GSEB Class 12 Maths Solutions Chapter 3 શ્રેણિક Exercise 3.2

GSEB Solutions Class 12 Maths Chapter 3 શ્રેણિક Ex 3.2

Question 1. A = \(\left[\begin{array}{ll} 2 & 4 \\ 3 & 2 \end{array}\right]\), B = \(\left[\begin{array}{cc} 1 & 3 \\ -2 & 5 \end{array}\right]\), C = \(\left[\begin{array}{cc} -2 & 5 \\ 3 & 4 \end{array}\right]\) હોય, તો નીચેના પૈકી પ્રત્યેક શ્રેણિક શોધો :
(i) A + B
(ii) A - B
(iii) 3A - C
(iv) AB
(v) BA
Answer:
(i) To find A + B, we add the corresponding elements of matrix A and matrix B.
\[ A + B = \left[\begin{array}{ll} 2 & 4 \\ 3 & 2 \end{array}\right] + \left[\begin{array}{cc} 1 & 3 \\ -2 & 5 \end{array}\right] \] \[ = \left[\begin{array}{cc} 2+1 & 4+3 \\ 3+(-2) & 2+5 \end{array}\right] \] \[ = \left[\begin{array}{cc} 3 & 7 \\ 1 & 7 \end{array}\right] \]
(ii) To find A - B, we subtract the corresponding elements of matrix B from matrix A.
\[ A - B = \left[\begin{array}{ll} 2 & 4 \\ 3 & 2 \end{array}\right] - \left[\begin{array}{cc} 1 & 3 \\ -2 & 5 \end{array}\right] \] \[ = \left[\begin{array}{cc} 2-1 & 4-3 \\ 3-(-2) & 2-5 \end{array}\right] \] \[ = \left[\begin{array}{cc} 1 & 1 \\ 5 & -3 \end{array}\right] \]
(iii) First, multiply matrix A by the scalar 3. Then, subtract matrix C from the result.
\[ 3A - C = 3\left[\begin{array}{ll} 2 & 4 \\ 3 & 2 \end{array}\right] - \left[\begin{array}{cc} -2 & 5 \\ 3 & 4 \end{array}\right] \] \[ = \left[\begin{array}{cc} 3 \times 2 & 3 \times 4 \\ 3 \times 3 & 3 \times 2 \end{array}\right] - \left[\begin{array}{cc} -2 & 5 \\ 3 & 4 \end{array}\right] \] \[ = \left[\begin{array}{cc} 6 & 12 \\ 9 & 6 \end{array}\right] - \left[\begin{array}{cc} -2 & 5 \\ 3 & 4 \end{array}\right] \] \[ = \left[\begin{array}{cc} 6-(-2) & 12-5 \\ 9-3 & 6-4 \end{array}\right] \] \[ = \left[\begin{array}{cc} 8 & 7 \\ 6 & 2 \end{array}\right] \]
(iv) To find the product AB, we multiply the rows of matrix A by the columns of matrix B.
\[ AB = \left[\begin{array}{ll} 2 & 4 \\ 3 & 2 \end{array}\right] \left[\begin{array}{cc} 1 & 3 \\ -2 & 5 \end{array}\right] \] \[ = \left[\begin{array}{cc} (2)(1)+(4)(-2) & (2)(3)+(4)(5) \\ (3)(1)+(2)(-2) & (3)(3)+(2)(5) \end{array}\right] \] \[ = \left[\begin{array}{cc} 2-8 & 6+20 \\ 3-4 & 9+10 \end{array}\right] \] \[ = \left[\begin{array}{cc} -6 & 26 \\ -1 & 19 \end{array}\right] \]
(v) To find the product BA, we multiply the rows of matrix B by the columns of matrix A.
\[ BA = \left[\begin{array}{cc} 1 & 3 \\ -2 & 5 \end{array}\right] \left[\begin{array}{ll} 2 & 4 \\ 3 & 2 \end{array}\right] \] \[ = \left[\begin{array}{cc} (1)(2)+(3)(3) & (1)(4)+(3)(2) \\ (-2)(2)+(5)(3) & (-2)(4)+(5)(2) \end{array}\right] \] \[ = \left[\begin{array}{cc} 2+9 & 4+6 \\ -4+15 & -8+10 \end{array}\right] \] \[ = \left[\begin{array}{cc} 11 & 10 \\ 11 & 2 \end{array}\right] \]In simple words: This question showed how to do basic math with matrices, like adding, subtracting, multiplying by a number, and multiplying two matrices together. Each operation combines elements in a specific way to get a new matrix.

🎯 Exam Tip: Pay close attention to the order of matrix multiplication (AB is not always equal to BA) and remember to perform scalar multiplication before addition or subtraction for accuracy.

Question 2. નીચેનાની ગણતરી કરો :
(i) \(\left[\begin{array}{cc} a & b \\ -b & a \end{array}\right] + \left[\begin{array}{cc} a & b \\ b & a \end{array}\right]\)
(ii) \(\left[\begin{array}{cc} a^2+b^2 & b^2+c^2 \\ a^2+c^2 & a^2+b^2 \end{array}\right] + \left[\begin{array}{cc} 2ab & 2bc \\ -2ac & -2ab \end{array}\right]\)
(iii) \(\left[\begin{array}{ccc} -1 & 4 & -6 \\ 8 & 5 & 16 \\ 2 & 8 & 5 \end{array}\right] + \left[\begin{array}{ccc} 12 & 7 & 6 \\ 8 & 0 & 5 \\ 3 & 2 & 4 \end{array}\right]\)
(iv) \(\left[\begin{array}{cc} \cos^2 x & \sin^2 x \\ \sin^2 x & \cos^2 x \end{array}\right] + \left[\begin{array}{cc} \sin^2 x & \cos^2 x \\ \cos^2 x & \sin^2 x \end{array}\right]\)
Answer:
(i) To add these two matrices, combine the elements that are in the same position in each matrix.
\[ \left[\begin{array}{cc} a & b \\ -b & a \end{array}\right] + \left[\begin{array}{cc} a & b \\ b & a \end{array}\right] = \left[\begin{array}{cc} a+a & b+b \\ -b+b & a+a \end{array}\right] \] \[ = \left[\begin{array}{cc} 2a & 2b \\ 0 & 2a \end{array}\right] \]
(ii) Add the corresponding elements of the matrices. Use algebraic identities to simplify the results.
\[ \left[\begin{array}{cc} a^2+b^2 & b^2+c^2 \\ a^2+c^2 & a^2+b^2 \end{array}\right] + \left[\begin{array}{cc} 2ab & 2bc \\ -2ac & -2ab \end{array}\right] \] \[ = \left[\begin{array}{cc} a^2+b^2+2ab & b^2+c^2+2bc \\ a^2+c^2-2ac & a^2+b^2-2ab \end{array}\right] \] \[ = \left[\begin{array}{cc} (a+b)^2 & (b+c)^2 \\ (a-c)^2 & (a-b)^2 \end{array}\right] \]
(iii) Sum the elements at the identical positions in both matrices.
\[ \left[\begin{array}{ccc} -1 & 4 & -6 \\ 8 & 5 & 16 \\ 2 & 8 & 5 \end{array}\right] + \left[\begin{array}{ccc} 12 & 7 & 6 \\ 8 & 0 & 5 \\ 3 & 2 & 4 \end{array}\right] \] \[ = \left[\begin{array}{ccc} -1+12 & 4+7 & -6+6 \\ 8+8 & 5+0 & 16+5 \\ 2+3 & 8+2 & 5+4 \end{array}\right] \] \[ = \left[\begin{array}{ccc} 11 & 11 & 0 \\ 16 & 5 & 21 \\ 5 & 10 & 9 \end{array}\right] \]
(iv) Add the corresponding elements and use the trigonometric identity \( \cos^2 x + \sin^2 x = 1 \).
\[ \left[\begin{array}{cc} \cos^2 x & \sin^2 x \\ \sin^2 x & \cos^2 x \end{array}\right] + \left[\begin{array}{cc} \sin^2 x & \cos^2 x \\ \cos^2 x & \sin^2 x \end{array}\right] \] \[ = \left[\begin{array}{cc} \cos^2 x + \sin^2 x & \sin^2 x + \cos^2 x \\ \sin^2 x + \cos^2 x & \cos^2 x + \sin^2 x \end{array}\right] \] \[ = \left[\begin{array}{cc} 1 & 1 \\ 1 & 1 \end{array}\right] \]In simple words: This question shows how to add matrices by adding elements in the same spot. It also reminds us to simplify expressions using math rules like \((a+b)^2\) or \(\sin^2x + \cos^2x = 1\).

🎯 Exam Tip: Remember to combine elements in the *exact* corresponding positions when adding matrices. Also, keep an eye out for algebraic or trigonometric identities that can simplify the resulting matrix elements.

