GSEB Class 12 Maths Solutions Chapter 3 શ્રેણિક Exercise 3.3

GSEB Solutions Class 12 Maths Chapter 3 श्रेषिङ Ex 3.3

Gujarat Board Textbook Solutions Class 12 Maths Chapter 3 श्रेषिङ Ex 3.3

Question 1. નીચેના પૈકી પ્રત્યેક શ્રેણિકનો પરિવર્ત શ્રેણિક મેળવો :
(i) \(\left[\begin{array}{c} 5 \\ \frac{1}{2} \\ -1 \end{array}\right]\)
(ii) \(\left[\begin{array}{cc} 1 & -1 \\ 2 & 3 \end{array}\right]\)
(iii) \(\left[\begin{array}{ccc} -1 & 5 & 6 \\ \sqrt{3} & 5 & 6 \\ 2 & 3 & -1 \end{array}\right]\)
Answer:
(i) If matrix A is given by \(A = \left[\begin{array}{c} 5 \\ \frac{1}{2} \\ -1 \end{array}\right]\), then its transpose \(A'\) is found by changing rows to columns. So, \(A' = \left[5 \quad \frac{1}{2} \quad -1\right]\).
(ii) If matrix A is \(A = \left[\begin{array}{cc} 1 & -1 \\ 2 & 3 \end{array}\right]\), its transpose \(A'\) is obtained by swapping rows and columns. Thus, \(A' = \left[\begin{array}{cc} 1 & 2 \\ -1 & 3 \end{array}\right]\).
(iii) For matrix A defined as \(A = \left[\begin{array}{ccc} -1 & 5 & 6 \\ \sqrt{3} & 5 & 6 \\ 2 & 3 & -1 \end{array}\right]\), the transpose \(A'\) is calculated by converting its rows into columns. Therefore, \(A' = \left[\begin{array}{ccc} -1 & \sqrt{3} & 2 \\ 5 & 5 & 3 \\ 6 & 6 & -1 \end{array}\right]\).
In simple words: To find the transpose of a matrix, you just swap its rows and columns. This means the first row becomes the first column, the second row becomes the second column, and so on.

🎯 Exam Tip: Understanding how to find the transpose of a matrix is a fundamental skill. Pay attention to the dimensions: if a matrix is \(m \times n\), its transpose will be \(n \times m\).

Question 2. જો \(A = \left[\begin{array}{ccc} -1 & 2 & 3 \\ 5 & 7 & 9 \\ -2 & 1 & 1 \end{array}\right]\) અને \(B = \left[\begin{array}{ccc} -4 & 1 & -5 \\ 1 & 2 & 0 \\ 1 & 3 & 1 \end{array}\right]\) હોય, તો (i) \((A + B)' = A' + B'\) (ii) \((A – B)' = A' – B'\) ચકાસો.
Answer:
Given matrices are:
\(A = \left[\begin{array}{ccc} -1 & 2 & 3 \\ 5 & 7 & 9 \\ -2 & 1 & 1 \end{array}\right]\) and \(B = \left[\begin{array}{ccc} -4 & 1 & -5 \\ 1 & 2 & 0 \\ 1 & 3 & 1 \end{array}\right]\)
First, find the transposes of A and B:
\(A' = \left[\begin{array}{ccc} -1 & 5 & -2 \\ 2 & 7 & 1 \\ 3 & 9 & 1 \end{array}\right]\)
\(B' = \left[\begin{array}{ccc} -4 & 1 & 1 \\ 1 & 2 & 3 \\ -5 & 0 & 1 \end{array}\right]\)

(i) Verification of \((A + B)' = A' + B'\):
First, calculate \(A + B\):
\(A + B = \left[\begin{array}{ccc} -1 & 2 & 3 \\ 5 & 7 & 9 \\ -2 & 1 & 1 \end{array}\right] + \left[\begin{array}{ccc} -4 & 1 & -5 \\ 1 & 2 & 0 \\ 1 & 3 & 1 \end{array}\right] = \left[\begin{array}{ccc} -1 + (-4) & 2 + 1 & 3 + (-5) \\ 5 + 1 & 7 + 2 & 9 + 0 \\ -2 + 1 & 1 + 3 & 1 + 1 \end{array}\right] = \left[\begin{array}{ccc} -5 & 3 & -2 \\ 6 & 9 & 9 \\ -1 & 4 & 2 \end{array}\right]\)
Now, find \((A + B)'\):
\((A + B)' = \left[\begin{array}{ccc} -5 & 6 & -1 \\ 3 & 9 & 4 \\ -2 & 9 & 2 \end{array}\right]\) .........(i)

Next, calculate \(A' + B'\):
\(A' + B' = \left[\begin{array}{ccc} -1 & 5 & -2 \\ 2 & 7 & 1 \\ 3 & 9 & 1 \end{array}\right] + \left[\begin{array}{ccc} -4 & 1 & 1 \\ 1 & 2 & 3 \\ -5 & 0 & 1 \end{array}\right] = \left[\begin{array}{ccc} -1 + (-4) & 5 + 1 & -2 + 1 \\ 2 + 1 & 7 + 2 & 1 + 3 \\ 3 + (-5) & 9 + 0 & 1 + 1 \end{array}\right] = \left[\begin{array}{ccc} -5 & 6 & -1 \\ 3 & 9 & 4 \\ -2 & 9 & 2 \end{array}\right]\) .........(ii)
From (i) and (ii), we can see that \((A + B)' = A' + B'\).

