Exercise 3.2 Practice
Matrix Operations: Addition, Multiplication, and Applications
Q1
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Let \( A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \), \( B = \begin{bmatrix} 2 & -1 \\ 0 & 3 \end{bmatrix}
\), \( C = \begin{bmatrix} 5 & 1 \\ 2 & 0 \end{bmatrix} \).
Find each of the following:
(i) \( A + B \) (ii) \( A - B \) (iii) \( 3A - C \) (iv) \( AB \) (v) \( BA \)
Find each of the following:
(i) \( A + B \) (ii) \( A - B \) (iii) \( 3A - C \) (iv) \( AB \) (v) \( BA \)
(i) Addition
\( A + B = \begin{bmatrix} 1+2 & 2+(-1) \\ 3+0 & 4+3 \end{bmatrix} = \begin{bmatrix} 3 & 1 \\ 3 & 7
\end{bmatrix} \)
(iii) Scalar Mult & Subtraction
\( 3A = \begin{bmatrix} 3 & 6 \\ 9 & 12 \end{bmatrix} \).
\( 3A - C = \begin{bmatrix} 3-5 & 6-1 \\ 9-2 & 12-0 \end{bmatrix} = \begin{bmatrix} -2 & 5 \\ 7 & 12 \end{bmatrix} \)
\( 3A - C = \begin{bmatrix} 3-5 & 6-1 \\ 9-2 & 12-0 \end{bmatrix} = \begin{bmatrix} -2 & 5 \\ 7 & 12 \end{bmatrix} \)
(iv) Multiplication AB
\( \begin{bmatrix} 1(2)+2(0) & 1(-1)+2(3) \\ 3(2)+4(0) & 3(-1)+4(3) \end{bmatrix} = \begin{bmatrix} 2 & 5 \\ 6
& 9 \end{bmatrix} \)
Q2
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Compute the following:
(i) \( \begin{bmatrix} x^2+y^2 & y^2+z^2 \\ x^2+z^2 & a^2+b^2 \end{bmatrix} + \begin{bmatrix} 2xy & 2yz \\ -2xz & -2ab \end{bmatrix} \)
(ii) \( \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} + \begin{bmatrix} -1 & 1 \\ 1 & -1 \end{bmatrix} \)
(i) \( \begin{bmatrix} x^2+y^2 & y^2+z^2 \\ x^2+z^2 & a^2+b^2 \end{bmatrix} + \begin{bmatrix} 2xy & 2yz \\ -2xz & -2ab \end{bmatrix} \)
(ii) \( \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} + \begin{bmatrix} -1 & 1 \\ 1 & -1 \end{bmatrix} \)
Part (i)
Add corresponding elements:
\( \begin{bmatrix} x^2+y^2+2xy & y^2+z^2+2yz \\ x^2+z^2-2xz & a^2+b^2-2ab \end{bmatrix} = \begin{bmatrix} (x+y)^2 & (y+z)^2 \\ (x-z)^2 & (a-b)^2 \end{bmatrix} \)
\( \begin{bmatrix} x^2+y^2+2xy & y^2+z^2+2yz \\ x^2+z^2-2xz & a^2+b^2-2ab \end{bmatrix} = \begin{bmatrix} (x+y)^2 & (y+z)^2 \\ (x-z)^2 & (a-b)^2 \end{bmatrix} \)
Part (ii)
\( \begin{bmatrix} 1-1 & -1+1 \\ -1+1 & 1-1 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \)
(Zero Matrix)
Q3
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Compute the indicated products:
(i) \( \begin{bmatrix} 2 & 1 \end{bmatrix} \begin{bmatrix} 3 \\ 4 \end{bmatrix} \)
(ii) \( \begin{bmatrix} 1 & -1 \\ 2 & 0 \end{bmatrix} \begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \end{bmatrix} \)
(i) \( \begin{bmatrix} 2 & 1 \end{bmatrix} \begin{bmatrix} 3 \\ 4 \end{bmatrix} \)
(ii) \( \begin{bmatrix} 1 & -1 \\ 2 & 0 \end{bmatrix} \begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \end{bmatrix} \)
Part (i)
\( [2(3) + 1(4)] = [6+4] = [10] \) (Order \( 1 \times 1 \))
Part (ii)
\( \begin{bmatrix} 1(1)+(-1)(2) & 1(2)+(-1)(3) & 1(3)+(-1)(1) \\ 2(1)+0(2) & 2(2)+0(3) & 2(3)+0(1)
\end{bmatrix} \)
\( = \begin{bmatrix} -1 & -1 & 2 \\ 2 & 4 & 6 \end{bmatrix} \)
\( = \begin{bmatrix} -1 & -1 & 2 \\ 2 & 4 & 6 \end{bmatrix} \)
Q4
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If \( A = \begin{bmatrix} 2 & 1 & 1 \\ 0 & 2 & 3 \\ 1 & -1 & 2 \end{bmatrix} \), \( B = \begin{bmatrix} 1 & 0 &
2 \\ 2 & 1 & 0 \\ 3 & 1 & 2 \end{bmatrix} \), \( C = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 2 \\ 2 & 0 & 1
\end{bmatrix} \), calculate \( A+B \) and \( B-C \). Also verify that \( A+(B-C) = (A+B)-C \).
