Miscellaneous Exercise Practice
Matrices: Proofs, Mathematical Induction, & Applications
Q1
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Let \( A = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \). Show that \( (pI + qA)^n = p^n I + n p^{n-1} q A
\), where \( I \) is the identity matrix of order 2 and \( n \in N \).
Base Case (n=1)
LHS: \( (pI + qA)^1 = pI + qA \)
RHS: \( p^1 I + 1 p^0 q A = pI + qA \)
True for n=1.
RHS: \( p^1 I + 1 p^0 q A = pI + qA \)
True for n=1.
Inductive Step
Assume true for n=k: \( (pI + qA)^k = p^k I + k p^{k-1} q A \).
For n=k+1:
\( (pI + qA)^{k+1} = (pI + qA)^k (pI + qA) \)
\( = (p^k I + k p^{k-1} q A)(pI + qA) \)
\( = p^{k+1} I + p^k q A + k p^k q A + k p^{k-1} q^2 A^2 \)
Note \( A^2 = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O \).
So last term vanishes.
\( = p^{k+1} I + (k+1) p^k q A \).
For n=k+1:
\( (pI + qA)^{k+1} = (pI + qA)^k (pI + qA) \)
\( = (p^k I + k p^{k-1} q A)(pI + qA) \)
\( = p^{k+1} I + p^k q A + k p^k q A + k p^{k-1} q^2 A^2 \)
Note \( A^2 = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O \).
So last term vanishes.
\( = p^{k+1} I + (k+1) p^k q A \).
Verified by PMI for all \( n \in N \).
Q2
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If \( A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix} \), prove that \( A^n = 3^{n-1} A
\) for every positive integer n.
Base Case (n=1)
\( A^1 = 3^0 A = 1 \cdot A = A \). True.
Induction
Assume \( A^k = 3^{k-1} A \).
\( A^{k+1} = A^k \cdot A = 3^{k-1} A \cdot A = 3^{k-1} A^2 \).
Calculate \( A^2 \):
Element (1,1): \( 1(1)+1(1)+1(1) = 3 \). All elements are 3.
So \( A^2 = 3A \).
Substitute back: \( 3^{k-1} (3A) = 3^k A \).
\( A^{k+1} = A^k \cdot A = 3^{k-1} A \cdot A = 3^{k-1} A^2 \).
Calculate \( A^2 \):
Element (1,1): \( 1(1)+1(1)+1(1) = 3 \). All elements are 3.
So \( A^2 = 3A \).
Substitute back: \( 3^{k-1} (3A) = 3^k A \).
Verified: \( A^n = 3^{n-1} A \).
Q3
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If \( A \) and \( B \) are symmetric matrices, show that \( AB - BA \) is a skew-symmetric matrix.
Given
\( A' = A \) and \( B' = B \).
Transpose
Let \( X = AB - BA \).
\( X' = (AB - BA)' = (AB)' - (BA)' \)
\( = B'A' - A'B' \) (Reversal Law)
\( = BA - AB \) (Using symmetry)
\( = -(AB - BA) = -X \).
\( X' = (AB - BA)' = (AB)' - (BA)' \)
\( = B'A' - A'B' \) (Reversal Law)
\( = BA - AB \) (Using symmetry)
\( = -(AB - BA) = -X \).
\( X' = -X \), so Skew-Symmetric.
Q4
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Find the matrix \( A \) such that \( \begin{bmatrix} 2 & 1 \\ 3 & 2 \end{bmatrix} A \begin{bmatrix} -3 & 2 \\
5 & -3 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \).
Concept
Let \( P A Q = I \). Then \( A = P^{-1} I Q^{-1} = P^{-1} Q^{-1} \).
Find P Inverse
\( P = \begin{bmatrix} 2 & 1 \\ 3 & 2 \end{bmatrix} \). Det(P) = 4-3 = 1.