(ii) Verification of \((A – B)' = A' – B'\):
First, calculate \(A - B\):
\(A - B = \left[\begin{array}{ccc} -1 & 2 & 3 \\ 5 & 7 & 9 \\ -2 & 1 & 1 \end{array}\right] - \left[\begin{array}{ccc} -4 & 1 & -5 \\ 1 & 2 & 0 \\ 1 & 3 & 1 \end{array}\right] = \left[\begin{array}{ccc} -1 - (-4) & 2 - 1 & 3 - (-5) \\ 5 - 1 & 7 - 2 & 9 - 0 \\ -2 - 1 & 1 - 3 & 1 - 1 \end{array}\right] = \left[\begin{array}{ccc} 3 & 1 & 8 \\ 4 & 5 & 9 \\ -3 & -2 & 0 \end{array}\right]\)
Now, find \((A - B)'\):
\((A - B)' = \left[\begin{array}{ccc} 3 & 4 & -3 \\ 1 & 5 & -2 \\ 8 & 9 & 0 \end{array}\right]\) .........(i)

Next, calculate \(A' - B'\):
\(A' - B' = \left[\begin{array}{ccc} -1 & 5 & -2 \\ 2 & 7 & 1 \\ 3 & 9 & 1 \end{array}\right] - \left[\begin{array}{ccc} -4 & 1 & 1 \\ 1 & 2 & 3 \\ -5 & 0 & 1 \end{array}\right] = \left[\begin{array}{ccc} -1 - (-4) & 5 - 1 & -2 - 1 \\ 2 - 1 & 7 - 2 & 1 - 3 \\ 3 - (-5) & 9 - 0 & 1 - 1 \end{array}\right] = \left[\begin{array}{ccc} 3 & 4 & -3 \\ 1 & 5 & -2 \\ 8 & 9 & 0 \end{array}\right]\) .........(ii)
From (i) and (ii), we can confirm that \((A – B)' = A' – B'\).
In simple words: This problem shows two important properties of matrix transposes. First, if you add two matrices and then find the transpose, it's the same as finding the transpose of each matrix and then adding them. Second, if you subtract two matrices and then find the transpose, it's the same as finding the transpose of each matrix and then subtracting them.

🎯 Exam Tip: These transpose properties, \((A + B)' = A' + B'\) and \((A - B)' = A' - B'\), are crucial for matrix algebra. Make sure to perform matrix addition/subtraction and transposition steps accurately.

Question 3. જો \(A' = \left[\begin{array}{cc} 3 & 4 \\ -1 & 2 \\ 0 & 1 \end{array}\right]\) અને \(B = \left[\begin{array}{ccc} -1 & 2 & 1 \\ 1 & 2 & 3 \end{array}\right]\) હોય, તો (i) \((A + B)' = A' + B'\) (ii) \((A – B)' = A' – B'\) ચકાસો.
Answer:
Given matrices:
\(A' = \left[\begin{array}{cc} 3 & 4 \\ -1 & 2 \\ 0 & 1 \end{array}\right]\) and \(B = \left[\begin{array}{ccc} -1 & 2 & 1 \\ 1 & 2 & 3 \end{array}\right]\)

First, find matrix A from \(A'\) and then find \(B'\) from B:
\(A = (A')' = \left[\begin{array}{ccc} 3 & -1 & 0 \\ 4 & 2 & 1 \end{array}\right]\)
\(B' = \left[\begin{array}{cc} -1 & 1 \\ 2 & 2 \\ 1 & 3 \end{array}\right]\)

(i) Verification of \((A + B)' = A' + B'\):
First, calculate \(A + B\):
\(A + B = \left[\begin{array}{ccc} 3 & -1 & 0 \\ 4 & 2 & 1 \end{array}\right] + \left[\begin{array}{ccc} -1 & 2 & 1 \\ 1 & 2 & 3 \end{array}\right] = \left[\begin{array}{ccc} 3 + (-1) & -1 + 2 & 0 + 1 \\ 4 + 1 & 2 + 2 & 1 + 3 \end{array}\right] = \left[\begin{array}{ccc} 2 & 1 & 1 \\ 5 & 4 & 4 \end{array}\right]\)
Now, find \((A + B)'\):
\((A + B)' = \left[\begin{array}{cc} 2 & 5 \\ 1 & 4 \\ 1 & 4 \end{array}\right]\) .........(i)

Next, calculate \(A' + B'\):
\(A' + B' = \left[\begin{array}{cc} 3 & 4 \\ -1 & 2 \\ 0 & 1 \end{array}\right] + \left[\begin{array}{cc} -1 & 1 \\ 2 & 2 \\ 1 & 3 \end{array}\right] = \left[\begin{array}{cc} 3 + (-1) & 4 + 1 \\ -1 + 2 & 2 + 2 \\ 0 + 1 & 1 + 3 \end{array}\right] = \left[\begin{array}{cc} 2 & 5 \\ 1 & 4 \\ 1 & 4 \end{array}\right]\) .........(ii)
From (i) and (ii), it is proven that \((A + B)' = A' + B'\).

(ii) Verification of \((A – B)' = A' – B'\):
First, calculate \(A - B\):
\(A - B = \left[\begin{array}{ccc} 3 & -1 & 0 \\ 4 & 2 & 1 \end{array}\right] - \left[\begin{array}{ccc} -1 & 2 & 1 \\ 1 & 2 & 3 \end{array}\right] = \left[\begin{array}{ccc} 3 - (-1) & -1 - 2 & 0 - 1 \\ 4 - 1 & 2 - 2 & 1 - 3 \end{array}\right] = \left[\begin{array}{ccc} 4 & -3 & -1 \\ 3 & 0 & -2 \end{array}\right]\)
Now, find \((A - B)'\):
\((A - B)' = \left[\begin{array}{cc} 4 & 3 \\ -3 & 0 \\ -1 & -2 \end{array}\right]\) .........(i)

Next, calculate \(A' - B'\):
\(A' - B' = \left[\begin{array}{cc} 3 & 4 \\ -1 & 2 \\ 0 & 1 \end{array}\right] - \left[\begin{array}{cc} -1 & 1 \\ 2 & 2 \\ 1 & 3 \end{array}\right] = \left[\begin{array}{cc} 3 - (-1) & 4 - 1 \\ -1 - 2 & 2 - 2 \\ 0 - 1 & 1 - 3 \end{array}\right] = \left[\begin{array}{cc} 4 & 3 \\ -3 & 0 \\ -1 & -2 \end{array}\right]\) .........(ii)
From (i) and (ii), it is confirmed that \((A – B)' = A' – B'\).
In simple words: This problem reconfirms the properties of transposes of sums and differences of matrices. Even if you start with the transpose of one matrix and the original of another, the rules still hold true: the transpose of a sum/difference is the sum/difference of the transposes.

