JEE Mathematics Matrices and Determinants MCQs Set 04

Question. If \( \begin{vmatrix} 1 & a^2 & a^4 \\ 1 & b^2 & b^4 \\ 1 & c^2 & c^4 \end{vmatrix} = k \begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{vmatrix} \) then k is
(a) (a + b) (b + c) (c + a)
(b) ab + bc + ac
(c) \( a^2b^2c^2 \)
(d) \( a^2 + b^2 + c^2 \)
Answer: (a) (a + b) (b + c) (c + a)
Solution: The determinant on the left expands to \( (b^2 - a^2) (c^2 - a^2) (c^2 - b^2) \)
The determinant on the right expands to \( (b - a)(c - a)(c - b) \)
\( (b - a)(b + a)(c - a)(c + a)(c - b)(c + b) = k(b - a)(c - a)(c - b) \)
\( k = (a + b) (b + c) (c + a) \)

Question. If a \(\neq\) b, then the system of equations ax+by+bz=0, bx + ay + bz = 0, bx + by + ax = 0 will have a non-trivial solution if
(a) a + b = 0
(b) a + 2b = 0
(c) 2a + b = 0
(d) a + 4b = 0
Answer: (b) a + 2b = 0
Solution: Given \( a \neq b \)
ax + by + bz = 0
bx + ay + bz = 0
bx + by + ax = 0
For non trivial solution
\( D = 0 \Rightarrow \begin{vmatrix} a & b & b \\ b & a & b \\ b & b & a \end{vmatrix} = 0 \)
\( C_1 \rightarrow C_1 + C_2 + C_3 \)
\( \Rightarrow (a + 2b) \begin{vmatrix} 1 & b & b \\ 1 & a & b \\ 1 & b & a \end{vmatrix} = 0 \)
\( R_2 \rightarrow R_2 - R_1 \) & \( R_3 \rightarrow R_3 - R_1 \)
\( \Rightarrow (a + 2b) \begin{vmatrix} 1 & b & b \\ 0 & a - b & 0 \\ 0 & 0 & a - b \end{vmatrix} = 0 \Rightarrow a + 2b = 0 \)

Question. Value of \( \Delta = \begin{vmatrix} \sin(2\alpha) & \sin(\alpha+\beta) & \sin(\alpha+\gamma) \\ \sin(\beta+\alpha) & \sin(2\beta) & \sin(\gamma+\beta) \\ \sin(\gamma+\alpha) & \sin(\gamma+\beta) & \sin(2\gamma) \end{vmatrix} \) is
(a) \( \Delta = 0 \)
(b) \( \Delta = \sin^2\alpha + \sin^2\beta + \sin^2\gamma \)
(c) \( \Delta = 3/2 \)
(d) None of the options
Answer: (a) \( \Delta = 0 \)
Solution: Expanding sine of sums:
\( \Delta = \begin{vmatrix} \sin\alpha\cos\alpha + \cos\alpha\sin\alpha & \sin\alpha\cos\beta + \cos\alpha\sin\beta & \sin\alpha\cos\gamma + \cos\alpha\sin\gamma \\ \sin\beta\cos\alpha + \cos\beta\sin\alpha & \sin\beta\cos\beta + \sin\beta\cos\beta & \sin\gamma\cos\beta + \cos\gamma\sin\beta \\ \sin\gamma\cos\alpha + \cos\gamma\sin\alpha & \sin\gamma\cos\beta + \cos\gamma\sin\beta & \sin\gamma\cos\gamma + \sin\gamma\cos\gamma \end{vmatrix} \)
Which factors into the product of two matrices:
\( = \begin{vmatrix} \sin\alpha & \cos\alpha & 0 \\ \sin\beta & \cos\beta & 0 \\ \sin\gamma & \cos\gamma & 0 \end{vmatrix} \times \begin{vmatrix} \cos\alpha & \sin\alpha & 0 \\ \cos\beta & \sin\beta & 0 \\ \cos\gamma & \sin\gamma & 0 \end{vmatrix} = 0 \times 0 = 0 \)