\( A+B = \begin{bmatrix} 3 & 1 & 3 \\ 2 & 3 & 3 \\ 4 & 0 & 4 \end{bmatrix} \), \( B-C = \begin{bmatrix} 0 & -2
& -1 \\ 2 & 0 & -2 \\ 1 & 1 & 1 \end{bmatrix} \)
Verification: Both sides equal \( \begin{bmatrix} 2 & -1 & 0 \\ 2 & 2 & 1 \\ 2 & 0 & 3
\end{bmatrix} \).
Q5
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If \( A = \begin{bmatrix} 1/2 & 3/2 & 1 \\ 1 & 5/2 & 2 \end{bmatrix} \) and \( B = \begin{bmatrix} 3/5 & 1/5 & 1
\\ 4/5 & 2/5 & 2 \end{bmatrix} \), compute \( 2A - 5B \).
\( 2A = \begin{bmatrix} 1 & 3 & 2 \\ 2 & 5 & 4 \end{bmatrix} \)
\( 5B = \begin{bmatrix} 3 & 1 & 5 \\ 4 & 2 & 10 \end{bmatrix} \)
\( 5B = \begin{bmatrix} 3 & 1 & 5 \\ 4 & 2 & 10 \end{bmatrix} \)
\( 2A - 5B = \begin{bmatrix} -2 & 2 & -3 \\ -2 & 3 & -6 \end{bmatrix} \)
Q6
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Simplify: \( \sin \theta \begin{bmatrix} \sin \theta & -\cos \theta \\ \cos \theta & \sin \theta \end{bmatrix} +
\cos \theta \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix} \)
First term: \( \begin{bmatrix} \sin^2\theta & -\sin\theta\cos\theta \\ \sin\theta\cos\theta & \sin^2\theta
\end{bmatrix} \)
Second term: \( \begin{bmatrix} \cos^2\theta & \sin\theta\cos\theta \\ -\sin\theta\cos\theta & \cos^2\theta \end{bmatrix} \)
Second term: \( \begin{bmatrix} \cos^2\theta & \sin\theta\cos\theta \\ -\sin\theta\cos\theta & \cos^2\theta \end{bmatrix} \)
Sum = \( \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I \)
Q7
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Find matrices \( X \) and \( Y \) if:
(i) \( X + Y = \begin{bmatrix} 5 & 0 \\ 1 & 4 \end{bmatrix} \) and \( X - Y = \begin{bmatrix} 3 & 6 \\ 1 & -2 \end{bmatrix} \)
(i) \( X + Y = \begin{bmatrix} 5 & 0 \\ 1 & 4 \end{bmatrix} \) and \( X - Y = \begin{bmatrix} 3 & 6 \\ 1 & -2 \end{bmatrix} \)
Adding equations: \( 2X = \begin{bmatrix} 8 & 6 \\ 2 & 2 \end{bmatrix} \Rightarrow X = \begin{bmatrix} 4 & 3
\\ 1 & 1 \end{bmatrix} \)
Subtracting equations: \( 2Y = \begin{bmatrix} 2 & -6 \\ 0 & 6 \end{bmatrix} \Rightarrow Y = \begin{bmatrix} 1 & -3 \\ 0 & 3 \end{bmatrix} \)
Subtracting equations: \( 2Y = \begin{bmatrix} 2 & -6 \\ 0 & 6 \end{bmatrix} \Rightarrow Y = \begin{bmatrix} 1 & -3 \\ 0 & 3 \end{bmatrix} \)
Q8
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Find \( X \) if \( Y = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \) and \( 2X + Y = \begin{bmatrix} 5 & 6 \\
7 & 8 \end{bmatrix} \)
\( 2X = \begin{bmatrix} 5 & 6 \\ 7 & 8 \end{bmatrix} - \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} =
\begin{bmatrix} 4 & 4 \\ 4 & 4 \end{bmatrix} \)
\( X = \begin{bmatrix} 2 & 2 \\ 2 & 2 \end{bmatrix} \)
Q9
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Find \( x \) and \( y \) if: \( 2 \begin{bmatrix} 1 & 3 \\ 0 & x \end{bmatrix} + \begin{bmatrix} y & 0 \\ 1 & 2
\end{bmatrix} = \begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix} \)
LHS: \( \begin{bmatrix} 2+y & 6 \\ 1 & 2x+2 \end{bmatrix} \)
Equate to RHS: \( 2+y=5 \Rightarrow y=3 \) and \( 2x+2=8 \Rightarrow 2x=6 \Rightarrow x=3 \).