🎯 Exam Tip: When given \(A'\), remember to find A by taking the transpose of \(A'\) before performing any operations with other matrices like B. Always ensure matrix dimensions are compatible for addition/subtraction.

Question 4. જો \(A' = \left[\begin{array}{cc} -2 & 3 \\ 1 & 2 \end{array}\right]\) અને \(B = \left[\begin{array}{cc} -1 & 0 \\ 1 & 2 \end{array}\right]\) હોય, તો \((A + 2B)'\) શોધો.
Answer:
Given:
\(A' = \left[\begin{array}{cc} -2 & 3 \\ 1 & 2 \end{array}\right]\) and \(B = \left[\begin{array}{cc} -1 & 0 \\ 1 & 2 \end{array}\right]\)

First, find matrix A from \(A'\):
\(A = (A')' = \left[\begin{array}{cc} -2 & 1 \\ 3 & 2 \end{array}\right]\)

Next, calculate \(2B\):
\(2B = 2 \left[\begin{array}{cc} -1 & 0 \\ 1 & 2 \end{array}\right] = \left[\begin{array}{cc} 2 \times -1 & 2 \times 0 \\ 2 \times 1 & 2 \times 2 \end{array}\right] = \left[\begin{array}{cc} -2 & 0 \\ 2 & 4 \end{array}\right]\)

Now, calculate \(A + 2B\):
\(A + 2B = \left[\begin{array}{cc} -2 & 1 \\ 3 & 2 \end{array}\right] + \left[\begin{array}{cc} -2 & 0 \\ 2 & 4 \end{array}\right] = \left[\begin{array}{cc} -2 + (-2) & 1 + 0 \\ 3 + 2 & 2 + 4 \end{array}\right] = \left[\begin{array}{cc} -4 & 1 \\ 5 & 6 \end{array}\right]\)

Finally, find \((A + 2B)'\):
\((A + 2B)' = \left[\begin{array}{cc} -4 & 5 \\ 1 & 6 \end{array}\right]\)
In simple words: To solve this, first get the original matrix A from its transpose. Then, multiply matrix B by 2. After that, add matrix A and the scaled matrix \(2B\). Finally, find the transpose of this new sum. This involves simple matrix operations like scaling, addition, and transposition.

🎯 Exam Tip: Remember that \((kA)' = kA'\) and \((A')' = A\). Ensure scalar multiplication is applied to all elements of the matrix. Accuracy in basic arithmetic is key.

Question 5. નીચે આપેલ A અને B માટે, ચકાસો કે \((AB)' = B'A'\) :
(i) \(A = \left[\begin{array}{c} 1 \\ -4 \\ 3 \end{array}\right], B = [-1 \quad 2 \quad 1]\)
(ii) \(A = \left[\begin{array}{l} 0 \\ 1 \\ 2 \end{array}\right], B = [1 \quad 5 \quad 7]\)
Answer:
(i) Given:
\(A = \left[\begin{array}{c} 1 \\ -4 \\ 3 \end{array}\right]\) and \(B = [-1 \quad 2 \quad 1]\)

First, find \(A'\) and \(B'\):
\(A' = [1 \quad -4 \quad 3]\)
\(B' = \left[\begin{array}{c} -1 \\ 2 \\ 1 \end{array}\right]\)

Now, calculate \(AB\):
\(AB = \left[\begin{array}{c} 1 \\ -4 \\ 3 \end{array}\right] [-1 \quad 2 \quad 1]\)
\(AB = \left[\begin{array}{ccc} 1 \times (-1) & 1 \times 2 & 1 \times 1 \\ -4 \times (-1) & -4 \times 2 & -4 \times 1 \\ 3 \times (-1) & 3 \times 2 & 3 \times 1 \end{array}\right] = \left[\begin{array}{ccc} -1 & 2 & 1 \\ 4 & -8 & -4 \\ -3 & 6 & 3 \end{array}\right]\)
Then, find \((AB)'\):
\((AB)' = \left[\begin{array}{ccc} -1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3 \end{array}\right]\) .........(i)

Next, calculate \(B'A'\):
\(B'A' = \left[\begin{array}{c} -1 \\ 2 \\ 1 \end{array}\right] [1 \quad -4 \quad 3]\)
\(B'A' = \left[\begin{array}{ccc} -1 \times 1 & -1 \times (-4) & -1 \times 3 \\ 2 \times 1 & 2 \times (-4) & 2 \times 3 \\ 1 \times 1 & 1 \times (-4) & 1 \times 3 \end{array}\right] = \left[\begin{array}{ccc} -1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3 \end{array}\right]\) .........(ii)
From (i) and (ii), it is verified that \((AB)' = B'A'\).