\( P^{-1} = \begin{bmatrix} 2 & -1 \\ -3 & 2 \end{bmatrix} \).
\( P^{-1} = \begin{bmatrix} 2 & -1 \\ -3 & 2 \end{bmatrix} \).
Find Q Inverse
\( Q = \begin{bmatrix} -3 & 2 \\ 5 & -3 \end{bmatrix} \). Det(Q) = 9-10 = -1.
Adj(Q) = \( \begin{bmatrix} -3 & -2 \\ -5 & -3 \end{bmatrix} \).
\( Q^{-1} = -1 \begin{bmatrix} -3 & -2 \\ -5 & -3 \end{bmatrix} = \begin{bmatrix} 3 & 2 \\ 5 & 3 \end{bmatrix} \).
Adj(Q) = \( \begin{bmatrix} -3 & -2 \\ -5 & -3 \end{bmatrix} \).
\( Q^{-1} = -1 \begin{bmatrix} -3 & -2 \\ -5 & -3 \end{bmatrix} = \begin{bmatrix} 3 & 2 \\ 5 & 3 \end{bmatrix} \).
Multiply
\( A = \begin{bmatrix} 2 & -1 \\ -3 & 2 \end{bmatrix} \begin{bmatrix} 3 & 2 \\ 5 & 3 \end{bmatrix} =
\begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix} \).
\( A = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix} \)
Q5
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If \( A = \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix} \), using Mathematical Induction prove that \( A^n =
\begin{bmatrix} 1+2n & -4n \\ n & 1-2n \end{bmatrix} \).
Base Case n=1
LHS: A. RHS: \( \begin{bmatrix} 1+2 & -4 \\ 1 & 1-2 \end{bmatrix} = \begin{bmatrix} 3 & -4 \\ 1 & -1
\end{bmatrix} \). True.
Inductive Step
\( A^{k+1} = \begin{bmatrix} 1+2k & -4k \\ k & 1-2k \end{bmatrix} \begin{bmatrix} 3 & -4 \\ 1 & -1
\end{bmatrix} \)
(1,1): \( (1+2k)3 + (-4k)1 = 3+6k-4k = 1+2(k+1) \).
Matches form.
(1,1): \( (1+2k)3 + (-4k)1 = 3+6k-4k = 1+2(k+1) \).
Matches form.
Proved for all \( n \in N \).
Q6
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Find \( x \), if \( \begin{bmatrix} x & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 2 & 5 \end{bmatrix}
\begin{bmatrix} 2 \\ 5 \end{bmatrix} = 0 \). (Order changed to 2x2 for practice).
First Product
\( [x+2 \quad 2x+5] \).
Second Product
\( [x+2 \quad 2x+5] \begin{bmatrix} 2 \\ 5 \end{bmatrix} = [2(x+2) + 5(2x+5)] \)
\( = [2x+4 + 10x+25] = [12x + 29] \).
\( = [2x+4 + 10x+25] = [12x + 29] \).
Solve
\( 12x + 29 = 0 \Rightarrow x = -29/12 \).
\( x = -29/12 \)
Q7
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A manufacturer sells products P and Q in two cities A and B. Sales: City A (500 P, 2000 Q), City B (3000 P,
1000 Q). Sale prices: P=₹5, Q=₹10. Find total revenue in each city.
Matrices
Quantity \( S = \begin{bmatrix} 500 & 2000 \\ 3000 & 1000 \end{bmatrix} \). Price \( P = \begin{bmatrix} 5
\\ 10 \end{bmatrix} \).
Calculation
\( R = S \times P = \begin{bmatrix} 500(5)+2000(10) \\ 3000(5)+1000(10) \end{bmatrix} = \begin{bmatrix}
2500+20000 \\ 15000+10000 \end{bmatrix} \).
City A: ₹22,500. City B: ₹25,000.
Q8
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Find matrix \( X \) such that \( X \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 3 & 5 \\ 5 &
9 \end{bmatrix} \).