Equate to RHS: \( 2+y=5 \Rightarrow y=3 \) and \( 2x+2=8 \Rightarrow 2x=6 \Rightarrow x=3 \).
\( x=3, y=3 \)
Q10
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Solve for \( x, y, z, t \): \( 2 \begin{bmatrix} x & z \\ y & t \end{bmatrix} + 3 \begin{bmatrix} 1 & -1 \\ 0 &
2 \end{bmatrix} = 3 \begin{bmatrix} 3 & 5 \\ 4 & 6 \end{bmatrix} \)
Eq 1: \( 2x + 3 = 9 \Rightarrow 2x=6 \Rightarrow x=3 \)
Eq 2: \( 2z - 3 = 15 \Rightarrow 2z=18 \Rightarrow z=9 \)
Eq 3: \( 2y + 0 = 12 \Rightarrow y=6 \)
Eq 4: \( 2t + 6 = 18 \Rightarrow 2t=12 \Rightarrow t=6 \)
Eq 2: \( 2z - 3 = 15 \Rightarrow 2z=18 \Rightarrow z=9 \)
Eq 3: \( 2y + 0 = 12 \Rightarrow y=6 \)
Eq 4: \( 2t + 6 = 18 \Rightarrow 2t=12 \Rightarrow t=6 \)
Q11
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If \( x \begin{bmatrix} 2 \\ 3 \end{bmatrix} + y \begin{bmatrix} -1 \\ 1 \end{bmatrix} = \begin{bmatrix} 10 \\ 5
\end{bmatrix} \), find the values of \( x \) and \( y \).
\( 2x - y = 10 \)
\( 3x + y = 5 \)
Add them: \( 5x = 15 \Rightarrow x=3 \).
Substitute: \( 6 - y = 10 \Rightarrow y = -4 \).
\( 3x + y = 5 \)
Add them: \( 5x = 15 \Rightarrow x=3 \).
Substitute: \( 6 - y = 10 \Rightarrow y = -4 \).
Q12
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Given \( 3 \begin{bmatrix} x & y \\ z & w \end{bmatrix} = \begin{bmatrix} x & 6 \\ -1 & 2w \end{bmatrix} +
\begin{bmatrix} 4 & x+y \\ z+w & 3 \end{bmatrix} \), find \( x, y, z, w \).
\( 3x = x+4 \Rightarrow 2x=4 \Rightarrow x=2 \)
\( 3y = 6+x+y \Rightarrow 2y = 6+2=8 \Rightarrow y=4 \)
\( 3w = 2w+3 \Rightarrow w=3 \)
\( 3z = -1+z+w \Rightarrow 2z = -1+3=2 \Rightarrow z=1 \)
\( 3y = 6+x+y \Rightarrow 2y = 6+2=8 \Rightarrow y=4 \)
\( 3w = 2w+3 \Rightarrow w=3 \)
\( 3z = -1+z+w \Rightarrow 2z = -1+3=2 \Rightarrow z=1 \)
Q13
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If \( f(x) = \begin{bmatrix} 1 & x \\ 0 & 1 \end{bmatrix} \), show that \( f(x)f(y) = f(x+y) \).
\( f(x)f(y) = \begin{bmatrix} 1 & x \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & y \\ 0 & 1 \end{bmatrix} \)
\( = \begin{bmatrix} 1(1)+x(0) & 1(y)+x(1) \\ 0(1)+1(0) & 0(y)+1(1) \end{bmatrix} = \begin{bmatrix} 1 & x+y \\ 0 & 1 \end{bmatrix} \)
\( = \begin{bmatrix} 1(1)+x(0) & 1(y)+x(1) \\ 0(1)+1(0) & 0(y)+1(1) \end{bmatrix} = \begin{bmatrix} 1 & x+y \\ 0 & 1 \end{bmatrix} \)
This exactly equals \( f(x+y) \).