Question 3. સૂચિત ગુણાકારની ગણતરી કરો :
(i) \(\left[\begin{array}{cc} a & b \\ -b & a \end{array}\right] \left[\begin{array}{cc} a & -b \\ b & a \end{array}\right]\)
(ii) \(\left[\begin{array}{c} 1 \\ 2 \\ 3 \end{array}\right] \left[\begin{array}{ccc} 2 & 3 & 4 \end{array}\right]\)
(iii) \(\left[\begin{array}{cc} 1 & -2 \\ 2 & 3 \end{array}\right] \left[\begin{array}{ccc} 1 & 2 & 3 \\ 2 & 3 & 1 \end{array}\right]\)
(iv) \(\left[\begin{array}{ccc} 2 & 3 & 4 \\ 3 & 4 & 5 \\ 4 & 5 & 6 \end{array}\right] \left[\begin{array}{ccc} 1 & -3 & 5 \\ 0 & 2 & 4 \\ 3 & 0 & 5 \end{array}\right]\)
(v) \(\left[\begin{array}{cc} 2 & 1 \\ 3 & 2 \\ -1 & 1 \end{array}\right] \left[\begin{array}{ccc} 1 & 0 & 1 \\ -1 & 2 & 1 \end{array}\right]\)
(vi) \(\left[\begin{array}{ccc} 3 & -1 & 3 \\ -1 & 0 & 2 \end{array}\right] \left[\begin{array}{cc} 2 & -3 \\ 1 & 0 \\ 3 & 1 \end{array}\right]\)
Answer:
(i) Multiply the rows of the first matrix by the columns of the second matrix.
\[ \left[\begin{array}{cc} a & b \\ -b & a \end{array}\right] \left[\begin{array}{cc} a & -b \\ b & a \end{array}\right] \] \[ = \left[\begin{array}{cc} (a)(a)+(b)(b) & (a)(-b)+(b)(a) \\ (-b)(a)+(a)(b) & (-b)(-b)+(a)(a) \end{array}\right] \] \[ = \left[\begin{array}{cc} a^2+b^2 & -ab+ab \\ -ab+ab & b^2+a^2 \end{array}\right] \] \[ = \left[\begin{array}{cc} a^2+b^2 & 0 \\ 0 & a^2+b^2 \end{array}\right] \]
(ii) Multiply the elements of the first matrix by the elements of the second matrix to form a new matrix.
\[ \left[\begin{array}{c} 1 \\ 2 \\ 3 \end{array}\right] \left[\begin{array}{ccc} 2 & 3 & 4 \end{array}\right] \] \[ = \left[\begin{array}{ccc} (1)(2) & (1)(3) & (1)(4) \\ (2)(2) & (2)(3) & (2)(4) \\ (3)(2) & (3)(3) & (3)(4) \end{array}\right] \] \[ = \left[\begin{array}{ccc} 2 & 3 & 4 \\ 4 & 6 & 8 \\ 6 & 9 & 12 \end{array}\right] \]
(iii) Multiply the rows of the first matrix by the columns of the second matrix.
\[ \left[\begin{array}{cc} 1 & -2 \\ 2 & 3 \end{array}\right] \left[\begin{array}{ccc} 1 & 2 & 3 \\ 2 & 3 & 1 \end{array}\right] \] \[ = \left[\begin{array}{ccc} (1)(1)+(-2)(2) & (1)(2)+(-2)(3) & (1)(3)+(-2)(1) \\ (2)(1)+(3)(2) & (2)(2)+(3)(3) & (2)(3)+(3)(1) \end{array}\right] \] \[ = \left[\begin{array}{ccc} 1-4 & 2-6 & 3-2 \\ 2+6 & 4+9 & 6+3 \end{array}\right] \] \[ = \left[\begin{array}{ccc} -3 & -4 & 1 \\ 8 & 13 & 9 \end{array}\right] \]
(iv) Multiply rows of the first matrix by columns of the second matrix.
\[ \left[\begin{array}{ccc} 2 & 3 & 4 \\ 3 & 4 & 5 \\ 4 & 5 & 6 \end{array}\right] \left[\begin{array}{ccc} 1 & -3 & 5 \\ 0 & 2 & 4 \\ 3 & 0 & 5 \end{array}\right] \] \[ = \left[\begin{array}{ccc} (2)(1)+(3)(0)+(4)(3) & (2)(-3)+(3)(2)+(4)(0) & (2)(5)+(3)(4)+(4)(5) \\ (3)(1)+(4)(0)+(5)(3) & (3)(-3)+(4)(2)+(5)(0) & (3)(5)+(4)(4)+(5)(5) \\ (4)(1)+(5)(0)+(6)(3) & (4)(-3)+(5)(2)+(6)(0) & (4)(5)+(5)(4)+(6)(5) \end{array}\right] \] \[ = \left[\begin{array}{ccc} 2+0+12 & -6+6+0 & 10+12+20 \\ 3+0+15 & -9+8+0 & 15+16+25 \\ 4+0+18 & -12+10+0 & 20+20+30 \end{array}\right] \] \[ = \left[\begin{array}{ccc} 14 & 0 & 42 \\ 18 & -1 & 56 \\ 22 & -2 & 70 \end{array}\right] \]
(v) Multiply rows by columns for the given matrices.
\[ \left[\begin{array}{cc} 2 & 1 \\ 3 & 2 \\ -1 & 1 \end{array}\right] \left[\begin{array}{ccc} 1 & 0 & 1 \\ -1 & 2 & 1 \end{array}\right] \] \[ = \left[\begin{array}{ccc} (2)(1)+(1)(-1) & (2)(0)+(1)(2) & (2)(1)+(1)(1) \\ (3)(1)+(2)(-1) & (3)(0)+(2)(2) & (3)(1)+(2)(1) \\ (-1)(1)+(1)(-1) & (-1)(0)+(1)(2) & (-1)(1)+(1)(1) \end{array}\right] \] \[ = \left[\begin{array}{ccc} 2-1 & 0+2 & 2+1 \\ 3-2 & 0+4 & 3+2 \\ -1-1 & 0+2 & -1+1 \end{array}\right] \] \[ = \left[\begin{array}{ccc} 1 & 2 & 3 \\ 1 & 4 & 5 \\ -2 & 2 & 0 \end{array}\right] \]
(vi) Multiply rows by columns for the given matrices.
\[ \left[\begin{array}{ccc} 3 & -1 & 3 \\ -1 & 0 & 2 \end{array}\right] \left[\begin{array}{cc} 2 & -3 \\ 1 & 0 \\ 3 & 1 \end{array}\right] \] \[ = \left[\begin{array}{cc} (3)(2)+(-1)(1)+(3)(3) & (3)(-3)+(-1)(0)+(3)(1) \\ (-1)(2)+(0)(1)+(2)(3) & (-1)(-3)+(0)(0)+(2)(1) \end{array}\right] \] \[ = \left[\begin{array}{cc} 6-1+9 & -9+0+3 \\ -2+0+6 & 3+0+2 \end{array}\right] \] \[ = \left[\begin{array}{cc} 14 & -6 \\ 4 & 5 \end{array}\right] \]In simple words: This question demonstrates how to multiply different types of matrices. Remember that for matrix multiplication, the number of columns in the first matrix must match the number of rows in the second matrix.

🎯 Exam Tip: Matrix multiplication involves a row-by-column operation. Be careful with signs and perform each multiplication and addition step accurately to avoid errors. Ensure the resulting matrix has the correct dimensions.

Question 4. ૪ A = \(\left[\begin{array}{ccc} 1 & 2 & -3 \\ 5 & 0 & 2 \\ 1 & -1 & 1 \end{array}\right]\), B = \(\left[\begin{array}{ccc} 3 & -1 & 2 \\ 4 & 2 & 5 \\ 2 & 0 & 3 \end{array}\right]\) ને C = \(\left[\begin{array}{ccc} 4 & 1 & 2 \\ 0 & 3 & 2 \\ 1 & -2 & 3 \end{array}\right]\) હોય, તો A + (B – C) = (A + B) – C.
Answer:
To verify the given matrix property, we first calculate A + B and then subtract C from the result to get the right-hand side. Next, we calculate B - C and then add A to this result to get the left-hand side. Both sides should yield the same matrix, proving the identity.
First, find A + B:
\[ A + B = \left[\begin{array}{ccc} 1 & 2 & -3 \\ 5 & 0 & 2 \\ 1 & -1 & 1 \end{array}\right] + \left[\begin{array}{ccc} 3 & -1 & 2 \\ 4 & 2 & 5 \\ 2 & 0 & 3 \end{array}\right] \] \[ = \left[\begin{array}{ccc} 1+3 & 2-1 & -3+2 \\ 5+4 & 0+2 & 2+5 \\ 1+2 & -1+0 & 1+3 \end{array}\right] \] \[ = \left[\begin{array}{ccc} 4 & 1 & -1 \\ 9 & 2 & 7 \\ 3 & -1 & 4 \end{array}\right] \] Now, calculate (A + B) - C:
\[ (A+B) - C = \left[\begin{array}{ccc} 4 & 1 & -1 \\ 9 & 2 & 7 \\ 3 & -1 & 4 \end{array}\right] - \left[\begin{array}{ccc} 4 & 1 & 2 \\ 0 & 3 & 2 \\ 1 & -2 & 3 \end{array}\right] \] \[ = \left[\begin{array}{ccc} 4-4 & 1-1 & -1-2 \\ 9-0 & 2-3 & 7-2 \\ 3-1 & -1-(-2) & 4-3 \end{array}\right] \] \[ = \left[\begin{array}{ccc} 0 & 0 & -3 \\ 9 & -1 & 5 \\ 2 & 1 & 1 \end{array}\right] \quad ...(ii) \] Next, find B - C:
\[ B - C = \left[\begin{array}{ccc} 3 & -1 & 2 \\ 4 & 2 & 5 \\ 2 & 0 & 3 \end{array}\right] - \left[\begin{array}{ccc} 4 & 1 & 2 \\ 0 & 3 & 2 \\ 1 & -2 & 3 \end{array}\right] \] \[ = \left[\begin{array}{ccc} 3-4 & -1-1 & 2-2 \\ 4-0 & 2-3 & 5-2 \\ 2-1 & 0-(-2) & 3-3 \end{array}\right] \] \[ = \left[\begin{array}{ccc} -1 & -2 & 0 \\ 4 & -1 & 3 \\ 1 & 2 & 0 \end{array}\right] \] Now, calculate A + (B - C):
\[ A + (B-C) = \left[\begin{array}{ccc} 1 & 2 & -3 \\ 5 & 0 & 2 \\ 1 & -1 & 1 \end{array}\right] + \left[\begin{array}{ccc} -1 & -2 & 0 \\ 4 & -1 & 3 \\ 1 & 2 & 0 \end{array}\right] \] \[ = \left[\begin{array}{ccc} 1+(-1) & 2+(-2) & -3+0 \\ 5+4 & 0+(-1) & 2+3 \\ 1+1 & -1+2 & 1+0 \end{array}\right] \] \[ = \left[\begin{array}{ccc} 0 & 0 & -3 \\ 9 & -1 & 5 \\ 2 & 1 & 1 \end{array}\right] \quad ...(i) \] From (i) and (ii), we see that both results are the same.
Therefore, A + (B - C) = (A + B) - C is verified.
In simple words: This problem showed us that when you add and subtract matrices, the way you group them with parentheses does not change the final answer. This is like how \( (2+3)-1 \) is the same as \( 2+(3-1) \) with numbers.