Logic
Let \( A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \) and \( B = \begin{bmatrix} 3 & 5 \\ 5 & 9
\end{bmatrix} \).
\( X A = B \Rightarrow X = B A^{-1} \).
\( X A = B \Rightarrow X = B A^{-1} \).
Inverse of A
Det = 4-6 = -2.
\( A^{-1} = \frac{-1}{2} \begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix} \).
\( A^{-1} = \frac{-1}{2} \begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix} \).
Solve X
\( X = \frac{-1}{2} \begin{bmatrix} 3 & 5 \\ 5 & 9 \end{bmatrix} \begin{bmatrix} 4 & -2 \\ -3 & 1
\end{bmatrix} \)
\( = \frac{-1}{2} \begin{bmatrix} 12-15 & -6+5 \\ 20-27 & -10+9 \end{bmatrix} = \frac{-1}{2} \begin{bmatrix} -3 & -1 \\ -7 & -1 \end{bmatrix} \).
\( = \frac{-1}{2} \begin{bmatrix} 12-15 & -6+5 \\ 20-27 & -10+9 \end{bmatrix} = \frac{-1}{2} \begin{bmatrix} -3 & -1 \\ -7 & -1 \end{bmatrix} \).
\( X = \begin{bmatrix} 1.5 & 0.5 \\ 3.5 & 0.5 \end{bmatrix} \)
Q9
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If \( A = \begin{bmatrix} \alpha & 0 \\ 1 & 1 \end{bmatrix} \) and \( B = \begin{bmatrix} 1 & 0 \\ 5 & 1
\end{bmatrix} \) are such that \( A^2 = B \), then \( \alpha \) is:
(A) 1 (B) -1 (C) 4 (D) No real value
(A) 1 (B) -1 (C) 4 (D) No real value
Calculate A²
\( \begin{bmatrix} \alpha & 0 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} \alpha & 0 \\ 1 & 1 \end{bmatrix} =
\begin{bmatrix} \alpha^2 & 0 \\ \alpha+1 & 1 \end{bmatrix} \).
Equate to B
\( \alpha^2 = 1 \Rightarrow \alpha = \pm 1 \).
\( \alpha + 1 = 5 \Rightarrow \alpha = 4 \).
No common value exists.
\( \alpha + 1 = 5 \Rightarrow \alpha = 4 \).
No common value exists.
Correct Option: (D) No real value
Q10
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If \( A \) is a symmetric matrix, then \( A^3 \) is:
(A) Skew symmetric (B) Symmetric (C) Identity (D) Zero
(A) Skew symmetric (B) Symmetric (C) Identity (D) Zero
Check Transpose
Given \( A' = A \).
\( (A^3)' = (AAA)' = A'A'A' = AAA = A^3 \).
\( (A^3)' = (AAA)' = A'A'A' = AAA = A^3 \).
Correct Option: (B) Symmetric
Q11
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If \( A^2 = I \), then \( (A-I)^3 + (A+I)^3 - 7A \) is equal to:
(A) \( A \) (B) \( I-A \) (C) \( I+A \) (D) \( 3A \)
(A) \( A \) (B) \( I-A \) (C) \( I+A \) (D) \( 3A \)
Expand
\( (A^3 - 3A^2I + 3AI^2 - I^3) + (A^3 + 3A^2I + 3AI^2 + I^3) - 7A \)
Simplify terms (\( A^2=I \implies A^3=A \)):
\( (A - 3I + 3A - I) + (A + 3I + 3A + I) - 7A \)
\( = 4A - 4I + 4A + 4I - 7A = 8A - 7A = A \).
Simplify terms (\( A^2=I \implies A^3=A \)):
\( (A - 3I + 3A - I) + (A + 3I + 3A + I) - 7A \)
\( = 4A - 4I + 4A + 4I - 7A = 8A - 7A = A \).
Correct Option: (A) \( A \)