Q14
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Show that \( \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \begin{bmatrix} 5 & 6 \\ 7 & 8 \end{bmatrix} \neq
\begin{bmatrix} 5 & 6 \\ 7 & 8 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \)
LHS: \( \begin{bmatrix} 1(5)+2(7) & 1(6)+2(8) \\ 3(5)+4(7) & 3(6)+4(8) \end{bmatrix} = \begin{bmatrix} 19 & 22
\\ 43 & 50 \end{bmatrix} \)
RHS: \( \begin{bmatrix} 5(1)+6(3) & 5(2)+6(4) \\ 7(1)+8(3) & 7(2)+8(4) \end{bmatrix} = \begin{bmatrix} 23 & 34
\\ 31 & 46 \end{bmatrix} \)
LHS \( \neq \) RHS. Proved.
Q15
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Find \( A^2 - 4A + 3I \) if \( A = \begin{bmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{bmatrix} \).
First calculate \( A^2 = A \times A \).
\( A^2 = \begin{bmatrix} 5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2 \end{bmatrix} \)
\( A^2 = \begin{bmatrix} 5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2 \end{bmatrix} \)
Substitute into equation: \( \begin{bmatrix} 5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2 \end{bmatrix} -
\begin{bmatrix} 8 & 0 & 4 \\ 8 & 4 & 12 \\ 4 & -4 & 0 \end{bmatrix} + \begin{bmatrix} 3 & 0 & 0 \\ 0 & 3 & 0
\\ 0 & 0 & 3 \end{bmatrix} \)
\( = \begin{bmatrix} 0 & -1 & -2 \\ 1 & -3 & -7 \\ -4 & 3 & 1 \end{bmatrix} \)
Q16
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If \( A = \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix} \), prove that \( A^3 - 6A^2 + 7A +
2I = O \).
Calculate \( A^2 \) first, then \( A^3 = A^2 \times A \).
\( A^2 = \begin{bmatrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{bmatrix} \), \( A^3 = \begin{bmatrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{bmatrix} \)
\( A^2 = \begin{bmatrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{bmatrix} \), \( A^3 = \begin{bmatrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{bmatrix} \)
Substitute: \( \begin{bmatrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{bmatrix} - \begin{bmatrix} 30 &
0 & 48 \\ 12 & 24 & 30 \\ 48 & 0 & 78 \end{bmatrix} + \begin{bmatrix} 7 & 0 & 14 \\ 0 & 14 & 7 \\ 14 & 0 & 21
\end{bmatrix} + \begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix} \)
Sum of all terms is the Zero Matrix \( O \). Proved.
Q17
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If \( A = \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix} \) and \( I = \begin{bmatrix} 1 & 0 \\ 0 & 1
\end{bmatrix} \), find \( k \) so that \( A^2 = kA - 2I \).
\( A^2 = \begin{bmatrix} 1 & -2 \\ 4 & -4 \end{bmatrix} \)
\( kA - 2I = \begin{bmatrix} 3k-2 & -2k \\ 4k & -2k-2 \end{bmatrix} \)
\( kA - 2I = \begin{bmatrix} 3k-2 & -2k \\ 4k & -2k-2 \end{bmatrix} \)
Compare elements: \( 4 = 4k \Rightarrow k=1 \).
Check others: \( 1 = 3(1)-2 \) (Correct).
Check others: \( 1 = 3(1)-2 \) (Correct).
\( k = 1 \)
Q18
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If \( A = \begin{bmatrix} 0 & -x \\ x & 0 \end{bmatrix} \), show that \( (I + A) = (I - A) \begin{bmatrix} \cos
2\alpha & -\sin 2\alpha \\ \sin 2\alpha & \cos 2\alpha \end{bmatrix} \) where \( x = \tan \alpha \).
Let \( x = \tan \alpha \).
LHS: \( I+A = \begin{bmatrix} 1 & -x \\ x & 1 \end{bmatrix} \).
RHS: \( (I-A) \times \text{Rotation Matrix} = \begin{bmatrix} 1 & x \\ -x & 1 \end{bmatrix} \begin{bmatrix} \frac{1-x^2}{1+x^2} & \frac{-2x}{1+x^2} \\ \frac{2x}{1+x^2} & \frac{1-x^2}{1+x^2} \end{bmatrix} \)
LHS: \( I+A = \begin{bmatrix} 1 & -x \\ x & 1 \end{bmatrix} \).
RHS: \( (I-A) \times \text{Rotation Matrix} = \begin{bmatrix} 1 & x \\ -x & 1 \end{bmatrix} \begin{bmatrix} \frac{1-x^2}{1+x^2} & \frac{-2x}{1+x^2} \\ \frac{2x}{1+x^2} & \frac{1-x^2}{1+x^2} \end{bmatrix} \)
Multiply RHS matrices. The result simplifies exactly to \( \begin{bmatrix} 1 & -x \\ x & 1 \end{bmatrix} \).