🎯 Exam Tip: When verifying matrix properties, clearly show each step of the calculation for both sides of the equation. Accuracy in addition and subtraction of corresponding elements is crucial for a successful verification.

Question 5. જો A = \(\left[\begin{array}{ccc} 2/3 & 3/3 & 5/3 \\ 1/3 & 2/3 & 4/3 \\ 7/3 & 6/3 & 2/3 \end{array}\right]\) અને B = \(\left[\begin{array}{ccc} 2/5 & 3/5 & 1/5 \\ -1/5 & 2/5 & 4/5 \\ 7/5 & 6/5 & 2/5 \end{array}\right]\) હોય, તો 3A – 5Bની ગણતરી કરો.
Answer:
To calculate 3A - 5B, first, multiply each element of matrix A by 3 and each element of matrix B by 5. Then, subtract the resulting matrix from the first one.
\[ 3A - 5B = 3 \left[\begin{array}{ccc} 2/3 & 3/3 & 5/3 \\ 1/3 & 2/3 & 4/3 \\ 7/3 & 6/3 & 2/3 \end{array}\right] - 5 \left[\begin{array}{ccc} 2/5 & 3/5 & 1/5 \\ -1/5 & 2/5 & 4/5 \\ 7/5 & 6/5 & 2/5 \end{array}\right] \] \[ = \left[\begin{array}{ccc} 3 \times (2/3) & 3 \times (3/3) & 3 \times (5/3) \\ 3 \times (1/3) & 3 \times (2/3) & 3 \times (4/3) \\ 3 \times (7/3) & 3 \times (6/3) & 3 \times (2/3) \end{array}\right] - \left[\begin{array}{ccc} 5 \times (2/5) & 5 \times (3/5) & 5 \times (1/5) \\ 5 \times (-1/5) & 5 \times (2/5) & 5 \times (4/5) \\ 5 \times (7/5) & 5 \times (6/5) & 5 \times (2/5) \end{array}\right] \] \[ = \left[\begin{array}{ccc} 2 & 3 & 5 \\ 1 & 2 & 4 \\ 7 & 6 & 2 \end{array}\right] - \left[\begin{array}{ccc} 2 & 3 & 1 \\ -1 & 2 & 4 \\ 7 & 6 & 2 \end{array}\right] \] \[ = \left[\begin{array}{ccc} 2-2 & 3-3 & 5-1 \\ 1-(-1) & 2-2 & 4-4 \\ 7-7 & 6-6 & 2-2 \end{array}\right] \] \[ = \left[\begin{array}{ccc} 0 & 0 & 4 \\ 2 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \]In simple words: This problem shows how to do two steps of matrix math: first, multiply a matrix by a number (scalar multiplication), and then subtract one matrix from another. We multiply each number in the matrix by the scalar before adding or subtracting.

🎯 Exam Tip: Remember to perform scalar multiplication on *every* element of the matrix before proceeding with addition or subtraction. Pay attention to signs, especially during subtraction, to get the correct final matrix.

Question 6. સાદુંરૂપ આપો :
\(\cos\theta \left[\begin{array}{cc} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{array}\right] + \sin\theta \left[\begin{array}{cc} \sin\theta & -\cos\theta \\ \cos\theta & \sin\theta \end{array}\right]\)
Answer:
To simplify the expression, first multiply each element of the first matrix by \(\cos\theta\) and each element of the second matrix by \(\sin\theta\). Then, add the resulting matrices. Use the identity \(\sin^2\theta + \cos^2\theta = 1\) to simplify the final matrix.
\[ \cos\theta \left[\begin{array}{cc} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{array}\right] + \sin\theta \left[\begin{array}{cc} \sin\theta & -\cos\theta \\ \cos\theta & \sin\theta \end{array}\right] \] \[ = \left[\begin{array}{cc} \cos^2\theta & \sin\theta\cos\theta \\ -\sin\theta\cos\theta & \cos^2\theta \end{array}\right] + \left[\begin{array}{cc} \sin^2\theta & -\sin\theta\cos\theta \\ \sin\theta\cos\theta & \sin^2\theta \end{array}\right] \] \[ = \left[\begin{array}{cc} \cos^2\theta + \sin^2\theta & \sin\theta\cos\theta - \sin\theta\cos\theta \\ -\sin\theta\cos\theta + \sin\theta\cos\theta & \cos^2\theta + \sin^2\theta \end{array}\right] \] \[ = \left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right] \]In simple words: This problem shows how to simplify a matrix expression by first multiplying matrices by a single number (like \(\cos\theta\) or \(\sin\theta\)), and then adding them. The key step is using the math rule that \(\sin^2\theta + \cos^2\theta\) always equals 1.

🎯 Exam Tip: When simplifying matrix expressions involving trigonometric functions, remember to apply scalar multiplication correctly to all elements. Also, recognize and use fundamental trigonometric identities (like \(\sin^2\theta + \cos^2\theta = 1\)) to arrive at the simplest form of the matrix.