(ii) Given:
\(A = \left[\begin{array}{l} 0 \\ 1 \\ 2 \end{array}\right]\) and \(B = [1 \quad 5 \quad 7]\)

First, find \(A'\) and \(B'\):
\(A' = [0 \quad 1 \quad 2]\)
\(B' = \left[\begin{array}{c} 1 \\ 5 \\ 7 \end{array}\right]\)

Now, calculate \(AB\):
\(AB = \left[\begin{array}{l} 0 \\ 1 \\ 2 \end{array}\right] [1 \quad 5 \quad 7]\)
\(AB = \left[\begin{array}{ccc} 0 \times 1 & 0 \times 5 & 0 \times 7 \\ 1 \times 1 & 1 \times 5 & 1 \times 7 \\ 2 \times 1 & 2 \times 5 & 2 \times 7 \end{array}\right] = \left[\begin{array}{ccc} 0 & 0 & 0 \\ 1 & 5 & 7 \\ 2 & 10 & 14 \end{array}\right]\)
Then, find \((AB)'\):
\((AB)' = \left[\begin{array}{ccc} 0 & 1 & 2 \\ 0 & 5 & 10 \\ 0 & 7 & 14 \end{array}\right]\) .........(i)

Next, calculate \(B'A'\):
\(B'A' = \left[\begin{array}{c} 1 \\ 5 \\ 7 \end{array}\right] [0 \quad 1 \quad 2]\)
\(B'A' = \left[\begin{array}{ccc} 1 \times 0 & 1 \times 1 & 1 \times 2 \\ 5 \times 0 & 5 \times 1 & 5 \times 2 \\ 7 \times 0 & 7 \times 1 & 7 \times 2 \end{array}\right] = \left[\begin{array}{ccc} 0 & 1 & 2 \\ 0 & 5 & 10 \\ 0 & 7 & 14 \end{array}\right]\) .........(ii)
From (i) and (ii), it is verified that \((AB)' = B'A'\).
In simple words: This problem demonstrates a key property of matrix transposes for multiplication: the transpose of a product of two matrices (AB) is equal to the product of their transposes in reverse order (B'A'). You multiply the matrices, then transpose the result. Then, you transpose each matrix individually and multiply them in reverse order, and both results should match.

🎯 Exam Tip: The property \((AB)' = B'A'\) is known as the "reversal rule" for transposes and is extremely important. Be careful with matrix multiplication order; it is not commutative, meaning \(AB \neq BA\) in general.

Question 6.
(i) જો \(A = \left[\begin{array}{cc} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{array}\right]\) હોય, તો ચકાસો કે, \(A'A = I\).
(ii) જો \(A = \left[\begin{array}{cc} \sin \alpha & \cos \alpha \\ -\cos \alpha & \sin \alpha \end{array}\right]\) હોય, તો ચકાસો કે, \(A'A = I\).
Answer:
(i) Given matrix:
\(A = \left[\begin{array}{cc} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{array}\right]\)
First, find its transpose \(A'\):
\(A' = \left[\begin{array}{cc} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right]\)

Now, calculate \(A'A\):
\(A'A = \left[\begin{array}{cc} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right] \left[\begin{array}{cc} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{array}\right]\)
\(A'A = \left[\begin{array}{cc} \cos^2 \alpha + (-\sin \alpha)(-\sin \alpha) & \cos \alpha \sin \alpha + (-\sin \alpha)\cos \alpha \\ \sin \alpha \cos \alpha + \cos \alpha (-\sin \alpha) & \sin^2 \alpha + \cos^2 \alpha \end{array}\right]\)
\(A'A = \left[\begin{array}{cc} \cos^2 \alpha + \sin^2 \alpha & \cos \alpha \sin \alpha - \sin \alpha \cos \alpha \\ \sin \alpha \cos \alpha - \cos \alpha \sin \alpha & \sin^2 \alpha + \cos^2 \alpha \end{array}\right]\)
Using the identity \(\sin^2 \alpha + \cos^2 \alpha = 1\):
\(A'A = \left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right] = I\)
Thus, it is verified that \(A'A = I\).

(ii) Given matrix:
\(A = \left[\begin{array}{cc} \sin \alpha & \cos \alpha \\ -\cos \alpha & \sin \alpha \end{array}\right]\)
First, find its transpose \(A'\):
\(A' = \left[\begin{array}{cc} \sin \alpha & -\cos \alpha \\ \cos \alpha & \sin \alpha \end{array}\right]\)

Now, calculate \(A'A\):
\(A'A = \left[\begin{array}{cc} \sin \alpha & -\cos \alpha \\ \cos \alpha & \sin \alpha \end{array}\right] \left[\begin{array}{cc} \sin \alpha & \cos \alpha \\ -\cos \alpha & \sin \alpha \end{array}\right]\)
\(A'A = \left[\begin{array}{cc} \sin^2 \alpha + (-\cos \alpha)(-\cos \alpha) & \sin \alpha \cos \alpha + (-\cos \alpha)\sin \alpha \\ \cos \alpha \sin \alpha + \sin \alpha (-\cos \alpha) & \cos^2 \alpha + \sin^2 \alpha \end{array}\right]\)
\(A'A = \left[\begin{array}{cc} \sin^2 \alpha + \cos^2 \alpha & \sin \alpha \cos \alpha - \cos \alpha \sin \alpha \\ \cos \alpha \sin \alpha - \sin \alpha \cos \alpha & \cos^2 \alpha + \sin^2 \alpha \end{array}\right]\)
Using the identity \(\sin^2 \alpha + \cos^2 \alpha = 1\):
\(A'A = \left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right] = I\)
Thus, it is verified that \(A'A = I\).
In simple words: This problem asks us to check if multiplying a matrix by its transpose results in an identity matrix. For both given matrices, we first find their transposes. Then, we perform matrix multiplication \(A'A\). By using the trigonometric identity \(\sin^2 \alpha + \cos^2 \alpha = 1\), we show that the resulting matrix is indeed the identity matrix. This means these matrices are orthogonal matrices.