LHS = RHS. Proved.
Q19
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A fund has ₹50,000 to invest in two bonds. The first pays 4% interest per year, and the second pays 6% interest.
Using matrix multiplication, determine how to divide the money to get a total annual interest of ₹2,600.
Let investment in first bond be \( x \), second be \( 50000 - x \).
Matrix: \( \begin{bmatrix} x & 50000-x \end{bmatrix} \begin{bmatrix} 0.04 \\ 0.06 \end{bmatrix} = [2600] \)
Matrix: \( \begin{bmatrix} x & 50000-x \end{bmatrix} \begin{bmatrix} 0.04 \\ 0.06 \end{bmatrix} = [2600] \)
Equation: \( 0.04x + 0.06(50000-x) = 2600 \)
\( 0.04x + 3000 - 0.06x = 2600 \Rightarrow -0.02x = -400 \).
\( 0.04x + 3000 - 0.06x = 2600 \Rightarrow -0.02x = -400 \).
\( x = 20,000 \). Invest ₹20,000 in 4% bond, ₹30,000 in 6% bond.
Q20
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A school stationery shop has 12 dozen pencils, 5 dozen erasers, and 10 dozen rulers. Selling prices are ₹5, ₹2,
and ₹10 each respectively. Find total amount received.
Quantity Matrix \( Q = \begin{bmatrix} 144 & 60 & 120 \end{bmatrix} \) (converted dozens to units).
Price Matrix \( P = \begin{bmatrix} 5 \\ 2 \\ 10 \end{bmatrix} \).
Price Matrix \( P = \begin{bmatrix} 5 \\ 2 \\ 10 \end{bmatrix} \).
Total = \( QP = 144(5) + 60(2) + 120(10) = 720 + 120 + 1200 \).
Total Amount = ₹2,040
Q21
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Assume matrices \( A, B, C \) are of order \( 2 \times n, 3 \times k, 2 \times p \) respectively. The
restriction on \( n, k, p \) so that \( AB + CB \) is defined is:
(A) \( k=3, p=n \) (B) \( k=2, p=3 \) (C) \( k=n, p \) is arbitrary (D) \( p=2, k=3 \)
(A) \( k=3, p=n \) (B) \( k=2, p=3 \) (C) \( k=n, p \) is arbitrary (D) \( p=2, k=3 \)
For \( AB \): Cols of A (n) = Rows of B (3) \( \Rightarrow n=3 \). Result size \( 2 \times k \).
For \( CB \): Cols of C (p) = Rows of B (3) \( \Rightarrow p=3 \). Result size \( 2 \times k \).
Since \( n=3 \) and \( p=3 \) allows definition, look for options matching these constraints or variable relations. Actually, for \( AB \) to be defined: \( n=3 \). For \( CB \) to be defined: \( p=3 \). For addition \( AB+CB \), orders must match: \( 2 \times k = 2 \times k \) (Always true). However, usually this question involves specific variable matching like \( p=n \). Let's assume the question meant \( PY + WY \) style. In our case, \( n=3 \) is required.
For \( CB \): Cols of C (p) = Rows of B (3) \( \Rightarrow p=3 \). Result size \( 2 \times k \).
Since \( n=3 \) and \( p=3 \) allows definition, look for options matching these constraints or variable relations. Actually, for \( AB \) to be defined: \( n=3 \). For \( CB \) to be defined: \( p=3 \). For addition \( AB+CB \), orders must match: \( 2 \times k = 2 \times k \) (Always true). However, usually this question involves specific variable matching like \( p=n \). Let's assume the question meant \( PY + WY \) style. In our case, \( n=3 \) is required.
Answer depends on specific variable setup, typically \( n=k \) or similar.
Q22
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If \( n = p \), then the order of the matrix \( 7A - 5C \) (from Q21 data) is:
(A) \( p \times 2 \) (B) \( 2 \times n \) (C) \( n \times 3 \) (D) \( p \times n \)
(A) \( p \times 2 \) (B) \( 2 \times n \) (C) \( n \times 3 \) (D) \( p \times n \)
Matrix A is \( 2 \times n \). Matrix C is \( 2 \times p \).
Since \( n=p \), both are effectively \( 2 \times n \) matrices.
Subtracting them results in a matrix of the same order.
Since \( n=p \), both are effectively \( 2 \times n \) matrices.
Subtracting them results in a matrix of the same order.
Correct Option: (B) \( 2 \times n \)