Question. The determinant \( D = \begin{vmatrix} a^2(1 + x) & ab & ac \\ ab & b^2(1 + x) & bc \\ ac & bc & c^2(1 + x) \end{vmatrix} \) is divisible by
(a) 1 + x
(b) \( (1 + x)^2 \)
(c) \( x^2 \)
(d) \( x^2 + 1 \)
Answer: (c) \( x^2 \)
Solution: \( D = \begin{vmatrix} a^2(1 + x) & ab & ac \\ ab & b^2(1 + x) & bc \\ ac & bc & c^2(1 + x) \end{vmatrix} \)
Take common a,b,c from R1, R2, R3 and multiply to C1, C2, C3 respectively:
\( = abc \begin{vmatrix} a(1 + x) & a & a \\ b & b(1 + x) & b \\ c & c & c(1 + x) \end{vmatrix} \)
\( = a^2b^2c^2 \begin{vmatrix} 1 + x & 1 & 1 \\ 1 & 1 + x & 1 \\ 1 & 1 & 1 + x \end{vmatrix} \)
\( C_1 \rightarrow C_1 + C_2 + C_3 \);
\( = a^2b^2c^2 (3 + x) \begin{vmatrix} 1 & 1 & 1 \\ 1 & 1 + x & 1 \\ 1 & 1 & 1 + x \end{vmatrix} \)
\( R_2 \rightarrow R_2 - R_1 \) & \( R_3 \rightarrow R_3 - R_1 \);
\( = a^2b^2c^2 (x + 3) \begin{vmatrix} 1 & 1 & 1 \\ 0 & x & 0 \\ 0 & 0 & x \end{vmatrix} \Rightarrow a^2b^2c^2 x^2 (x + 3) \)
Thus it is divisible by \( x^2 \).

Question. If A, B, C are angles of a triangle ABC, then \( \begin{vmatrix} \sin\frac{A}{2} & \sin\frac{B}{2} & \sin\frac{C}{2} \\ \sin(A + B + C) & \sin\frac{B}{2} & \sin\frac{A}{2} \\ \cos\frac{A+B+C}{2} & \tan(A + B + C) & \sin\frac{C}{2} \end{vmatrix} \) is less than or equal to
(a) \( \frac{3\sqrt{3}}{8} \)
(b) \( \frac{1}{8} \)
(c) \( 2\sqrt{2} \)
(d) 2
Answer: (b) \( \frac{1}{8} \)
Solution: \( A + B + C = \pi \)
\( \sin(A + B + C) = \sin(\pi) = 0 \)
\( \cos(\frac{A + B + C}{2}) = \cos(\pi/2) = 0 \)
\( \tan(A + B + C) = \tan(\pi) = 0 \)
Substitute these into the determinant:
\( D = \begin{vmatrix} \sin\frac{A}{2} & \sin\frac{B}{2} & \sin\frac{C}{2} \\ 0 & \sin\frac{B}{2} & \sin\frac{A}{2} \\ 0 & 0 & \sin\frac{C}{2} \end{vmatrix} \)
The matrix is upper triangular, so the determinant is the product of diagonal elements.
\( D = \sin\frac{A}{2} \sin\frac{B}{2} \sin\frac{C}{2} \)
For any triangle, the maximum value of \( \sin\frac{A}{2} \sin\frac{B}{2} \sin\frac{C}{2} \) is achieved when the triangle is equilateral (\( A = B = C = \pi/3 \)).
\( D \le \sin(\frac{\pi}{6}) \sin(\frac{\pi}{6}) \sin(\frac{\pi}{6}) = (\frac{1}{2})(\frac{1}{2})(\frac{1}{2}) = \frac{1}{8} \)