🎯 Exam Tip: Recognizing that these matrices are orthogonal (where \(A'A = I\)) is important. Remember the trigonometric identity \(\sin^2 \alpha + \cos^2 \alpha = 1\) as it's frequently used in such problems. Careful calculation during matrix multiplication is crucial to avoid errors.

Question 7.
(i) સાબિત કરો કે શ્રેણિક \(A = \left[\begin{array}{ccc} 1 & -1 & 5 \\ -1 & 2 & 1 \\ 5 & 1 & 3 \end{array}\right]\) સંમિત શ્રેણિક છે.
(ii) સાબિત કરો કે શ્રેણિક \(A = \left[\begin{array}{ccc} 0 & 1 & -1 \\ -1 & 0 & 1 \\ 1 & -1 & 0 \end{array}\right]\) વિસંમિત શ્રેણિક છે.
Answer:
(i) To prove that matrix A is symmetric, we need to show that \(A' = A\).
Given matrix \(A = \left[\begin{array}{ccc} 1 & -1 & 5 \\ -1 & 2 & 1 \\ 5 & 1 & 3 \end{array}\right]\)
Find its transpose \(A'\):
\(A' = \left[\begin{array}{ccc} 1 & -1 & 5 \\ -1 & 2 & 1 \\ 5 & 1 & 3 \end{array}\right]\)
Since \(A' = A\), the matrix A is a symmetric matrix.

(ii) To prove that matrix A is skew-symmetric (વિસંમિત શ્રેણિક), we need to show that \(A' = -A\).
Given matrix \(A = \left[\begin{array}{ccc} 0 & 1 & -1 \\ -1 & 0 & 1 \\ 1 & -1 & 0 \end{array}\right]\)
Find its transpose \(A'\):
\(A' = \left[\begin{array}{ccc} 0 & -1 & 1 \\ 1 & 0 & -1 \\ -1 & 1 & 0 \end{array}\right]\)
Now, find \(-A\):
\(-A = -1 \times \left[\begin{array}{ccc} 0 & 1 & -1 \\ -1 & 0 & 1 \\ 1 & -1 & 0 \end{array}\right] = \left[\begin{array}{ccc} 0 & -1 & 1 \\ 1 & 0 & -1 \\ -1 & 1 & 0 \end{array}\right]\)
Since \(A' = -A\), the matrix A is a skew-symmetric matrix.
In simple words: A symmetric matrix is like a mirror image: if you flip it over its main diagonal (transpose it), it looks exactly the same. A skew-symmetric matrix is similar, but when you transpose it, all its numbers become their opposites (negative versions). The numbers on the main diagonal of a skew-symmetric matrix are always zero.

🎯 Exam Tip: Remember the definitions: A matrix A is symmetric if \(A' = A\), and skew-symmetric if \(A' = -A\). For skew-symmetric matrices, all diagonal elements must be zero. Clearly state the condition being checked for each type of matrix.

Question 8. શ્રેણિક \(A = \left[\begin{array}{ll} 1 & 5 \\ 6 & 7 \end{array}\right]\) માટે, ચકાસો કે
(i) \((A + A')\) સંમિત શ્રેણિક છે.
(ii) \((A – A')\) વિસંમિત શ્રેણિક છે.
Answer:
Given matrix \(A = \left[\begin{array}{ll} 1 & 5 \\ 6 & 7 \end{array}\right]\)
First, find its transpose \(A'\):
\(A' = \left[\begin{array}{ll} 1 & 6 \\ 5 & 7 \end{array}\right]\)

(i) To check if \((A + A')\) is a symmetric matrix, we need to show that \((A + A')' = (A + A')\).
Calculate \(A + A'\):
\(A + A' = \left[\begin{array}{ll} 1 & 5 \\ 6 & 7 \end{array}\right] + \left[\begin{array}{ll} 1 & 6 \\ 5 & 7 \end{array}\right] = \left[\begin{array}{cc} 1+1 & 5+6 \\ 6+5 & 7+7 \end{array}\right] = \left[\begin{array}{cc} 2 & 11 \\ 11 & 14 \end{array}\right]\)
Now, find the transpose of \((A + A')\):
\((A + A')' = \left[\begin{array}{cc} 2 & 11 \\ 11 & 14 \end{array}\right]\)
Since \((A + A')' = (A + A')\), it is verified that \((A + A')\) is a symmetric matrix.

(ii) To check if \((A – A')\) is a skew-symmetric matrix, we need to show that \((A – A')' = -(A – A')\).
Calculate \(A - A'\):
\(A - A' = \left[\begin{array}{ll} 1 & 5 \\ 6 & 7 \end{array}\right] - \left[\begin{array}{ll} 1 & 6 \\ 5 & 7 \end{array}\right] = \left[\begin{array}{cc} 1-1 & 5-6 \\ 6-5 & 7-7 \end{array}\right] = \left[\begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array}\right]\)
Now, find the transpose of \((A – A')\):
\((A – A')' = \left[\begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array}\right]\)
Next, find \(-(A – A')\):
\( -(A – A') = - \left[\begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array}\right] = \left[\begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array}\right]\)
Since \((A – A')' = -(A – A')\), it is verified that \((A – A')\) is a skew-symmetric matrix.
In simple words: This problem shows that for any square matrix, adding it to its transpose always gives a symmetric matrix. Also, subtracting its transpose from the original matrix always gives a skew-symmetric matrix. This is a general property for square matrices.

🎯 Exam Tip: Remember these fundamental properties: \(A+A'\) is always symmetric, and \(A-A'\) is always skew-symmetric for any square matrix A. These properties are often used to express any square matrix as the sum of a symmetric and a skew-symmetric matrix.