Question 7.
(i) X + Y = \(\left[\begin{array}{ll} 7 & 0 \\ 2 & 5 \end{array}\right]\) ને X - Y = \(\left[\begin{array}{ll} 3 & 0 \\ 0 & 3 \end{array}\right]\) હોય, તો X અને Y શોધો.
(ii) 2X + 3Y = \(\left[\begin{array}{ll} 2 & 3 \\ 4 & 0 \end{array}\right]\) ને 3X + 2Y = \(\left[\begin{array}{cc} 2 & -2 \\ -1 & 5 \end{array}\right]\) હોય, તો X અને Y શોધો.
Answer:
(i) To find matrices X and Y, we treat the given matrix equations like algebraic equations. We can add the two equations together to find X, and subtract one from the other to find Y.
First, add equations (i) and (ii):
\[ X + Y = \left[\begin{array}{ll} 7 & 0 \\ 2 & 5 \end{array}\right] \quad ...(i) \] \[ X - Y = \left[\begin{array}{ll} 3 & 0 \\ 0 & 3 \end{array}\right] \quad ...(ii) \] Adding (i) and (ii):
\[ (X+Y) + (X-Y) = \left[\begin{array}{ll} 7 & 0 \\ 2 & 5 \end{array}\right] + \left[\begin{array}{ll} 3 & 0 \\ 0 & 3 \end{array}\right] \] \[ 2X = \left[\begin{array}{cc} 7+3 & 0+0 \\ 2+0 & 5+3 \end{array}\right] = \left[\begin{array}{cc} 10 & 0 \\ 2 & 8 \end{array}\right] \] \[ X = \frac{1}{2} \left[\begin{array}{cc} 10 & 0 \\ 2 & 8 \end{array}\right] = \left[\begin{array}{cc} 5 & 0 \\ 1 & 4 \end{array}\right] \] Now, subtract equation (ii) from equation (i):
\[ (X+Y) - (X-Y) = \left[\begin{array}{ll} 7 & 0 \\ 2 & 5 \end{array}\right] - \left[\begin{array}{ll} 3 & 0 \\ 0 & 3 \end{array}\right] \] \[ 2Y = \left[\begin{array}{cc} 7-3 & 0-0 \\ 2-0 & 5-3 \end{array}\right] = \left[\begin{array}{cc} 4 & 0 \\ 2 & 2 \end{array}\right] \] \[ Y = \frac{1}{2} \left[\begin{array}{cc} 4 & 0 \\ 2 & 2 \end{array}\right] = \left[\begin{array}{cc} 2 & 0 \\ 1 & 1 \end{array}\right] \]
(ii) To find matrices X and Y from these equations, we use a method similar to solving simultaneous equations. We multiply the equations by scalars to make the coefficients of X or Y equal, and then add or subtract them.
Given:
\[ 2X + 3Y = \left[\begin{array}{ll} 2 & 3 \\ 4 & 0 \end{array}\right] \quad ...(i) \] \[ 3X + 2Y = \left[\begin{array}{cc} 2 & -2 \\ -1 & 5 \end{array}\right] \quad ...(ii) \] To eliminate Y, multiply equation (i) by 2 and equation (ii) by 3:
\[ 2(2X + 3Y) = 2\left[\begin{array}{ll} 2 & 3 \\ 4 & 0 \end{array}\right] \implies 4X + 6Y = \left[\begin{array}{cc} 4 & 6 \\ 8 & 0 \end{array}\right] \quad ...(iii) \] \[ 3(3X + 2Y) = 3\left[\begin{array}{cc} 2 & -2 \\ -1 & 5 \end{array}\right] \implies 9X + 6Y = \left[\begin{array}{cc} 6 & -6 \\ -3 & 15 \end{array}\right] \quad ...(iv) \] Subtract equation (iii) from equation (iv):
\[ (9X+6Y) - (4X+6Y) = \left[\begin{array}{cc} 6 & -6 \\ -3 & 15 \end{array}\right] - \left[\begin{array}{cc} 4 & 6 \\ 8 & 0 \end{array}\right] \] \[ 5X = \left[\begin{array}{cc} 6-4 & -6-6 \\ -3-8 & 15-0 \end{array}\right] = \left[\begin{array}{cc} 2 & -12 \\ -11 & 15 \end{array}\right] \] \[ X = \frac{1}{5} \left[\begin{array}{cc} 2 & -12 \\ -11 & 15 \end{array}\right] = \left[\begin{array}{cc} 2/5 & -12/5 \\ -11/5 & 3 \end{array}\right] \] To eliminate X, multiply equation (i) by 3 and equation (ii) by 2:
\[ 3(2X + 3Y) = 3\left[\begin{array}{ll} 2 & 3 \\ 4 & 0 \end{array}\right] \implies 6X + 9Y = \left[\begin{array}{cc} 6 & 9 \\ 12 & 0 \end{array}\right] \quad ...(v) \] \[ 2(3X + 2Y) = 2\left[\begin{array}{cc} 2 & -2 \\ -1 & 5 \end{array}\right] \implies 6X + 4Y = \left[\begin{array}{cc} 4 & -4 \\ -2 & 10 \end{array}\right] \quad ...(vi) \] Subtract equation (vi) from equation (v):
\[ (6X+9Y) - (6X+4Y) = \left[\begin{array}{cc} 6 & 9 \\ 12 & 0 \end{array}\right] - \left[\begin{array}{cc} 4 & -4 \\ -2 & 10 \end{array}\right] \] \[ 5Y = \left[\begin{array}{cc} 6-4 & 9-(-4) \\ 12-(-2) & 0-10 \end{array}\right] = \left[\begin{array}{cc} 2 & 13 \\ 14 & -10 \end{array}\right] \] \[ Y = \frac{1}{5} \left[\begin{array}{cc} 2 & 13 \\ 14 & -10 \end{array}\right] = \left[\begin{array}{cc} 2/5 & 13/5 \\ 14/5 & -2 \end{array}\right] \]In simple words: These problems are like solving puzzles with matrices instead of simple numbers. We use addition and subtraction for part (i), and multiplication by numbers followed by addition or subtraction for part (ii), just like we solve equations for 'x' and 'y' in algebra.

🎯 Exam Tip: When solving matrix equations, treat the matrices as variables. Use elimination or substitution methods, carefully performing scalar multiplication, matrix addition, and subtraction step-by-step. Double-check all calculations, especially when dealing with negative signs and fractions.

Question 8. જો Y = \(\left[\begin{array}{ll} 3 & 2 \\ 1 & 4 \end{array}\right]\) ને 2X + Y = \(\left[\begin{array}{cc} 1 & 0 \\ -3 & 2 \end{array}\right]\) હોય, તો X શોધો.
Answer:
To find matrix X, we can substitute the given matrix Y into the second equation. Then, we solve for X by moving Y to the other side and performing matrix subtraction, followed by scalar division.
Given:
\[ Y = \left[\begin{array}{ll} 3 & 2 \\ 1 & 4 \end{array}\right] \] \[ 2X + Y = \left[\begin{array}{cc} 1 & 0 \\ -3 & 2 \end{array}\right] \] Substitute the value of Y into the second equation:
\[ 2X + \left[\begin{array}{ll} 3 & 2 \\ 1 & 4 \end{array}\right] = \left[\begin{array}{cc} 1 & 0 \\ -3 & 2 \end{array}\right] \] Subtract matrix Y from both sides:
\[ 2X = \left[\begin{array}{cc} 1 & 0 \\ -3 & 2 \end{array}\right] - \left[\begin{array}{ll} 3 & 2 \\ 1 & 4 \end{array}\right] \] \[ 2X = \left[\begin{array}{cc} 1-3 & 0-2 \\ -3-1 & 2-4 \end{array}\right] \] \[ 2X = \left[\begin{array}{cc} -2 & -2 \\ -4 & -2 \end{array}\right] \] Divide both sides by 2 (scalar multiplication by \( \frac{1}{2} \)):
\[ X = \frac{1}{2} \left[\begin{array}{cc} -2 & -2 \\ -4 & -2 \end{array}\right] \] \[ X = \left[\begin{array}{cc} -1 & -1 \\ -2 & -1 \end{array}\right] \]In simple words: This problem is about finding an unknown matrix, X. We use simple algebra steps, but with matrices. First, we move the known matrix Y to the other side by subtracting it, and then we divide by the number 2 to find X.

🎯 Exam Tip: Treat matrix equations like algebraic equations. Isolate the unknown matrix by performing inverse operations (subtraction for addition, division for scalar multiplication). Ensure each element is correctly operated upon.

Question 9. જો 2\(\left[\begin{array}{ll} 1 & 3 \\ 0 & x \end{array}\right]\)+\(\left[\begin{array}{ll} y & 0 \\ 1 & 2 \end{array}\right]\)=\(\left[\begin{array}{ll} 5 & 6 \\ 1 & 8 \end{array}\right]\) હોય, તો x અને y શોધો.
Answer:
To find the values of x and y, we first multiply the first matrix by 2, then add the two matrices together. Since the resulting matrix is equal to the matrix on the right side, their corresponding elements must be equal. This gives us two simple equations to solve for x and y.
Given:
\[ 2\left[\begin{array}{ll} 1 & 3 \\ 0 & x \end{array}\right]+\left[\begin{array}{ll} y & 0 \\ 1 & 2 \end{array}\right]=\left[\begin{array}{ll} 5 & 6 \\ 1 & 8 \end{array}\right] \] First, multiply the first matrix by 2:
\[ \left[\begin{array}{cc} 2 \times 1 & 2 \times 3 \\ 2 \times 0 & 2 \times x \end{array}\right]+\left[\begin{array}{ll} y & 0 \\ 1 & 2 \end{array}\right]=\left[\begin{array}{ll} 5 & 6 \\ 1 & 8 \end{array}\right] \] \[ \left[\begin{array}{cc} 2 & 6 \\ 0 & 2x \end{array}\right]+\left[\begin{array}{ll} y & 0 \\ 1 & 2 \end{array}\right]=\left[\begin{array}{ll} 5 & 6 \\ 1 & 8 \end{array}\right] \] Next, add the two matrices on the left side:
\[ \left[\begin{array}{cc} 2+y & 6+0 \\ 0+1 & 2x+2 \end{array}\right]=\left[\begin{array}{ll} 5 & 6 \\ 1 & 8 \end{array}\right] \] \[ \left[\begin{array}{cc} 2+y & 6 \\ 1 & 2x+2 \end{array}\right]=\left[\begin{array}{ll} 5 & 6 \\ 1 & 8 \end{array}\right] \] Since the matrices are equal, their corresponding elements must be equal:
\( 2+y = 5 \)
\( y = 5-2 \)
\( y = 3 \)
\( 2x+2 = 8 \)
\( 2x = 8-2 \)
\( 2x = 6 \)
\( x = 3 \)
Thus, x = 3 and y = 3.
In simple words: We solved this matrix puzzle by doing the math step-by-step. First, we multiplied a matrix by 2, then we added the matrices. Finally, we matched the numbers in the same spots to find what x and y were.

🎯 Exam Tip: Remember that two matrices are equal only if their dimensions are the same and all corresponding elements are equal. This principle is key to setting up and solving algebraic equations for unknown variables within matrices.

