Question 1: If A and B are symmetric matrices, prove that AB-BA is a skew‑symmetric matrix.
Solution:
Because A and B are symmetric, their transposes equal themselves: A' = A and B' = B.
Consider the transpose of AB-BA:
(AB-BA)' = B'A' - A'B'
(AB-BA)' = BA - AB
(AB-BA)' = -\,(AB-BA).
By definition, a matrix C is skew‑symmetric when C' = -C.
Hence AB-BA is skew‑symmetric.
Question 2: Show that B'AB is symmetric or skew‑symmetric according as A is symmetric or skew‑symmetric.
Solution:
Step 1 Recall the key definitions
- A square matrix A is symmetric ⇔ A' = A.
- A square matrix A is skew-symmetric ⇔ A' = -A.
- For any conformable matrices,
- (QP)' = Q'P' (transpose of a product reverses the order).
Take the transpose of B'AB
\bigl(B'AB\bigr)' = B'A'B
(Because latex‘ = B[/latex] and the order reverses.)
Case I When A is symmetric
If A is symmetric, then A' = A.
Substitute this into the expression above:
\bigl(B'AB\bigr)' = B'A'B = B'AB
The transpose equals the original matrix, so B'AB is symmetric.
Case II When A is skew-symmetric
If A is skew-symmetric, then A' = -A.
Substitute:
\bigl(B'AB\bigr)' = B'A'B = B'(-A)B = -(B'AB)
The transpose is the negative of the original matrix, so B'AB is skew-symmetric.
Final Conclusion
B'AB inherits its symmetry type directly from A:
- If A is symmetric → B'AB is symmetric.
- If A is skew-symmetric → B'AB is skew-symmetric.
Question 3: Find the values of x, y, z if the matrix
A = \begin{bmatrix}0 & 2y & z\\ x & y & -z\\ x & -y & z\end{bmatrix} satisfies A'A = I.
Solution:
Compute the transpose A':**
A' = \begin{bmatrix}0 & x & x\\ 2y & y & -y\\ z & -z & z\end{bmatrix}.
Form the product A'A = A = \begin{bmatrix}0 & x & x\\ 2y & y & -y\\ z & -z & z\end{bmatrix} \begin{bmatrix}0 & 2y & z\\ x & y & -z\\ x & -y & z\end{bmatrix}
We multiply row-by-column.
• Entry (1,1): 0^{2} + x^{2} + x^{2} = 2x^{2}.
• Entry (2,2): (2y)^{2} + y^{2} + (-y)^{2} = 4y^{2} + y^{2} + y^{2} = 6y^{2}.
• Entry (3,3): z^{2} + (-z)^{2} + z^{2} = 3z^{2}.
• A representative for non–diagonal entry, say (1,2):
0\cdot 2y + x\cdot y + x(-y) = xy - xy = 0.
A similar cancellation occurs for the other non–diagonal positions, so all are zero.
Thus
A'A = \begin{bmatrix}2x^{2} & 0 & 0\\ 0 & 6y^{2} & 0\\ 0 & 0 & 3z^{2}\end{bmatrix}.
Impose the identity condition A'A = I_{3}. Hence each diagonal element equals 1:
2x^{2} = 1 \;\Rightarrow\; x^{2} = \dfrac12.
6y^{2} = 1 \;\Rightarrow\; y^{2} = \dfrac16.
3z^{2} = 1 \;\Rightarrow\; z^{2} = \dfrac13.
Extract the values: Each variable can be positive or negative because the squares were equated.
\boxed{\displaystyle x = \pm\frac{1}{\sqrt2},\; y = \pm\frac{1}{\sqrt6},\; z = \pm\frac{1}{\sqrt3}.}
Question 4: Find the value(s) of x that satisfy
\begin{bmatrix}1 & 2 & 1\end{bmatrix}\begin{bmatrix}1 & 2 & 0\\2 & 0 & 1\\1 & 0 & 2\end{bmatrix}\begin{bmatrix}0\\2\\x\end{bmatrix} = 0.Solution:
Multiply the row vector with the 3×3 matrix:
\begin{bmatrix}1 & 2 & 1\end{bmatrix}\begin{bmatrix}1 & 2 & 0\\2 & 0 & 1\\1 & 0 & 2\end{bmatrix}= \begin{bmatrix}(1)(1)+(2)(2)+(1)(1) & (1)(2)+(2)(0)+(1)(0) & (1)(0)+(2)(1)+(1)(2)\end{bmatrix}
\begin{bmatrix}1 & 2 & 1\end{bmatrix}\begin{bmatrix}1 & 2 & 0\\2 & 0 & 1\\1 & 0 & 2\end{bmatrix} = \begin{bmatrix}6 & 2 & 4\end{bmatrix}.
Now Form the dot–product with the column vector:
\begin{bmatrix}6 & 2 & 4\end{bmatrix}\begin{bmatrix}0\\2\\x\end{bmatrix} = 6(0) + 2(2) + 4x = 4 + 4x.