Question. Let \( f(x) = \begin{vmatrix} 1+\sin^2x & \cos^2x & 4\sin 2x \\ \sin^2x & 1+\cos^2x & 4\sin 2x \\ \sin^2x & \cos^2x & 1+4\sin 2x \end{vmatrix} \) then the maximum value of f(x) is
(a) 4
(b) 6
(c) 8
(d) 12
Answer: (b) 6
Solution: \( C_1 \rightarrow C_1 + C_2 \)
\( f(x) = \begin{vmatrix} 2 & \cos^2x & 4\sin 2x \\ 2 & 1+\cos^2x & 4\sin 2x \\ 1 & \cos^2x & 1+4\sin 2x \end{vmatrix} \)
\( R_2 \rightarrow R_2 - R_1 \) & \( R_3 \rightarrow R_3 - \frac{1}{2} R_1 \)
\( f(x) = \begin{vmatrix} 2 & \cos^2x & 4\sin 2x \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix} \)
(Wait, the simpler row ops is \( R_2 \rightarrow R_2 - R_1 \) and \( R_3 \rightarrow R_3 - R_1 \)):
Let's first do \( C_1 \rightarrow C_1 + C_2 + C_3 \) from original? No, original soln uses \( C_1 \rightarrow C_1 + C_2 \). Then we have 2, 2, 1 in column 1.
From the solution provided: \( f(x) = \begin{vmatrix} 1 & \cos^2x & 4\sin 2x \\ 1 & 1+\cos^2x & 4\sin 2x \\ 1 & \cos^2x & 1+4\sin 2x \end{vmatrix} \) when taking out (2+4sin2x)? No, solution says \( f(x) = 2 + 4 \sin 2x \).
Wait, expanding \( \begin{vmatrix} 1+\sin^2x & \cos^2x & 4\sin 2x \\ \sin^2x & 1+\cos^2x & 4\sin 2x \\ \sin^2x & \cos^2x & 1+4\sin 2x \end{vmatrix} \):
\( R_1 \rightarrow R_1 - R_2 \) gives \( \begin{vmatrix} 1 & -1 & 0 \\ \sin^2x & 1+\cos^2x & 4\sin 2x \\ \sin^2x & \cos^2x & 1+4\sin 2x \end{vmatrix} \)
\( R_2 \rightarrow R_2 - R_3 \) gives \( \begin{vmatrix} 1 & -1 & 0 \\ 0 & 1 & -1 \\ \sin^2x & \cos^2x & 1+4\sin 2x \end{vmatrix} \)
Expanding: \( 1(1 + 4\sin 2x + \cos^2x) - (-1)(0 - (-\sin^2x)) = 1 + 4\sin 2x + \cos^2x + \sin^2x = 2 + 4\sin 2x \).
\( f(x) = 2 + 4 \sin 2x \)
Maximum value of \( \sin 2x \) is 1, so the maximum value of f(x) is 2 + 4(1) = 6.

Question. Value of the \( D = \begin{vmatrix} a^3 - x & a^4 - x & a^5 - x \\ a^5 - x & a^6 - x & a^7 - x \\ a^7 - x & a^8 - x & a^9 - x \end{vmatrix} \) is
(a) 0
(b) \( (a^3 - 1) (a^6 - 1) (a^9 - 1) \)
(c) \( (a^3 + 1) (a^6 + 1) (a^9 + 1) \)
(d) \( a^{15} - 1 \)
Answer: (a) 0
Solution: \( R_2 \rightarrow R_2 - R_1 \) & \( R_3 \rightarrow R_3 - R_1 \)
\( D = \begin{vmatrix} a^3 - x & a^4 - x & a^5 - x \\ a^5 - a^3 & a^6 - a^4 & a^7 - a^5 \\ a^7 - a^3 & a^8 - a^4 & a^9 - a^5 \end{vmatrix} \)
\( D = \begin{vmatrix} a^3 - x & a^4 - x & a^5 - x \\ a^2(a^3 - a) & a^2(a^4 - a^2) & a^2(a^5 - a^3) \\ a^4(a^3 - a^{-1}) & a^4(a^4 - 1) & a^4(a^5 - a) \end{vmatrix} \)
Actually, pulling common factors from rows:
\( R_2 = a^2(a^3 - a), a^4(a^2 - 1) \dots \)
The simpler way: Since the powers are in geometric progression. Rows 2 and 3 are linearly dependent after subtraction. Let's see: \( R_2 - R_1 = a^3(a^2 - 1), a^4(a^2 - 1), a^5(a^2 - 1) \). So row 2 is proportional to \( [a^3, a^4, a^5] \).
\( R_3 - R_1 = a^3(a^4 - 1), a^4(a^4 - 1), a^5(a^4 - 1) \). So row 3 is proportional to \( [a^3, a^4, a^5] \).
Since two rows are proportional, the determinant is 0.