Question. જો શ્રેણિક \(A = \left[\begin{array}{ccc} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{array}\right]\) હોય, તો \(\frac{1}{2}(A + A')\) અને \(\frac{1}{2}(A – A')\) શોધો.
Answer:
Given matrix \(A = \left[\begin{array}{ccc} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{array}\right]\)
First, find its transpose \(A'\):
\(A' = \left[\begin{array}{ccc} 0 & -a & -b \\ a & 0 & -c \\ b & c & 0 \end{array}\right]\)

Now, calculate \(A + A'\):
\(A + A' = \left[\begin{array}{ccc} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{array}\right] + \left[\begin{array}{ccc} 0 & -a & -b \\ a & 0 & -c \\ b & c & 0 \end{array}\right]\)
\(A + A' = \left[\begin{array}{ccc} 0+0 & a-a & b-b \\ -a+a & 0+0 & c-c \\ -b+b & -c+c & 0+0 \end{array}\right] = \left[\begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right]\)
Then, calculate \(\frac{1}{2}(A + A')\):
\(\frac{1}{2}(A + A') = \frac{1}{2} \left[\begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] = \left[\begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right]\)

Next, calculate \(A - A'\):
\(A - A' = \left[\begin{array}{ccc} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{array}\right] - \left[\begin{array}{ccc} 0 & -a & -b \\ a & 0 & -c \\ b & c & 0 \end{array}\right]\)
\(A - A' = \left[\begin{array}{ccc} 0-0 & a-(-a) & b-(-b) \\ -a-a & 0-0 & c-(-c) \\ -b-b & -c-c & 0-0 \end{array}\right] = \left[\begin{array}{ccc} 0 & 2a & 2b \\ -2a & 0 & 2c \\ -2b & -2c & 0 \end{array}\right]\)
Then, calculate \(\frac{1}{2}(A – A')\):
\(\frac{1}{2}(A – A') = \frac{1}{2} \left[\begin{array}{ccc} 0 & 2a & 2b \\ -2a & 0 & 2c \\ -2b & -2c & 0 \end{array}\right] = \left[\begin{array}{ccc} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{array}\right]\)
In simple words: For the given matrix, which is already a skew-symmetric matrix, adding it to its transpose results in a zero matrix. Half of a zero matrix is still a zero matrix. Subtracting its transpose from the original matrix results in twice the original matrix. So, half of this difference just brings us back to the original matrix. This shows that a skew-symmetric matrix can be represented as \(0 + A\), where 0 is symmetric and A is skew-symmetric.

🎯 Exam Tip: Notice that for a skew-symmetric matrix A, \(A + A'\) will always be a zero matrix, and \(A - A'\) will be \(2A\). This problem highlights how skew-symmetric matrices behave under these operations.

Question 10. નીચેના પ્રત્યેક શ્રેણિકને એક સંમિત અને એક વિસંમિત શ્રેણિકના સરવાળા તરીકે અભિવ્યક્ત કરો :
(i) \(\left[\begin{array}{cc} 3 & 5 \\ 1 & -1 \end{array}\right]\)
(ii) \(\left[\begin{array}{ccc} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{array}\right]\)
(iii) \(\left[\begin{array}{ccc} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \end{array}\right]\)
(iv) \(\left[\begin{array}{cc} 1 & 5 \\ -1 & 2 \end{array}\right]\)
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 \(Q = \frac{1}{2}(A – A')\).
So, \(A = P + Q\).

(i) Given matrix \(A = \left[\begin{array}{cc} 3 & 5 \\ 1 & -1 \end{array}\right]\)
Find its transpose \(A'\): \(A' = \left[\begin{array}{cc} 3 & 1 \\ 5 & -1 \end{array}\right]\)

Calculate \(A + A'\):
\(A + A' = \left[\begin{array}{cc} 3 & 5 \\ 1 & -1 \end{array}\right] + \left[\begin{array}{cc} 3 & 1 \\ 5 & -1 \end{array}\right] = \left[\begin{array}{cc} 3+3 & 5+1 \\ 1+5 & -1-1 \end{array}\right] = \left[\begin{array}{cc} 6 & 6 \\ 6 & -2 \end{array}\right]\)
Calculate \(P = \frac{1}{2}(A + A')\):
\(P = \frac{1}{2} \left[\begin{array}{cc} 6 & 6 \\ 6 & -2 \end{array}\right] = \left[\begin{array}{cc} 3 & 3 \\ 3 & -1 \end{array}\right]\) (This is a symmetric matrix)

Calculate \(A - A'\):
\(A - A' = \left[\begin{array}{cc} 3 & 5 \\ 1 & -1 \end{array}\right] - \left[\begin{array}{cc} 3 & 1 \\ 5 & -1 \end{array}\right] = \left[\begin{array}{cc} 3-3 & 5-1 \\ 1-5 & -1-(-1) \end{array}\right] = \left[\begin{array}{cc} 0 & 4 \\ -4 & 0 \end{array}\right]\)
Calculate \(Q = \frac{1}{2}(A – A')\):
\(Q = \frac{1}{2} \left[\begin{array}{cc} 0 & 4 \\ -4 & 0 \end{array}\right] = \left[\begin{array}{cc} 0 & 2 \\ -2 & 0 \end{array}\right]\) (This is a skew-symmetric matrix)

Thus, \(A = P + Q = \left[\begin{array}{cc} 3 & 3 \\ 3 & -1 \end{array}\right] + \left[\begin{array}{cc} 0 & 2 \\ -2 & 0 \end{array}\right]\)

(ii) Given matrix \(A = \left[\begin{array}{ccc} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{array}\right]\)
Find its transpose \(A'\): \(A' = \left[\begin{array}{ccc} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{array}\right]\)