Set the scalar result equal to 0 and solve:
4 + 4x = 0 \;\Longrightarrow\; x = -1.
Answer: \boxed{x = -1}
Question 5: Let A = \begin{bmatrix}3 & 1\\ -1 & 2\end{bmatrix}. Show that A^{2} - 5A + 7I = 0.
Solution:
Compute A^{2}:
A^{2} = A\cdot A= \begin{bmatrix}3 & 1\\ -1 & 2\end{bmatrix}\begin{bmatrix}3 & 1\\ -1 & 2\end{bmatrix}
A^{2} = \begin{bmatrix}3(3)+1(-1) & 3(1)+1(2)\\ -1(3)+2(-1) & -1(1)+2(2)\end{bmatrix}= \begin{bmatrix}8 & 5\\ -5 & 3\end{bmatrix}.
Compute each term of the expression:
-5A = -5\begin{bmatrix}3 & 1\\ -1 & 2\end{bmatrix}= \begin{bmatrix}-15 & -5\\ 5 & -10\end{bmatrix}.
7I = 7\begin{bmatrix}1 & 0\\ 0 & 1\end{bmatrix}= \begin{bmatrix}7 & 0\\ 0 & 7\end{bmatrix}.
Add the three matrices:
A^{2} - 5A + 7I= \begin{bmatrix}8 & 5\\ -5 & 3\end{bmatrix}+ \begin{bmatrix}-15 & -5\\ 5 & -10\end{bmatrix}+ \begin{bmatrix}7 & 0\\ 0 & 7\end{bmatrix}
A^{2} - 5A + 7I= \begin{bmatrix}8-15+7 & 5-5+0\\ -5+5+0 & 3-10+7\end{bmatrix}= \begin{bmatrix}0 & 0\\ 0 & 0\end{bmatrix}.
Because the result is the zero (null) matrix, we have
\boxed{A^{2} - 5A + 7I = O}.
Question 6: Find x if
\bigl[\,x\; -5\; -1\,\bigr]\begin{bmatrix}1 & 0 & 2\\0 & 2 & 1\\2 & 0 & 3\end{bmatrix}\begin{bmatrix}x\\4\\1\end{bmatrix}=0.Solution:
First multiply the row vector with the 3×3 matrix.
\begin{aligned}\bigl[\,x\; -5\; -1\,\bigr]\begin{bmatrix}1 & 0 & 2\\0 & 2 & 1\\2 & 0 & 3\end{bmatrix}&= \bigl[\,x(1)+(-5)(0)+(-1)(2),\; x(0)+(-5)(2)+(-1)(0),\; x(2)+(-5)(1)+(-1)(3)\bigr]\\[2pt]&\end{aligned}
\begin{aligned}\bigl[\,x\; -5\; -1\,\bigr]\begin{bmatrix}1 & 0 & 2\\0 & 2 & 1\\2 & 0 & 3\end{bmatrix}&=\bigl[\,x-2,\;-10,\;2x-8\bigr].\end{aligned}
Now multiply this row with the column vector \begin{bmatrix}x\\4\\1\end{bmatrix}:
\bigl[\,x-2,\;-10,\;2x-8\bigr].\begin{bmatrix}x\\4\\1\end{bmatrix}
(x-2)(x) + (-10)(4) + (2x-8)(1)= x^{2} - 2x - 40 + 2x - 8= x^{2} - 48.
Set the scalar result equal to 0: x^{2} - 48 = 0
⇒ x^{2} = 48 \;\Longrightarrow\; x = \pm 4\sqrt{3}.
\boxed{\displaystyle x = 4\sqrt3 \;\text{or}\; x = -4\sqrt3}
Question 7: A manufacturer sells three products x, y, z in two markets. Annual sales (units) are
S = \begin{bmatrix}10000 & 2000 & 18000\\6000 & 20000 & 8000\end{bmatrix}, where rows correspond to Market I and II, and columns to x,y,z.
Solution:
(a) Total revenue in each market (prices ₹ 2.50, 1.50, 1.00):
Let the price column vector be P = \begin{bmatrix}2.50\\1.50\\1.00\end{bmatrix}.
Revenue vector R = SP.
\begin{aligned}R &= \begin{bmatrix}10000 & 2000 & 18000\\6000 & 20000 & 8000\end{bmatrix}\begin{bmatrix}2.5\\1.5\\1.0\end{bmatrix}\\[4pt]&= \begin{bmatrix}10000(2.5)+2000(1.5)+18000(1.0)\\6000(2.5)+20000(1.5)+8000(1.0)\end{bmatrix}\\[4pt]&= \begin{bmatrix}46000\\53000\end{bmatrix}. \end{aligned}
Answer:
Revenue (Market I) = ₹ 46 000 and
Revenue (Market II) = ₹ 53 000.
(b) Gross profit given unit costs ₹ 2.00, 1.00, 0.50.