Question. If \( f(x) = \begin{vmatrix} a^{-x} & e^{x/\ln a} & x^2 \\ a^{-3x} & e^{3x/\ln a} & x^4 \\ a^{-5x} & e^{5x/\ln a} & 1 \end{vmatrix} \), then
(a) f(x) - f(- x) = 0
(b) f(x) . f(-x) = 0
(c) f(x) + f(-x) = 0
(d) f(x) = f(-x) = 0
Answer: (c) f(x) + f(-x) = 0
Solution: Notice that \( e^{kx/\ln a} = e^{kx \log_e e / \ln a} = \dots \) wait. The expression is \( e^{x \ln a} \)? OCR shows `x/na`, but math context implies \( x \ln a \), because \( e^{kx \ln a} = (e^{\ln a})^{kx} = a^{kx} \). Let's assume the term is \( a^x \).
\( f(x) = \begin{vmatrix} a^{-x} & a^x & x^2 \\ a^{-3x} & a^{3x} & x^4 \\ a^{-5x} & a^{5x} & 1 \end{vmatrix} \)
\( f(-x) = \begin{vmatrix} a^x & a^{-x} & x^2 \\ a^{3x} & a^{-3x} & x^4 \\ a^{5x} & a^{-5x} & 1 \end{vmatrix} \)
Swapping \( C_1 \) and \( C_2 \) in \( f(-x) \) introduces a negative sign:
\( f(-x) = - \begin{vmatrix} a^{-x} & a^x & x^2 \\ a^{-3x} & a^{3x} & x^4 \\ a^{-5x} & a^{5x} & 1 \end{vmatrix} = -f(x) \)
Thus, \( f(x) + f(-x) = 0 \).

Question. \( D = \begin{vmatrix} 1 & \frac{4\sin B}{b} & \cos A \\ 2a & 8\sin A & 1 \\ 3a & 12\sin A & \cos B \end{vmatrix} \) is (where a, b, c are the sides opposite to angles A, B, C respectively in a triangle)
(a) \( \frac{1}{2}\cos 2A \)
(b) 0
(c) \( \frac{1}{2}\sin 2A \)
(d) \( \frac{1}{2}(\cos^2A + \cos^2B) \)
Answer: (b) 0
Solution: From the sine rule in a triangle, \( \frac{a}{\sin A} = \frac{b}{\sin B} \Rightarrow \frac{\sin A}{a} = \frac{\sin B}{b} = k \)
\( D = \begin{vmatrix} 1 & 4k & \cos A \\ 2a & 8ak & 1 \\ 3a & 12ak & \cos B \end{vmatrix} \)
Taking out 4k from \( C_2 \):
\( D = 4k \begin{vmatrix} 1 & 1 & \cos A \\ 2a & 2a & 1 \\ 3a & 3a & \cos B \end{vmatrix} \)
Since Column 1 and Column 2 are identical, \( D = 0 \).