Question 10. જો 2\(\left[\begin{array}{ll} x & z \\ y & t \end{array}\right]\)+3\(\left[\begin{array}{cc} 1 & -1 \\ 0 & 2 \end{array}\right]\)=3\(\left[\begin{array}{ll} 3 & 5 \\ 4 & 6 \end{array}\right]\) હોય, તો x, y, z અને t માટે સમીકરણ ઉકેલો.
Answer:
To find the unknown values x, y, z, and t, we first multiply each matrix by its scalar number. Then, we add the matrices on the left side. Since the final matrix on the left is equal to the matrix on the right, we set their matching elements equal to each other to form simple equations, which we then solve.
Given:
\[ 2\left[\begin{array}{ll} x & z \\ y & t \end{array}\right]+3\left[\begin{array}{cc} 1 & -1 \\ 0 & 2 \end{array}\right]=3\left[\begin{array}{ll} 3 & 5 \\ 4 & 6 \end{array}\right] \] Perform scalar multiplication for each matrix:
\[ \left[\begin{array}{cc} 2x & 2z \\ 2y & 2t \end{array}\right]+\left[\begin{array}{cc} 3 \times 1 & 3 \times (-1) \\ 3 \times 0 & 3 \times 2 \end{array}\right]=\left[\begin{array}{cc} 3 \times 3 & 3 \times 5 \\ 3 \times 4 & 3 \times 6 \end{array}\right] \] \[ \left[\begin{array}{cc} 2x & 2z \\ 2y & 2t \end{array}\right]+\left[\begin{array}{cc} 3 & -3 \\ 0 & 6 \end{array}\right]=\left[\begin{array}{cc} 9 & 15 \\ 12 & 18 \end{array}\right] \] Add the matrices on the left side:
\[ \left[\begin{array}{cc} 2x+3 & 2z-3 \\ 2y+0 & 2t+6 \end{array}\right]=\left[\begin{array}{cc} 9 & 15 \\ 12 & 18 \end{array}\right] \] \[ \left[\begin{array}{cc} 2x+3 & 2z-3 \\ 2y & 2t+6 \end{array}\right]=\left[\begin{array}{cc} 9 & 15 \\ 12 & 18 \end{array}\right] \] By equating corresponding elements:
\( 2x+3 = 9 \)
\( 2x = 9-3 \)
\( 2x = 6 \)
\( x = 3 \)
\( 2y = 12 \)
\( y = 6 \)
\( 2z-3 = 15 \)
\( 2z = 15+3 \)
\( 2z = 18 \)
\( z = 9 \)
\( 2t+6 = 18 \)
\( 2t = 18-6 \)
\( 2t = 12 \)
\( t = 6 \)
Thus, x = 3, y = 6, z = 9, and t = 6.
In simple words: This problem asks us to find four unknown numbers hidden inside matrices. We do this by following the matrix math rules, like multiplying by a number and adding. Then, we set the numbers in the same spots of equal matrices to find our answers.

🎯 Exam Tip: Systematically perform scalar multiplication and matrix addition before equating corresponding elements. This approach simplifies the problem into a set of easily solvable linear equations, minimizing calculation errors.

Question 12. જો 3\(\left[\begin{array}{cc} x & y \\ z & w \end{array}\right]\)=\(\left[\begin{array}{cc} x & 6 \\ -1 & 2w \end{array}\right]\)+\(\left[\begin{array}{cc} 4 & x+y \\ z+w & 3 \end{array}\right]\) હોય, તો x, y, z અને w નાં મૂલ્ય શોધો.
Answer:
To find x, y, z, and w, first multiply the matrix on the left by 3. Then, add the two matrices on the right side. Since the resulting matrices are equal, we can set their corresponding elements equal to each other to form equations. We solve these equations one by one to find the values of all variables.
Given:
\[ 3\left[\begin{array}{cc} x & y \\ z & w \end{array}\right]=\left[\begin{array}{cc} x & 6 \\ -1 & 2w \end{array}\right]+\left[\begin{array}{cc} 4 & x+y \\ z+w & 3 \end{array}\right] \] First, perform scalar multiplication on the left side:
\[ \left[\begin{array}{cc} 3x & 3y \\ 3z & 3w \end{array}\right]=\left[\begin{array}{cc} x & 6 \\ -1 & 2w \end{array}\right]+\left[\begin{array}{cc} 4 & x+y \\ z+w & 3 \end{array}\right] \] Next, add the two matrices on the right side:
\[ \left[\begin{array}{cc} 3x & 3y \\ 3z & 3w \end{array}\right]=\left[\begin{array}{cc} x+4 & 6+x+y \\ -1+z+w & 2w+3 \end{array}\right] \] By equating corresponding elements:
1. \( 3x = x+4 \)
\( 3x - x = 4 \)
\( 2x = 4 \)
\( x = 2 \)
2. \( 3y = 6+x+y \)
\( 3y - y = 6+x \)
\( 2y = 6+x \)
Substitute \( x = 2 \):
\( 2y = 6+2 \)
\( 2y = 8 \)
\( y = 4 \)
3. \( 3w = 2w+3 \)
\( 3w - 2w = 3 \)
\( w = 3 \)
4. \( 3z = -1+z+w \)
\( 3z - z = -1+w \)
\( 2z = -1+w \)
Substitute \( w = 3 \):
\( 2z = -1+3 \)
\( 2z = 2 \)
\( z = 1 \)
Thus, x = 2, y = 4, z = 1, and w = 3.
In simple words: This problem involves solving for unknown letters inside matrices. We multiply by a number, add matrices, and then set the matching parts of the matrices equal to each other to solve for each letter one by one.

🎯 Exam Tip: When dealing with matrix equations involving multiple variables, carefully equate the corresponding elements to form a system of linear equations. Solve these equations systematically, substituting known values to find the unknowns.

Question 13. F(x) = \(\left[\begin{array}{ccc} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{array}\right]\) હોય, તો દર્શાવો કે, F(x) F(y) = F(x + y).
Answer:
To prove this, we need to multiply F(x) by F(y). First, write down F(x) and F(y) by replacing x with y in the F(x) matrix. Then, multiply these two matrices using the row-by-column method. After multiplying, use basic trigonometry rules like \(\cos(A+B)\) and \(\sin(A+B)\) to simplify the elements. The final matrix should look exactly like F(x+y).
Given:
\[ F(x) = \left[\begin{array}{ccc} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{array}\right] \] So, \( F(y) \) will be:
\[ F(y) = \left[\begin{array}{ccc} \cos y & -\sin y & 0 \\ \sin y & \cos y & 0 \\ 0 & 0 & 1 \end{array}\right] \] Now, let's calculate \( F(x) F(y) \):
\[ F(x) F(y) = \left[\begin{array}{ccc} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{array}\right] \left[\begin{array}{ccc} \cos y & -\sin y & 0 \\ \sin y & \cos y & 0 \\ 0 & 0 & 1 \end{array}\right] \] Performing the matrix multiplication, we get:
\[ F(x) F(y) = \left[\begin{array}{ccc} \cos x \cos y - \sin x \sin y & -\cos x \sin y - \sin x \cos y & 0 \\ \sin x \cos y + \cos x \sin y & -\sin x \sin y + \cos x \cos y & 0 \\ (0)(\cos y) + (0)(\sin y) + (1)(0) & (0)(-\sin y) + (0)(\cos y) + (1)(0) & (0)(0) + (0)(0) + (1)(1) \end{array}\right] \] Using the trigonometric identities:
\( \cos A \cos B - \sin A \sin B = \cos(A+B) \)
\( \sin A \cos B + \cos A \sin B = \sin(A+B) \)
We can simplify the elements of the matrix:
\[ F(x) F(y) = \left[\begin{array}{ccc} \cos(x+y) & -\sin(x+y) & 0 \\ \sin(x+y) & \cos(x+y) & 0 \\ 0 & 0 & 1 \end{array}\right] \] This result is exactly \( F(x+y) \).
Therefore, we have shown that \( F(x) F(y) = F(x+y) \).
In simple words: This problem asked us to prove that multiplying two special matrices, F(x) and F(y), is the same as getting one matrix F(x+y). We did this by multiplying the matrices and then using angle addition rules from trigonometry to simplify the answers in each spot of the matrix.

🎯 Exam Tip: This question tests matrix multiplication combined with trigonometric identities. Carefully perform each element's calculation in the product matrix. Knowledge of sum/difference identities for sine and cosine is critical for simplification.

