & 5/2 \\ 3/2 & 4 & 4 \\ 5/2 & 4 & 8 \end{bmatrix} \) To verify X is symmetric, check if \( X' = X \): \( X' = \begin{bmatrix} 3 & 3/2 & 5/2 \\ 3/2 & 4 & 4 \\ 5/2 & 4 & 8 \end{bmatrix} \). Since \( X' = X \), X is a symmetric matrix. Next, calculate Y, the skew-symmetric part: \( A - A' = \begin{bmatrix} 3 & 2 & 7 \\ 1 & 4 & 3 \\ -2 & 5 & 8 \end{bmatrix} - \begin{bmatrix} 3 & 1 & -2 \\ 2 & 4 & 5 \\ 7 & 3 & 8 \end{bmatrix} \)
\( \implies A - A' = \begin{bmatrix} 3-3 & 2-1 & 7-(-2) \\ 1-2 & 4-4 & 3-5 \\ -2-7 & 5-3 & 8-8 \end{bmatrix} \)
\( \implies A - A' = \begin{bmatrix} 0 & 1 & 9 \\ -1 & 0 & -2 \\ -9 & 2 & 0 \end{bmatrix} \)
\( \implies Y = \frac{1}{2}(A - A') = \frac{1}{2} \begin{bmatrix} 0 & 1 & 9 \\ -1 & 0 & -2 \\ -9 & 2 & 0 \end{bmatrix} \)
\( \implies Y = \begin{bmatrix} 0 & 1/2 & 9/2 \\ -1/2 & 0 & -1 \\ -9/2 & 1 & 0 \end{bmatrix} \) To verify Y is skew-symmetric, check if \( Y' = -Y \): \( Y' = \begin{bmatrix} 0 & -1/2 & -9/2 \\ 1/2 & 0 & 1 \\ 9/2 & -1 & 0 \end{bmatrix} \)
\( \implies -Y = - \begin{bmatrix} 0 & 1/2 & 9/2 \\ -1/2 & 0 & -1 \\ -9/2 & 1 & 0 \end{bmatrix} = \begin{bmatrix} 0 & -1/2 & -9/2 \\ 1/2 & 0 & 1 \\ 9/2 & -1 & 0 \end{bmatrix} \) Since \( Y' = -Y \), Y is a skew-symmetric matrix. Finally, verify that \( X + Y = A \): \( X + Y = \begin{bmatrix} 3 & 3/2 & 5/2 \\ 3/2 & 4 & 4 \\ 5/2 & 4 & 8 \end{bmatrix} + \begin{bmatrix} 0 & 1/2 & 9/2 \\ -1/2 & 0 & -1 \\ -9/2 & 1 & 0 \end{bmatrix} \)
\( \implies X + Y = \begin{bmatrix} 3+0 & 3/2+1/2 & 5/2+9/2 \\ 3/2-1/2 & 4+0 & 4-1 \\ 5/2-9/2 & 4+1 & 8+0 \end{bmatrix} \)
\( \implies X + Y = \begin{bmatrix} 3 & 4/2 & 14/2 \\ 2/2 & 4 & 3 \\ -4/2 & 5 & 8 \end{bmatrix} \)
\( \implies X + Y = \begin{bmatrix} 3 & 2 & 7 \\ 1 & 4 & 3 \\ -2 & 5 & 8 \end{bmatrix} = A \) So, we have found X and Y such that X is symmetric, Y is skew-symmetric, and their sum is A. This method helps analyze matrices by separating their 'even' and 'odd' components.In simple words: We took the matrix A and split it into two special parts: X and Y. X is a symmetric matrix, which means it looks the same even if you flip its rows and columns. Y is a skew-symmetric matrix, meaning it becomes its negative when you flip its rows and columns. When you add X and Y back together, you get the original matrix A.

🎯 Exam Tip: Remember the formulas \( X = \frac{1}{2}(A + A') \) for the symmetric part and \( Y = \frac{1}{2}(A - A') \) for the skew-symmetric part. Always verify your result by checking if \( X' = X \), \( Y' = -Y \), and \( X+Y = A \).