Calculate \(A + A'\):
\(A + A' = \left[\begin{array}{ccc} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{array}\right] + \left[\begin{array}{ccc} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{array}\right] = \left[\begin{array}{ccc} 12 & -4 & 4 \\ -4 & 6 & -2 \\ 4 & -2 & 6 \end{array}\right]\)
Calculate \(P = \frac{1}{2}(A + A')\):
\(P = \frac{1}{2} \left[\begin{array}{ccc} 12 & -4 & 4 \\ -4 & 6 & -2 \\ 4 & -2 & 6 \end{array}\right] = \left[\begin{array}{ccc} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{array}\right]\) (This is a symmetric matrix, which is A itself)

Calculate \(A - A'\):
\(A - A' = \left[\begin{array}{ccc} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{array}\right] - \left[\begin{array}{ccc} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{array}\right] = \left[\begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right]\)
Calculate \(Q = \frac{1}{2}(A – A')\):
\(Q = \frac{1}{2} \left[\begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] = \left[\begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right]\) (This is a skew-symmetric matrix)

Thus, \(A = P + Q = \left[\begin{array}{ccc} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{array}\right] + \left[\begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right]\)
*Note: The given matrix A is already symmetric, so \(A = A'\). This means \(Q = \frac{1}{2}(A - A') = \frac{1}{2}(A - A) = \frac{1}{2}(0) = 0\), and \(P = \frac{1}{2}(A + A') = \frac{1}{2}(A + A) = \frac{1}{2}(2A) = A\).*

(iii) Given matrix \(A = \left[\begin{array}{ccc} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \end{array}\right]\)
Find its transpose \(A'\): \(A' = \left[\begin{array}{ccc} 3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2 \end{array}\right]\)

Calculate \(A + A'\):
\(A + A' = \left[\begin{array}{ccc} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \end{array}\right] + \left[\begin{array}{ccc} 3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2 \end{array}\right] = \left[\begin{array}{ccc} 6 & 1 & -5 \\ 1 & -4 & -4 \\ -5 & -4 & 4 \end{array}\right]\)
Calculate \(P = \frac{1}{2}(A + A')\):
\(P = \frac{1}{2} \left[\begin{array}{ccc} 6 & 1 & -5 \\ 1 & -4 & -4 \\ -5 & -4 & 4 \end{array}\right] = \left[\begin{array}{ccc} 3 & 1/2 & -5/2 \\ 1/2 & -2 & -2 \\ -5/2 & -2 & 2 \end{array}\right]\) (This is a symmetric matrix)

Calculate \(A - A'\):
\(A - A' = \left[\begin{array}{ccc} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \end{array}\right] - \left[\begin{array}{ccc} 3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2 \end{array}\right] = \left[\begin{array}{ccc} 0 & 5 & 3 \\ -5 & 0 & 6 \\ -3 & -6 & 0 \end{array}\right]\)
Calculate \(Q = \frac{1}{2}(A – A')\):
\(Q = \frac{1}{2} \left[\begin{array}{ccc} 0 & 5 & 3 \\ -5 & 0 & 6 \\ -3 & -6 & 0 \end{array}\right] = \left[\begin{array}{ccc} 0 & 5/2 & 3/2 \\ -5/2 & 0 & 3 \\ -3/2 & -3 & 0 \end{array}\right]\) (This is a skew-symmetric matrix)

Thus, \(A = P + Q = \left[\begin{array}{ccc} 3 & 1/2 & -5/2 \\ 1/2 & -2 & -2 \\ -5/2 & -2 & 2 \end{array}\right] + \left[\begin{array}{ccc} 0 & 5/2 & 3/2 \\ -5/2 & 0 & 3 \\ -3/2 & -3 & 0 \end{array}\right]\)

(iv) Given matrix \(A = \left[\begin{array}{cc} 1 & 5 \\ -1 & 2 \end{array}\right]\)
Find its transpose \(A'\): \(A' = \left[\begin{array}{cc} 1 & -1 \\ 5 & 2 \end{array}\right]\)

Calculate \(A + A'\):
\(A + A' = \left[\begin{array}{cc} 1 & 5 \\ -1 & 2 \end{array}\right] + \left[\begin{array}{cc} 1 & -1 \\ 5 & 2 \end{array}\right] = \left[\begin{array}{cc} 1+1 & 5-1 \\ -1+5 & 2+2 \end{array}\right] = \left[\begin{array}{cc} 2 & 4 \\ 4 & 4 \end{array}\right]\)
Calculate \(P = \frac{1}{2}(A + A')\):
\(P = \frac{1}{2} \left[\begin{array}{cc} 2 & 4 \\ 4 & 4 \end{array}\right] = \left[\begin{array}{cc} 1 & 2 \\ 2 & 2 \end{array}\right]\) (This is a symmetric matrix)

Calculate \(A - A'\):
\(A - A' = \left[\begin{array}{cc} 1 & 5 \\ -1 & 2 \end{array}\right] - \left[\begin{array}{cc} 1 & -1 \\ 5 & 2 \end{array}\right] = \left[\begin{array}{cc} 1-1 & 5-(-1) \\ -1-5 & 2-2 \end{array}\right] = \left[\begin{array}{cc} 0 & 6 \\ -6 & 0 \end{array}\right]\)
Calculate \(Q = \frac{1}{2}(A – A')\):
\(Q = \frac{1}{2} \left[\begin{array}{cc} 0 & 6 \\ -6 & 0 \end{array}\right] = \left[\begin{array}{cc} 0 & 3 \\ -3 & 0 \end{array}\right]\) (This is a skew-symmetric matrix)

Thus, \(A = P + Q = \left[\begin{array}{cc} 1 & 2 \\ 2 & 2 \end{array}\right] + \left[\begin{array}{cc} 0 & 3 \\ -3 & 0 \end{array}\right]\)
In simple words: This problem shows how to break down any square matrix into two special parts: one symmetric matrix and one skew-symmetric matrix. You find the symmetric part by adding the matrix to its transpose and dividing by two. You find the skew-symmetric part by subtracting the transpose from the matrix and dividing by two. When you add these two parts back together, you get the original matrix.