Question 14. સાબિત કરો કે,
(i) \(\left[\begin{array}{cc} 5 & -1 \\ 6 & 7 \end{array}\right] \left[\begin{array}{cc} 2 & 1 \\ 3 & 4 \end{array}\right] \neq \left[\begin{array}{cc} 2 & 1 \\ 3 & 4 \end{array}\right] \left[\begin{array}{cc} 5 & -1 \\ 6 & 7 \end{array}\right]\)
(ii) \(\left[\begin{array}{ccc} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{array}\right] \left[\begin{array}{ccc} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4 \end{array}\right] \neq \left[\begin{array}{ccc} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4 \end{array}\right] \left[\begin{array}{ccc} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{array}\right]\)
Answer:
This problem asks us to show that matrix multiplication is not always commutative, meaning the order in which you multiply matrices matters (A times B is not always the same as B times A). To prove this for each part, we will calculate the product of the matrices in one order (Left Hand Side) and then calculate it in the reverse order (Right Hand Side). If the two resulting matrices are different, the statement is proven.
(i) First, let's calculate the Left Hand Side (LHS) product:
\[ \text{LHS} = \left[\begin{array}{cc} 5 & -1 \\ 6 & 7 \end{array}\right] \left[\begin{array}{cc} 2 & 1 \\ 3 & 4 \end{array}\right] \] \[ = \left[\begin{array}{cc} (5)(2)+(-1)(3) & (5)(1)+(-1)(4) \\ (6)(2)+(7)(3) & (6)(1)+(7)(4) \end{array}\right] \] \[ = \left[\begin{array}{cc} 10-3 & 5-4 \\ 12+21 & 6+28 \end{array}\right] = \left[\begin{array}{cc} 7 & 1 \\ 33 & 34 \end{array}\right] \] Next, calculate the Right Hand Side (RHS) product:
\[ \text{RHS} = \left[\begin{array}{cc} 2 & 1 \\ 3 & 4 \end{array}\right] \left[\begin{array}{cc} 5 & -1 \\ 6 & 7 \end{array}\right] \] \[ = \left[\begin{array}{cc} (2)(5)+(1)(6) & (2)(-1)+(1)(7) \\ (3)(5)+(4)(6) & (3)(-1)+(4)(7) \end{array}\right] \] \[ = \left[\begin{array}{cc} 10+6 & -2+7 \\ 15+24 & -3+28 \end{array}\right] = \left[\begin{array}{cc} 16 & 5 \\ 39 & 25 \end{array}\right] \] Since the LHS matrix \(\left[\begin{array}{cc} 7 & 1 \\ 33 & 34 \end{array}\right]\) is not equal to the RHS matrix \(\left[\begin{array}{cc} 16 & 5 \\ 39 & 25 \end{array}\right]\), the statement is proven.
(ii) First, calculate the Left Hand Side (LHS) product:
\[ \text{LHS} = \left[\begin{array}{ccc} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{array}\right] \left[\begin{array}{ccc} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4 \end{array}\right] \] \[ = \left[\begin{array}{ccc} (1)(-1)+(2)(0)+(3)(2) & (1)(1)+(2)(-1)+(3)(3) & (1)(0)+(2)(1)+(3)(4) \\ (0)(-1)+(1)(0)+(0)(2) & (0)(1)+(1)(-1)+(0)(3) & (0)(0)+(1)(1)+(0)(4) \\ (1)(-1)+(1)(0)+(0)(2) & (1)(1)+(1)(-1)+(0)(3) & (1)(0)+(1)(1)+(0)(4) \end{array}\right] \] \[ = \left[\begin{array}{ccc} -1+0+6 & 1-2+9 & 0+2+12 \\ 0+0+0 & 0-1+0 & 0+1+0 \\ -1+0+0 & 1-1+0 & 0+1+0 \end{array}\right] \] \[ = \left[\begin{array}{ccc} 5 & 8 & 14 \\ 0 & -1 & 1 \\ -1 & 0 & 1 \end{array}\right] \] Next, calculate the Right Hand Side (RHS) product:
\[ \text{RHS} = \left[\begin{array}{ccc} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4 \end{array}\right] \left[\begin{array}{ccc} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{array}\right] \] \[ = \left[\begin{array}{ccc} (-1)(1)+(1)(0)+(0)(1) & (-1)(2)+(1)(1)+(0)(1) & (-1)(3)+(1)(0)+(0)(0) \\ (0)(1)+(-1)(0)+(1)(1) & (0)(2)+(-1)(1)+(1)(1) & (0)(3)+(-1)(0)+(1)(0) \\ (2)(1)+(3)(0)+(4)(1) & (2)(2)+(3)(1)+(4)(1) & (2)(3)+(3)(0)+(4)(0) \end{array}\right] \] \[ = \left[\begin{array}{ccc} -1+0+0 & -2+1+0 & -3+0+0 \\ 0+0+1 & 0-1+1 & 0+0+0 \\ 2+0+4 & 4+3+4 & 6+0+0 \end{array}\right] \] \[ = \left[\begin{array}{ccc} -1 & -1 & -3 \\ 1 & 0 & 0 \\ 6 & 11 & 6 \end{array}\right] \] Since the LHS matrix \(\left[\begin{array}{ccc} 5 & 8 & 14 \\ 0 & -1 & 1 \\ -1 & 0 & 1 \end{array}\right]\) is not equal to the RHS matrix \(\left[\begin{array}{ccc} -1 & -1 & -3 \\ 1 & 0 & 0 \\ 6 & 11 & 6 \end{array}\right]\), the statement is proven.
In simple words: This problem shows that the order of multiplying matrices matters. We found that if you multiply matrix A by B, you often get a different answer than if you multiply B by A. This is unlike regular numbers where 2 times 3 is always the same as 3 times 2.

🎯 Exam Tip: When proving non-commutativity (AB ≠ BA), it's crucial to calculate both AB and BA separately and show their resulting matrices are different. Emphasize the detailed steps of matrix multiplication for each element to clearly demonstrate the distinct outcomes.

 

Question 15. જો \(A = \begin{bmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{bmatrix}\) હોય, તો \(A^2 - 5A + 6I\) શોધો.

Answer:

To find \(A^2\), we multiply matrix \(A\) by itself:

\[A^2 = A \cdot A = \begin{bmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{bmatrix} \begin{bmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{bmatrix}\]

\( (2)(2)+(0)(2)+(1)(1) \)	\( (2)(0)+(0)(1)+(1)(-1) \)	\( (2)(1)+(0)(3)+(1)(0) \)
\( (2)(2)+(1)(2)+(3)(1) \)	\( (2)(0)+(1)(1)+(3)(-1) \)	\( (2)(1)+(1)(3)+(3)(0) \)
\( (1)(2)+(-1)(2)+(0)(1) \)	\( (1)(0)+(-1)(1)+(0)(-1) \)	\( (1)(1)+(-1)(3)+(0)(0) \)

\( 4+0+1 \)	\( 0+0-1 \)	\( 2+0+0 \)
\( 4+2+3 \)	\( 0+1-3 \)	\( 2+3+0 \)
\( 2-2+0 \)	\( 0-1+0 \)	\( 1-3+0 \)

\[A^2 = \begin{bmatrix} 5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2 \end{bmatrix}\]

Now, we calculate \(A^2 - 5A + 6I\):

\[A^2 - 5A + 6I = \begin{bmatrix} 5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2 \end{bmatrix} - 5\begin{bmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{bmatrix} + 6\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\]

Perform scalar multiplication:

\[= \begin{bmatrix} 5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2 \end{bmatrix} - \begin{bmatrix} 10 & 0 & 5 \\ 10 & 5 & 15 \\ 5 & -5 & 0 \end{bmatrix} + \begin{bmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{bmatrix}\]

Perform the matrix addition and subtraction:

\[= \begin{bmatrix} 5-10+6 & -1-0+0 & 2-5+0 \\ 9-10+0 & -2-5+6 & 5-15+0 \\ 0-5+0 & -1-(-5)+0 & -2-0+6 \end{bmatrix}\]

\[= \begin{bmatrix} 5-10+6 & -1 & -3 \\ -1 & -1 & -10 \\ -5 & 4 & 4 \end{bmatrix}\]

\[= \begin{bmatrix} 1 & -1 & -3 \\ -1 & -1 & -10 \\ -5 & 4 & 4 \end{bmatrix}\]

In simple words: We first find A-squared by multiplying matrix A by itself. Then, we calculate the expression by subtracting five times matrix A and adding six times the identity matrix to A-squared, which gives the final matrix result.

🎯 Exam Tip: Remember to perform matrix multiplication carefully, element by element, and then apply scalar multiplication before summing or subtracting matrices. This is a common area for calculation errors.

 

Question 16. જો \(A = \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix}\) હોય, તો સાબિત કરો કે \(A^3 - 6A^2 + 7A + 2I = 0\).

Answer:

First, we calculate \(A^2\) by multiplying matrix \(A\) by itself:

\[A^2 = A \cdot A = \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix} \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix}\]

\( (1)(1)+(0)(0)+(2)(2) \)	\( (1)(0)+(0)(2)+(2)(0) \)	\( (1)(2)+(0)(1)+(2)(3) \)
\( (0)(1)+(2)(0)+(1)(2) \)	\( (0)(0)+(2)(2)+(1)(0) \)	\( (0)(2)+(2)(1)+(1)(3) \)
\( (2)(1)+(0)(0)+(3)(2) \)	\( (2)(0)+(0)(2)+(3)(0) \)	\( (2)(2)+(0)(1)+(3)(3) \)

\( 1+0+4 \)	\( 0+0+0 \)	\( 2+0+6 \)
\( 0+0+2 \)	\( 0+4+0 \)	\( 0+2+3 \)
\( 2+0+6 \)	\( 0+0+0 \)	\( 4+0+9 \)

\[A^2 = \begin{bmatrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{bmatrix}\]

Next, we calculate \(A^3\) by multiplying \(A^2\) by matrix \(A\):

\[A^3 = A^2 \cdot A = \begin{bmatrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{bmatrix} \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix}\]

\( (5)(1)+(0)(0)+(8)(2) \)	\( (5)(0)+(0)(2)+(8)(0) \)	\( (5)(2)+(0)(1)+(8)(3) \)
\( (2)(1)+(4)(0)+(5)(2) \)	\( (2)(0)+(4)(2)+(5)(0) \)	\( (2)(2)+(4)(1)+(5)(3) \)
\( (8)(1)+(0)(0)+(13)(2) \)	\( (8)(0)+(0)(2)+(13)(0) \)	\( (8)(2)+(0)(1)+(13)(3) \)

\( 5+0+16 \)	\( 0+0+0 \)	\( 10+0+24 \)
\( 2+0+10 \)	\( 0+8+0 \)	\( 4+4+15 \)
\( 8+0+26 \)	\( 0+0+0 \)	\( 16+0+39 \)

\[A^3 = \begin{bmatrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{bmatrix}\]

Now, we evaluate the left-hand side (LHS) of the expression \(A^3 - 6A^2 + 7A + 2I\):

\[A^3 - 6A^2 + 7A + 2I = \begin{bmatrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{bmatrix} - 6\begin{bmatrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{bmatrix} + 7\begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix} + 2\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\]

Perform scalar multiplications:

\[= \begin{bmatrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{bmatrix} - \begin{bmatrix} 30 & 0 & 48 \\ 12 & 24 & 30 \\ 48 & 0 & 78 \end{bmatrix} + \begin{bmatrix} 7 & 0 & 14 \\ 0 & 14 & 7 \\ 14 & 0 & 21 \end{bmatrix} + \begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix}\]

Perform the matrix additions and subtractions:

\[= \begin{bmatrix} 21-30+7+2 & 0-0+0+0 & 34-48+14+0 \\ 12-12+0+0 & 8-24+14+2 & 23-30+7+0 \\ 34-48+14+0 & 0-0+0+0 & 55-78+21+2 \end{bmatrix}\]

\[= \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} = 0 = \text{RHS}\]

In simple words: We first find A-squared, then A-cubed. After that, we substitute these matrices along with A and the identity matrix into the given expression. We perform all scalar multiplications, then add and subtract the resulting matrices. The final calculation shows that the expression evaluates to a zero matrix, which proves the given statement.