Question. If \( \Delta_1 = \begin{vmatrix} 2a & b & e \\ 2d & e & f \\ 4x & 2y & 2z \end{vmatrix} \), \( \Delta_2 = \begin{vmatrix} f & 2d & e \\ 2z & 4x & 2y \\ e & 2a & b \end{vmatrix} \), then the value of \( \Delta_1 - \Delta_2 \) is
(a) \( x + y/2 + z \)
(b) 2
(c) 0
(d) 3
Answer: (c) 0
Solution: \( \Delta_1 = \begin{vmatrix} 2a & b & e \\ 2d & e & f \\ 4x & 2y & 2z \end{vmatrix} \)
Taking 2 out of \( C_1 \) and 2 out of \( R_3 \):
\( \Delta_1 = 4 \begin{vmatrix} a & b & e \\ d & e & f \\ x & y & z \end{vmatrix} \)
For \( \Delta_2 = \begin{vmatrix} f & 2d & e \\ 2z & 4x & 2y \\ e & 2a & b \end{vmatrix} \), take 2 out of \( C_2 \) and 2 out of \( R_2 \):
\( \Delta_2 = 4 \begin{vmatrix} f & d & e \\ z & x & y \\ e & a & b \end{vmatrix} \)
Applying \( C_1 \leftrightarrow C_3 \) (introducing a negative sign), then \( C_1 \leftrightarrow C_2 \) (introducing another negative sign):
\( \Delta_2 = 4 \begin{vmatrix} d & e & f \\ x & y & z \\ a & b & e \end{vmatrix} \)
Now, swap \( R_1 \leftrightarrow R_3 \) (negative sign), then \( R_2 \leftrightarrow R_3 \) (negative sign):
\( \Delta_2 = 4 \begin{vmatrix} a & b & e \\ d & e & f \\ x & y & z \end{vmatrix} = \Delta_1 \)
Therefore, \( \Delta_1 - \Delta_2 = 0 \).

Question. If \( \begin{vmatrix} (b + c)^2 & a^2 & a^2 \\ b^2 & (c + a)^2 & b^2 \\ c^2 & c^2 & (a + b)^2 \end{vmatrix} = k abc(a + b + c)^3 \), then k is
(a) 1
(b) 2
(c) 0
(d) ab + bc + ac
Answer: (b) 2
Solution: Applying \( C_2 \rightarrow C_2 - C_1 \) and \( C_3 \rightarrow C_3 - C_1 \)
\( = \begin{vmatrix} (b + c)^2 & a^2 - (b + c)^2 & a^2 - (b + c)^2 \\ b^2 & (c + a)^2 - b^2 & 0 \\ c^2 & 0 & (a + b)^2 - c^2 \end{vmatrix} \)
Take out (a + b + c) common from \( C_2 \) and \( C_3 \)
\( = (a + b + c)^2 \begin{vmatrix} (b + c)^2 & a - b - c & a - b - c \\ b^2 & c + a - b & 0 \\ c^2 & 0 & a + b - c \end{vmatrix} \)
Applying \( R_1 \rightarrow R_1 - (R_2 + R_3) \)
\( = (a + b + c)^2 \begin{vmatrix} 2bc & -2c & -2b \\ b^2 & c + a - b & 0 \\ c^2 & 0 & a + b - c \end{vmatrix} \)
Applying \( C_2 \rightarrow C_2 + \frac{1}{b} C_1 \) and \( C_3 \rightarrow C_3 + \frac{1}{c} C_1 \)
\( = (a + b + c)^2 \begin{vmatrix} 2bc & 0 & 0 \\ b^2 & c + a + b^2/c & b^2/c \\ c^2 & c^2/b & a + b + c^2/b \end{vmatrix} \)
Expanding along \( R_1 \):
\( = 2bc (a + b + c)^2 [(a + c)(a + b) - bc] \)
\( = 2bc (a + b + c)^2 (a^2 + ab + ac) = 2abc(a + b + c)^3 \)
Hence, k = 2.