🎯 Exam Tip: This method is universal for expressing any square matrix as a sum of a symmetric and a skew-symmetric matrix. Ensure accurate calculation of \(A+A'\) and \(A-A'\), and then correct scalar multiplication by \(\frac{1}{2}\). Double-check that P is symmetric and Q is skew-symmetric by finding their transposes.

પ્રશ્ના 11 તથા 12 માં વિધાન સાચું બને તે રીતે આપેલ વિકલ્પોમાંથી યોગ્ય વિકલ્પ પસંદ કરો :

Question 11. જો A અને B સમાન ક્ષાવાળા સંમિત શ્રેણિક હોય, તો AB – BA એ.....
(A) વિસંમિત શ્રેણિક છે.
(B) સંમિત શ્રેણિક છે.
(C) શૂન્ય શ્રેણિક છે.
(D) એમ શ્રેણિક છે.
Answer: (A) વિસંમિત શ્રેણિક છે.
Explanation:
Given that A and B are symmetric matrices of the same order. This means \(A' = A\) and \(B' = B\).
We need to determine the nature of \((AB - BA)\). Let's find its transpose:
\((AB - BA)' = (AB)' - (BA)'\) (Property of transpose of difference)
\((AB - BA)' = B'A' - A'B'\) (Reversal rule for transpose of product)

Since A and B are symmetric, \(A' = A\) and \(B' = B\). Substitute these into the expression:
\((AB - BA)' = BA - AB\)
\((AB - BA)' = -(AB - BA)\)
Since the transpose of \((AB - BA)\) is equal to its negative, \((AB - BA)\) is a skew-symmetric matrix (વિસંમિત શ્રેણિક).
In simple words: If you have two symmetric matrices (meaning they are the same even after swapping rows and columns), then the matrix you get by doing \(AB - BA\) will always be a skew-symmetric matrix. This means if you transpose \(AB - BA\), you get the negative of the original matrix.

🎯 Exam Tip: This is a standard property of symmetric matrices. Remember the transpose properties: \((X-Y)' = X' - Y'\) and \((XY)' = Y'X'\). The substitution \(A'=A\) and \(B'=B\) is critical for solving this type of MCQ.

Question 12. જો \(\alpha\) નું મૂલ્ય ...... હોય, તો \(A + A' = I\) થાય, જ્યાં \(A = \left[\begin{array}{cc} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right]\)
(B) \(\frac{\pi}{3}\)
Answer: (B) \(\frac{\pi}{3}\)
Explanation:
Given matrix \(A = \left[\begin{array}{cc} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right]\)
First, find its transpose \(A'\):
\(A' = \left[\begin{array}{cc} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{array}\right]\)

Now, calculate \(A + A'\):
\(A + A' = \left[\begin{array}{cc} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right] + \left[\begin{array}{cc} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{array}\right]\)
\(A + A' = \left[\begin{array}{cc} \cos \alpha + \cos \alpha & -\sin \alpha + \sin \alpha \\ \sin \alpha - \sin \alpha & \cos \alpha + \cos \alpha \end{array}\right] = \left[\begin{array}{cc} 2\cos \alpha & 0 \\ 0 & 2\cos \alpha \end{array}\right]\)

We are given that \(A + A' = I\), where I is the identity matrix \(I = \left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right]\).
So, \(\left[\begin{array}{cc} 2\cos \alpha & 0 \\ 0 & 2\cos \alpha \end{array}\right] = \left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right]\)
Comparing the corresponding elements, we get:
\(2\cos \alpha = 1\)
\(\cos \alpha = \frac{1}{2}\)

For the principal value, \(\alpha = \cos^{-1}\left(\frac{1}{2}\right)\)
\(\alpha = \frac{\pi}{3}\)
In simple words: We are given a matrix with sine and cosine terms and told that when we add it to its transpose, we get the identity matrix. By performing the addition and comparing the result to the identity matrix, we find an equation involving \(\cos \alpha\). Solving this equation gives us the value of \(\alpha\) as \(\frac{\pi}{3}\).

🎯 Exam Tip: This problem tests your understanding of matrix addition, transposition, and basic trigonometric values. The identity matrix plays a key role here. Make sure to recall common angles for trigonometric ratios.

Question 12. જો \(\alpha\) નું મૂલ્ય ...... હોય, તો \(A + A' = I\) થાય, જ્યાં \(A = \left[\begin{array}{cc} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right]\)
Answer: (B) \(\frac{\pi}{3}\)
To solve this, we first determine the transpose of matrix \(A\), denoted as \(A'\). Then, we add matrix \(A\) and its transpose \(A'\). This sum results in a new matrix where the main diagonal elements are \(2\cos \alpha\) and the off-diagonal elements are \(0\). The problem states that this sum is equal to the identity matrix \(I\). By comparing the elements of the sum matrix with the identity matrix, we deduce that \(2\cos \alpha\) must be \(1\). This implies that \(\cos \alpha = \frac{1}{2}\). The value of \(\alpha\) that satisfies \(\cos \alpha = \frac{1}{2}\) is \(\frac{\pi}{3}\).
In simple words: To find the answer, we take matrix A and flip it to get A-transpose (A'). We then add A and A'. This new matrix should be the same as the identity matrix, which has 1s on its main line and 0s everywhere else. By matching these matrices, we find that 2 times cos(alpha) equals 1, meaning cos(alpha) is 1/2. The angle alpha that makes cos(alpha) equal to 1/2 is \(\frac{\pi}{3}\).

🎯 Exam Tip: When working with matrix transposes and identity matrices, clearly show how you find the transpose and perform matrix operations. Remember to equate corresponding elements of matrices to solve for unknown variables like \(\alpha\).