🎯 Exam Tip: This type of problem requires meticulous calculation of matrix powers and careful handling of scalar multiplication and matrix arithmetic. Ensure all elements are added/subtracted correctly to arrive at the zero matrix.

 

Question 17. જો \(A = \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix}\) અને \(I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\) હોય, તો એવો \(k\) શોધો કે જેથી \(A^2 = kA - 2I\) થાય.

Answer:

First, we find \(A^2\) by multiplying matrix \(A\) by itself:

\[A^2 = A \cdot A = \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix} \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix}\]

\( (3)(3)+(-2)(4) \)	\( (3)(-2)+(-2)(-2) \)
\( (4)(3)+(-2)(4) \)	\( (4)(-2)+(-2)(-2) \)

\( 9-8 \)	\( -6+4 \)
\( 12-8 \)	\( -8+4 \)

\[A^2 = \begin{bmatrix} 1 & -2 \\ 4 & -4 \end{bmatrix}\]

The problem states that \(A^2 = kA - 2I\). Substitute the matrices into the equation:

\[\begin{bmatrix} 1 & -2 \\ 4 & -4 \end{bmatrix} = k\begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix} - 2\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\]

Perform scalar multiplication on the right side:

\[\begin{bmatrix} 1 & -2 \\ 4 & -4 \end{bmatrix} = \begin{bmatrix} 3k & -2k \\ 4k & -2k \end{bmatrix} - \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}\]

Perform matrix subtraction on the right side:

\[\begin{bmatrix} 1 & -2 \\ 4 & -4 \end{bmatrix} = \begin{bmatrix} 3k-2 & -2k-0 \\ 4k-0 & -2k-2 \end{bmatrix}\]

\[\begin{bmatrix} 1 & -2 \\ 4 & -4 \end{bmatrix} = \begin{bmatrix} 3k-2 & -2k \\ 4k & -2k-2 \end{bmatrix}\]

By applying the properties of equal matrices, we compare the corresponding elements. Let's use the element in the second row, first column:

\(4k = 4 \implies k = 1\)

We can verify this with other elements: \(3k-2 = 1 \implies 3k = 3 \implies k = 1\) \(-2k = -2 \implies k = 1\) \(-2k-2 = -4 \implies -2k = -2 \implies k = 1\)

Thus, we find that \(k = 1\).

In simple words: We first find A-squared. Then, we substitute A-squared, k times A, and 2 times the identity matrix into the given equation. By comparing the elements of the resulting matrices, we find that the value of k is 1.

🎯 Exam Tip: When finding an unknown scalar like 'k' in a matrix equation, ensure you equate corresponding elements from both sides of the matrix equation. Any single consistent equation is sufficient to find 'k', but checking with other elements confirms the answer.

 

Question 18. જો \(A = \begin{bmatrix} 0 & -\tan\frac{\alpha}{2} \\ \tan\frac{\alpha}{2} & 0 \end{bmatrix}\) અને \(I\) એ 2 કલાવાળો એકમ શ્રેણિક હોય, તો સાબિત કરો કે \(I + A = (I - A) \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix}\)

Answer:

To solve this, we will use the trigonometric identities for \(\cos \alpha\) and \(\sin \alpha\) in terms of \(\tan(\frac{\alpha}{2})\):

We know that: \(\cos \alpha = \frac{1-\tan^2\frac{\alpha}{2}}{1+\tan^2\frac{\alpha}{2}}\) \(\sin \alpha = \frac{2\tan(\frac{\alpha}{2})}{1+\tan^2\frac{\alpha}{2}}\)

Let's simplify by substituting \(t = \tan(\frac{\alpha}{2})\).

So, the matrix \(A\) becomes: \[A = \begin{bmatrix} 0 & -t \\ t & 0 \end{bmatrix}\]

And the trigonometric terms become: \(\cos \alpha = \frac{1-t^2}{1+t^2}\) \(\sin \alpha = \frac{2t}{1+t^2}\)

First, calculate the Left Hand Side (LHS), \(I + A\):

\[I + A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + \begin{bmatrix} 0 & -t \\ t & 0 \end{bmatrix} = \begin{bmatrix} 1+0 & 0-t \\ 0+t & 1+0 \end{bmatrix} = \begin{bmatrix} 1 & -t \\ t & 1 \end{bmatrix} \quad \ldots(i)\]

Next, calculate the Right Hand Side (RHS), \((I - A) \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix}\):

First find \(I - A\): \[I - A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} - \begin{bmatrix} 0 & -t \\ t & 0 \end{bmatrix} = \begin{bmatrix} 1-0 & 0-(-t) \\ 0-t & 1-0 \end{bmatrix} = \begin{bmatrix} 1 & t \\ -t & 1 \end{bmatrix}\]

Now, substitute \(\cos \alpha\) and \(\sin \alpha\) in terms of \(t\) into the trigonometric matrix:

\[\begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix} = \begin{bmatrix} \frac{1-t^2}{1+t^2} & -\frac{2t}{1+t^2} \\ \frac{2t}{1+t^2} & \frac{1-t^2}{1+t^2} \end{bmatrix}\]

Now, multiply \((I - A)\) by this trigonometric matrix:

\[(I - A) \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix} = \begin{bmatrix} 1 & t \\ -t & 1 \end{bmatrix} \begin{bmatrix} \frac{1-t^2}{1+t^2} & -\frac{2t}{1+t^2} \\ \frac{2t}{1+t^2} & \frac{1-t^2}{1+t^2} \end{bmatrix}\]

\( (1)\frac{1-t^2}{1+t^2} + (t)\frac{2t}{1+t^2} \)	\( (1)(-\frac{2t}{1+t^2}) + (t)\frac{1-t^2}{1+t^2} \)
\( (-t)\frac{1-t^2}{1+t^2} + (1)\frac{2t}{1+t^2} \)	\( (-t)(-\frac{2t}{1+t^2}) + (1)\frac{1-t^2}{1+t^2} \)

\( \frac{1-t^2+2t^2}{1+t^2} \)	\( \frac{-2t+t-t^3}{1+t^2} \)
\( \frac{-t+t^3+2t}{1+t^2} \)	\( \frac{2t^2+1-t^2}{1+t^2} \)

\( \frac{1+t^2}{1+t^2} \)	\( \frac{-t-t^3}{1+t^2} \)
\( \frac{t+t^3}{1+t^2} \)	\( \frac{1+t^2}{1+t^2} \)

\( 1 \)	\( \frac{-t(1+t^2)}{1+t^2} \)
\( \frac{t(1+t^2)}{1+t^2} \)	\( 1 \)

\[= \begin{bmatrix} 1 & -t \\ t & 1 \end{bmatrix}\]

From equation (i), we see that this result is equal to \(I + A\).

Therefore, \(I + A = (I - A) \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix}\) is proven.

In simple words: We first express \(\cos \alpha\) and \(\sin \alpha\) using the half-angle tangent formula, substituting \(t\) for \(\tan(\frac{\alpha}{2})\). Then, we calculate the left side of the equation, \(I+A\). Next, we calculate the right side by first finding \(I-A\) and then multiplying it by the given trigonometric matrix. After simplifying both sides, we show that they are equal, thus proving the statement.

🎯 Exam Tip: This problem tests your knowledge of trigonometric identities and matrix multiplication. Clearly define your substitutions (like \(t = \tan(\frac{\alpha}{2})\)) and keep track of calculations for each element in the matrices.

 

Question 19. એક ટ્રસ્ટ પાસે Rs. 30,000 ભંડોળ છે. ટ્રસ્ટે આ ભંડોળ જુદા-જુદા પ્રકારના બૉન્ડમાં રોકવું છે. પ્રથમ બૉન્ડ પ્રતિ વર્ષ 5% વ્યાજ આપે છે અને બીજા બૉન્ડ પ્રતિ વર્ષ 7% વ્યાજ આપે છે. જો ટ્રસ્ટને વાર્ષિક વ્યાજ (a) Rs. 1,800, (b) Rs. 2,000 મેળવવું હોય, તો ટ્રસ્ટે Rs. 30,000 બે બૉન્ડમાં રોકવા માટે મૂડીના કેવા ભાગ કરવા પડશે, તે શ્રેણિક ગુણાકારના ઉપયોગથી નક્કી કરો.

Answer:

Let's assume the trust invests Rs. \(x\) in the first bond (5% interest) and the remaining Rs. (30,000 - \(x\)) in the second bond (7% interest).