Question. If \( U_n = \begin{vmatrix} n & 1 & 5 \\ n^2 & 2N + 1 & 2N + 1 \\ n^3 & 3N^2 & 3N + 1 \end{vmatrix} \), then \( \sum_{n=1}^N U_n \) is equal to
(a) \( 2 \sum_{n=1}^N n \)
(b) \( 2 \sum_{n=1}^N n^2 \)
(c) \( \frac{1}{2} \sum_{n=1}^N n^2 \)
(d) 0
Answer: (b) \( 2 \sum_{n=1}^N n^2 \)
Solution: \( \sum_{n=1}^N U_n = \begin{vmatrix} \sum n & 1 & 5 \\ \sum n^2 & 2N + 1 & 2N + 1 \\ \sum n^3 & 3N^2 & 3N + 1 \end{vmatrix} \)
\( = \begin{vmatrix} \frac{N(N+1)}{2} & 1 & 5 \\ \frac{N(N+1)(2N+1)}{6} & 2N + 1 & 2N + 1 \\ (\frac{N(N+1)}{2})^2 & 3N^2 & 3N + 1 \end{vmatrix} \)
Taking factors common and applying row/column operations simplifies to:
\( \sum_{n=1}^N U_n = \frac{N(N + 1)(2N + 1)}{3} = 2 \frac{N(N + 1)(2N + 1)}{6} = 2 \sum_{n=1}^N n^2 \)

Question. Matrix \( \begin{bmatrix} a & b & (a\alpha - b) \\ b & c & (b\alpha - c) \\ 2 & 1 & 0 \end{bmatrix} \) is non invertible if
(a) \( \alpha = 1/2 \)
(b) a, b, c are in A.P.
(c) a, b, c are in G.P.
(d) a, b, c are in H.P.
Answer: (a), (c)
Solution:
Let \( A = \begin{bmatrix} a & b & (a\alpha - b) \\ b & c & (b\alpha - c) \\ 2 & 1 & 0 \end{bmatrix} \)
\( |A| = 0 \)
\( \begin{vmatrix} a & b & a\alpha - b \\ b & c & b\alpha - c \\ 2 & 1 & 0 \end{vmatrix} = 0 \)
\( C_3 \rightarrow C_3 - \alpha C_1 + C_2 \)
\( \begin{vmatrix} a & b & a\alpha \\ b & c & b\alpha \\ 2 & 1 & 1 \end{vmatrix} = 0 \)
\( \Rightarrow a\{c - b\alpha\} - b\{b - 2b\alpha\} + a\alpha (b - 2c) = 0 \)
\( \Rightarrow ac - ab\alpha - b^2 + 2b^2\alpha + ab\alpha - 2ac\alpha = 0 \)
\( \Rightarrow (ac - b^2) - 2\alpha (ac - b^2) = 0 \)
\( \Rightarrow (1 - 2\alpha) (ac - b^2) = 0 \)
either \( 1 - 2\alpha = 0 \) or \( ac - b^2 = 0 \)
\( \alpha = \frac{1}{2} \) or \( b^2 = ac \)

Question. If A is a square matrix, then
(a) \( AA' \) is symmetric
(b) \( AA' \) is skew – symmetric
(c) \( A'A \) is symmetric
(d) \( A'A \) is skew – symmetric
Answer: (a), (c)
Solution:
Given A is a square matrix, Let it be equals to
\( A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \)
\( A' = \begin{bmatrix} a & c \\ b & d \end{bmatrix} \)
Now, \( AA' = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} a & c \\ b & d \end{bmatrix} = \begin{bmatrix} a^2 + b^2 & ac + bd \\ ac + bd & c^2 + d^2 \end{bmatrix} \)
\( AA' \) is symmetric.

Question. If D is a determinant of order three and \( \Delta \) is a determinant formed by the cofactors of determinant D then
(a) \( \Delta = D^2 \)
(b) \( D = 0 \) implies \( \Delta = 0 \)
(c) if \( D = 27 \), then \( \Delta \) is perfect cube
(d) None of the options
Answer: (a), (b), (c)
Solution:
Let \( D = \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} \) & \( \Delta = \begin{vmatrix} C_{11} & C_{12} & C_{13} \\ C_{21} & C_{22} & C_{23} \\ C_{31} & C_{32} & C_{33} \end{vmatrix} \)
\( D \cdot \Delta^T = \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} \begin{vmatrix} C_{11} & C_{21} & C_{31} \\ C_{12} & C_{22} & C_{32} \\ C_{13} & C_{23} & C_{33} \end{vmatrix} = \begin{vmatrix} D & 0 & 0 \\ 0 & D & 0 \\ 0 & 0 & D \end{vmatrix} = D^3 \)
\( \Delta^T = D^2 = \Delta \)
If \( D = 27 \)
\( \Delta = (3^3)^2 = 3^6 \)
\( \Rightarrow \Delta \) is perfect cube.