Cost vector C = \begin{bmatrix}2.00\\1.00\\0.50\end{bmatrix}.
Cost of goods sold \text{COGS} = SC.
\begin{aligned}\text{COGS} &= \begin{bmatrix}10000 & 2000 & 18000\\6000 & 20000 & 8000\end{bmatrix}\begin{bmatrix}2.0\\1.0\\0.5\end{bmatrix}\\[4pt]&= \begin{bmatrix}31000\\36000\end{bmatrix}. \end{aligned}
Gross profit vector G = R - \text{COGS} = \begin{bmatrix}15000\\17000\end{bmatrix}.
Answer:
Gross Profit (Market I) = ₹ 15 000 and
Gross Profit (Market II) = ₹ 17 000.
Question 8: Find matrix X such that
X\,\begin{bmatrix}1 & 2 & 3\\4 & 5 & 6\end{bmatrix}= \begin{bmatrix}-7 & -8 & -9\\2 & 4 & 6\end{bmatrix}.Solution:
Assume X = \begin{bmatrix}a & b\\ c & d\end{bmatrix} (a 2×2 matrix so that the product is defined).
Multiply:
X\begin{bmatrix}1 & 2 & 3\\4 & 5 & 6\end{bmatrix}= \begin{bmatrix}a+4b & 2a+5b & 3a+6b\\ c+4d & 2c+5d & 3c+6d\end{bmatrix}.
Set this equal to the given matrix and equate corresponding entries.
\begin{bmatrix}a+4b & 2a+5b & 3a+6b\\ c+4d & 2c+5d & 3c+6d\end{bmatrix} = \begin{bmatrix}-7 & -8 & -9\\2 & 4 & 6\end{bmatrix}.
First row equations
\begin{cases}a + 4b = -7\\[4pt] 2a + 5b = -8\end{cases}
Solving gives a = 1,\; b = -2.
The third first‑row equation 3a + 6b = -9 is automatically satisfied.
Second row equations
\begin{cases}c + 4d = 2\\[4pt] 2c + 5d = 4\end{cases}
Solving gives c = 2,\; d = 0. (Again, the third equation checks automatically.)
Answer: \boxed{X = \begin{bmatrix}1 & -2\\2 & 0\end{bmatrix}.}
Question 9: If A = \begin{bmatrix}\alpha & \beta\\ \gamma & -\alpha\end{bmatrix} satisfies A^{2}=I, which relation is true?
(A) 1 + \alpha^{2} + \beta\gamma = 0
(B) 1 - \alpha^{2} + \beta\gamma = 0
(C) 1 - \alpha^{2} - \beta\gamma = 0
(D) 1 + \alpha^{2} - \beta\gamma = 0
Solution:
Compute A^{2}:
A^{2} = \begin{bmatrix}\alpha & \beta\\ \gamma & -\alpha\end{bmatrix}\begin{bmatrix}\alpha & \beta\\ \gamma & -\alpha\end{bmatrix}= \begin{bmatrix}\alpha^{2}+\beta\gamma & \alpha\beta-\alpha\beta\\[2pt]\gamma\alpha-\alpha\gamma & \gamma\beta+\alpha^{2}\end{bmatrix}= \begin{bmatrix}\alpha^{2}+\beta\gamma & 0\\ 0 & \alpha^{2}+\beta\gamma\end{bmatrix}.
Equate to A^{2} = I
\begin{bmatrix}\alpha^{2}+\beta\gamma & 0\\ 0 & \alpha^{2}+\beta\gamma\end{bmatrix} = \begin{bmatrix}1&0\\0&1\end{bmatrix}
\alpha^{2}+\beta\gamma = 1 (from each diagonal entry).
Rearrange to match the options:
1 - \alpha^{2} - \beta\gamma = 0, which is option (C).
Answer: C
Question 10: If matrix A is simultaneously symmetric and skew‑symmetric, then –
(A) A is a diagonal matrix
(B) A is a zero matrix
(C) A is a square matrix
(D) None of these
Solution:
• Symmetric ⇒ A' = A.
• Skew‑symmetric ⇒ A' = -A.
Equating: A = -A \;\Rightarrow\; 2A = 0 \;\Rightarrow\; A = 0.
Thus A must be the zero matrix. Option (B) is correct.
Answer: B
Question 11: If A is a square matrix such that A^{2}=A. Then (I+A)^{3} - 7A is equal to
(A) A (B) I - A (C) I (D) 3A
Solution:
Expand using the binomial theorem (matrices commute with identity):
(I+A)^{3} = I + 3.I^{2}.A + 3A^{2}.I + A^{3}.
Replace higher powers using A^{2}=A:
A^{2}=A ⇒ A^{3}=A^{2}\!\cdot\!A = A\cdot A = A.
Hence (I+A)^{3} = I + 3A + 3A + A = I + 7A.
Subtract 7A:
(I+A)^{3} - 7A = (I + 7A) - 7A = I.
Therefore the expression equals the identity matrix and option (C) is correct.
Answer: B