(a) If the total annual interest is Rs. 1,800:

We can represent this using matrix multiplication:

\[\begin{bmatrix} x & 30000-x \end{bmatrix} \begin{bmatrix} 5\% \\ 7\% \end{bmatrix} = 1800\]

\[\begin{bmatrix} x & 30000-x \end{bmatrix} \begin{bmatrix} \frac{5}{100} \\ \frac{7}{100} \end{bmatrix} = 1800\]

Multiplying the matrices, we get a linear equation:

\(\frac{5x}{100} + \frac{7(30000-x)}{100} = 1800\)

\(5x + 7(30000 - x) = 1800 \times 100\)

\(5x + 210000 - 7x = 180000\)

\(-2x = 180000 - 210000\)

\(-2x = -30000\)

\(x = 15000\)

So, the amount invested in the first bond is Rs. 15,000.

The amount invested in the second bond is Rs. (30,000 - 15,000) = Rs. 15,000.

Therefore, to earn Rs. 1,800 interest, the trust should invest Rs. 15,000 in the 5% bond and Rs. 15,000 in the 7% bond.

(b) If the total annual interest is Rs. 2,000:

We use the same matrix setup:

\[\begin{bmatrix} x & 30000-x \end{bmatrix} \begin{bmatrix} 5\% \\ 7\% \end{bmatrix} = 2000\]

\[\begin{bmatrix} x & 30000-x \end{bmatrix} \begin{bmatrix} \frac{5}{100} \\ \frac{7}{100} \end{bmatrix} = 2000\]

Multiplying the matrices, we get a linear equation:

\(\frac{5x}{100} + \frac{7(30000-x)}{100} = 2000\)

\(5x + 7(30000 - x) = 2000 \times 100\)

\(5x + 210000 - 7x = 200000\)

\(-2x = 200000 - 210000\)

\(-2x = -10000\)

\(x = 5000\)

So, the amount invested in the first bond is Rs. 5,000.

The amount invested in the second bond is Rs. (30,000 - 5,000) = Rs. 25,000.

Therefore, to earn Rs. 2,000 interest, the trust should invest Rs. 5,000 in the 5% bond and Rs. 25,000 in the 7% bond.

In simple words: The trust divides Rs. 30,000 into two bonds with different interest rates. By setting up a matrix multiplication for the investments and their interest rates, we form a linear equation. Solving this equation for 'x' (the amount in the first bond) tells us how to split the investment to get the target annual interest of either Rs. 1,800 or Rs. 2,000.

🎯 Exam Tip: For problems involving investments and interest, clearly define variables and set up the matrix multiplication carefully. Ensure correct conversion of percentages to fractions (e.g., 5% to 5/100) and accurate algebraic manipulation to solve for the unknown amounts.

 

Question 20. એક સવિશેષ શાળાના પુસ્તકભંડારમાં 10 ડઝન રસાયણવિજ્ઞાનનાં પુસ્તકો, 8 ડઝન ભૌતિકવિજ્ઞાનનાં પુસ્તકો અને 10 ડઝન અર્થશાસ્ત્રનાં પુસ્તકો છે. તેમની વેચાણકિંમત અનુક્રમે Rs. 80, Rs. 60 અને Rs. 40 છે. પુસ્તકભંડાર બધાં જ પુસ્તકોનું વેચાણ કરી દે, તો શ્રેણિક બીજગણિતની મદદથી ભંડારને કેટલી રકમ મળશે તે શોધો.

Answer:

First, calculate the total number of books for each subject, knowing that one dozen equals 12 books:

Chemistry books: 10 dozen = \(10 \times 12 = 120\) books

Physics books: 8 dozen = \(8 \times 12 = 96\) books

Economics books: 10 dozen = \(10 \times 12 = 120\) books

Now, we can represent the number of books as a row matrix and their respective selling prices as a column matrix. To find the total revenue, we multiply these two matrices.

Number of books matrix (N): \([\begin{array}{ccc} 120 & 96 & 120 \end{array}]\) Price per book matrix (P): \(\begin{bmatrix} 80 \\ 60 \\ 40 \end{bmatrix}\)

Total Revenue = \(N \times P\)

\[= \begin{bmatrix} 120 & 96 & 120 \end{bmatrix} \begin{bmatrix} 80 \\ 60 \\ 40 \end{bmatrix}\]

\[= (120 \times 80) + (96 \times 60) + (120 \times 40)\]

\[= 9600 + 5760 + 4800\]

\[= 20160\]

The total amount the bookstore will receive is Rs. 20,160.

In simple words: We first find the total quantity of each type of book. Then, we use matrix multiplication by combining the quantities (as a row matrix) with their prices (as a column matrix). The result of this multiplication gives the total money the bookstore will earn.

🎯 Exam Tip: When using matrix multiplication for real-world problems like revenue calculation, ensure the dimensions of your matrices are compatible. The number of columns in the first matrix must match the number of rows in the second matrix. Also, double-check your arithmetic for accuracy.

 

Question 21. પ્રશ્ન 21 તથા 22 માં વિધાન સાચું બને તે રીતે આપેલ વિકલ્પોમાંથી સોગ્ય વિક્લ્પ પસંદ કરો :
ધારો કે X, Y, Z, W અને P અનુક્રમે \(2 \times n, 3 \times k, 2 \times p, n \times 3\) અને \(p \times k\) કક્ષાવાળા શ્રેણિક છે.
PY + WY વ્યાખ્યાયિત થાય તે રીતે n, k અને p પર પ્રતિબંધ મૂકવામાં આવે તો :
(A) \(k = 3, p = n\)
(B) \(k\) સ્વૈર, \(p = 2\)
(C) \(p\) સ્વૈર, \(k = 3\)
(D) \(k = 2, p = 3\)

Answer:

Given the dimensions of the matrices:

Matrix X: \(2 \times n\) Matrix Y: \(3 \times k\) Matrix Z: \(2 \times p\) Matrix W: \(n \times 3\) Matrix P: \(p \times k\)

For the expression \(PY + WY\) to be defined, two conditions must be met:

1. **Matrix products PY and WY must be defined.** * For \(PY\): The number of columns in P must equal the number of rows in Y. P is \(p \times k\), Y is \(3 \times k\). So, \(k\) must be equal to 3. (i.e., \(k=3\)). The dimension of \(PY\) will be \(p \times k\), which is \(p \times 3\) (since \(k=3\)). * For \(WY\): The number of columns in W must equal the number of rows in Y. W is \(n \times 3\), Y is \(3 \times k\). So, 3 must be equal to 3. This condition is already satisfied. The dimension of \(WY\) will be \(n \times k\), which is \(n \times 3\) (since \(k=3\)).

2. **Matrices PY and WY must have the same dimensions for addition.** * Dimension of \(PY\) is \(p \times 3\). * Dimension of \(WY\) is \(n \times 3\). * For \(PY + WY\) to be defined, their dimensions must be identical. So, \(p\) must be equal to \(n\). (i.e., \(p=n\)).

Combining these conditions, we have \(k=3\) and \(p=n\).

Comparing with the given options, option (A) matches these conditions.

Answer: (A) k = 3, p = nIn simple words: For two matrices to be multiplied, the first matrix's columns must match the second matrix's rows. For two matrices to be added, they must have the exact same dimensions. By applying these rules to PY and WY, we find that k must be 3 and p must be equal to n.

🎯 Exam Tip: Always remember the rules for matrix addition and multiplication: `(A_mxn) + (B_pxq)` is possible if `m=p` and `n=q`. `(A_mxn) * (B_pxq)` is possible if `n=p`. The resulting matrix will have dimension `m x q`.

 

Question 22. જો \(n = p\) હોય, તો શ્રેણિક \(7X - 5Z\) ની કક્ષા :
(A) \(P \times 2\)
(B) \(2 \times n\)
(C) \(n \times 3\)
(D) \(p \times n\)

Answer:

Given the dimensions of the matrices from the previous problem:

Matrix X: \(2 \times n\) Matrix Z: \(2 \times p\)

We are also given that \(n = p\).

We need to find the dimension of the matrix \(7X - 5Z\).

1. **Scalar multiplication:** When a matrix is multiplied by a scalar, its dimension does not change.

* The dimension of \(7X\) is the same as \(X\), which is \(2 \times n\).

* The dimension of \(5Z\) is the same as \(Z\), which is \(2 \times p\).

2. **Matrix subtraction:** For two matrices to be subtracted, they must have the same dimensions.

* Since \(n = p\), the dimension of \(5Z\) can also be written as \(2 \times n\).

* Therefore, both \(7X\) and \(5Z\) have the dimension \(2 \times n\).

* When two matrices of the same dimension are subtracted, the resulting matrix will have the same dimension.

So, the dimension of \(7X - 5Z\) is \(2 \times n\).

Comparing with the given options, option (B) matches this dimension.

Answer: (B) 2 × nIn simple words: Multiplying a matrix by a number does not change its size. For us to subtract two matrices, they must be the same size. Since matrix X is 2 by n, and matrix Z is 2 by p, and n equals p, both 7X and 5Z are 2 by n. So, when we subtract them, the new matrix will also be 2 by n.

🎯 Exam Tip: Scalar multiplication does not alter matrix dimensions. Matrix addition or subtraction is only possible if both matrices have identical dimensions, and the resulting matrix retains that same dimension.