Question 9. Express the following matrix as the sum of a symmetric and a skew symmetric matrix, and verify your result \( \begin{bmatrix} 3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2 \end{bmatrix} \)
Answer:Let the given matrix be \( A = \begin{bmatrix} 3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2 \end{bmatrix} \). Any square matrix A can be written as the sum of a symmetric matrix P and a skew-symmetric matrix Q, where \( P = \frac{1}{2}(A + A') \) and \( Q = \frac{1}{2}(A - A') \). First, find the transpose of A, which is \( A' \): \( A' = \begin{bmatrix} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \end{bmatrix} \) Now, calculate P, the symmetric part: \( A + A' = \begin{bmatrix} 3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2 \end{bmatrix} + \begin{bmatrix} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \end{bmatrix} \)
\( \implies A + A' = \begin{bmatrix} 3+3 & -2+3 & -4-1 \\ 3-2 & -2-2 & -5+1 \\ -1-4 & 1-5 & 2+2 \end{bmatrix} \)
\( \implies A + A' = \begin{bmatrix} 6 & 1 & -5 \\ 1 & -4 & -4 \\ -5 & -4 & 4 \end{bmatrix} \)
\( \implies P = \frac{1}{2}(A + A') = \frac{1}{2} \begin{bmatrix} 6 & 1 & -5 \\ 1 & -4 & -4 \\ -5 & -4 & 4 \end{bmatrix} \)
\( \implies P = \begin{bmatrix} 3 & 1/2 & -5/2 \\ 1/2 & -2 & -2 \\ -5/2 & -2 & 2 \end{bmatrix} \) To verify if P is symmetric, we check its transpose \( P' \): \( P' = \begin{bmatrix} 3 & 1/2 & -5/2 \\ 1/2 & -2 & -2 \\ -5/2 & -2 & 2 \end{bmatrix} \). Since \( P' = P \), P is a symmetric matrix. Next, calculate Q, the skew-symmetric part: \( A - A' = \begin{bmatrix} 3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2 \end{bmatrix} - \begin{bmatrix} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \end{bmatrix} \)
\( \implies A - A' = \begin{bmatrix} 3-3 & -2-3 & -4-(-1) \\ 3-(-2) & -2-(-2) & -5-1 \\ -1-(-4) & 1-(-5) & 2-2 \end{bmatrix} \)
\( \implies A - A' = \begin{bmatrix} 0 & -5 & -3 \\ 5 & 0 & -6 \\ 3 & 6 & 0 \end{bmatrix} \)
\( \implies Q = \frac{1}{2}(A - A') = \frac{1}{2} \begin{bmatrix} 0 & -5 & -3 \\ 5 & 0 & -6 \\ 3 & 6 & 0 \end{bmatrix} \)
\( \implies Q = \begin{bmatrix} 0 & -5/2 & -3/2 \\ 5/2 & 0 & -3 \\ 3/2 & 3 & 0 \end{bmatrix} \) To verify if Q is skew-symmetric, we check its transpose \( Q' \): \( Q' = \begin{bmatrix} 0 & 5/2 & 3/2 \\ -5/2 & 0 & 3 \\ -3/2 & -3 & 0 \end{bmatrix} \)
\( \implies -Q = - \begin{bmatrix} 0 & -5/2 & -3/2 \\ 5/2 & 0 & -3 \\ 3/2 & 3 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 5/2 & 3/2 \\ -5/2 & 0 & 3 \\ -3/2 & -3 & 0 \end{bmatrix} \) Since \( Q' = -Q \), Q is a skew-symmetric matrix. Finally, verify that \( P + Q = A \): \( P + Q = \begin{bmatrix} 3 & 1/2 & -5/2 \\ 1/2 & -2 & -2 \\ -5/2 & -2 & 2 \end{bmatrix} + \begin{bmatrix} 0 & -5/2 & -3/2 \\ 5/2 & 0 & -3 \\ 3/2 & 3 & 0 \end{bmatrix} \)
\( \implies P + Q = \begin{bmatrix} 3+0 & 1/2-5/2 & -5/2-3/2 \\ 1/2+5/2 & -2+0 & -2-3 \\ -5/2+3/2 & -2+3 & 2+0 \end{bmatrix} \)
\( \implies P + Q = \begin{bmatrix} 3 & -4/2 & -8/2 \\ 6/2 & -2 & -5 \\ -2/2 & 1 & 2 \end{bmatrix} \)
\( \implies P + Q = \begin{bmatrix} 3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2 \end{bmatrix} = A \) The matrix is expressed as the sum of a symmetric and a skew-symmetric matrix. This process shows that every square matrix can be seen as a combination of these two fundamental forms.In simple words: We took the given matrix and separated it into two parts. One part is 'symmetric', meaning its mirror image across the main diagonal is identical to itself. The other part is 'skew-symmetric', meaning its mirror image is the negative of itself. When you put these two pieces back together, you get the original matrix, which helps us understand its properties.

🎯 Exam Tip: Remember to clearly show all steps for finding both \( A' \) and then calculating \( P \) and \( Q \). Verification that \( P' = P \) and \( Q' = -Q \) is crucial for full marks. A final check that \( P+Q=A \) confirms your calculations.