Question. If B is an idempotent matrix, and A = I – B, then
(a) \( A^2 = A \)
(b) \( A^2 = I \)
(c) \( AB = 0 \)
(d) \( BA = 0 \)
Answer: (a), (c), (d)
Solution:
Given B is an idempotent matrix & A = I – B
Since B is idempotent so, \( B^2 = B \)
Consider A = I – B ....(1)
Post multiply both sides by 'B' in (1)
So, \( AB = IB - B^2 = B - B \quad (\because B^2 = B) \)
\( AB = 0 \)
Premultiply both sides by 'B' in (1) :
\( BA = BI - B^2 = B - B \)
\( BA = 0 \)
Premultiply both sides by A in (1)
\( A^2 = A \quad (\because AB = 0) \)

Question. A square matrix A with elements from the set of real numbers is said to be orthogonal if \( A' = A^{-1} \). If A is an orthogonal matrix, then
(a) \( A' \) is orthogonal
(b) \( A^{-1} \) is orthogonal
(c) Adj A = \( A' \)
(d) \( |A^{-1}| = 1 \)
Answer: (a), (b)
Solution:
\( A' = A^{-1} \)
\( AA' = I \Rightarrow (AA')^{-1} = I^{-1} \Rightarrow (A')^{-1}A^{-1} = I \)
\( (A^{-1})'A^{-1} = I \Rightarrow A^{-1} \) is orthogonal
Now, \( A' = A^{-1} \Rightarrow (A')' = (A^{-1})' \)
\( A = (A^{-1})' \Rightarrow A = (A')^{-1} \)
\( A^{-1} \cdot A = A^{-1} (A')^{-1} \)
\( I = A^{-1} (A^{-1})^{-1} \quad (\because A' = A^{-1}) \)
\( \Rightarrow A^{-1} \) is orthogonal

Question. If \( A^{-1} = \begin{bmatrix} 1 & -1 & 0 \\ 0 & -2 & 1 \\ 0 & 0 & -1 \end{bmatrix} \), then
(a) \( |A| = 2 \)
(b) A is non-singular
(c) Adj. \( A = \begin{bmatrix} 1/2 & -1/2 & 0 \\ 0 & -1 & 1/2 \\ 0 & 0 & -1/2 \end{bmatrix} \)
(d) A is skew symmetric matrix
Answer: (b), (c)
Solution:
Given \( A^{-1} = \begin{bmatrix} 1 & -1 & 0 \\ 0 & -2 & 1 \\ 0 & 0 & -1 \end{bmatrix} \)
As we know that \( |A^{-1}| = \frac{1}{|A|} \)
For \( |A| \neq 0 \), So,
\( |A^{-1}| = 1(2) + 1(0) + 0 = 2 \)
\( \frac{1}{|A|} = 2 \Rightarrow |A| = \frac{1}{2} \Rightarrow \) A is non singular
Now \( \text{Adj } A = |A| \cdot A^{-1} = \frac{1}{2} \begin{bmatrix} 1 & -1 & 0 \\ 0 & -2 & 1 \\ 0 & 0 & -1 \end{bmatrix} = \begin{bmatrix} 1/2 & -1/2 & 0 \\ 0 & -1 & 1/2 \\ 0 & 0 & -1/2 \end{bmatrix} \)

Question. Which of the following is true for matrix \( A = \begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix} \)
(a) \( A + 4I \) is a symmetric matrix
(b) \( A^2 - 4A + 5I_2 = 0 \)
(c) A – B is a diagonal matrix for any value of \( \alpha \) if \( B = \begin{bmatrix} \alpha & -1 \\ 2 & 5 \end{bmatrix} \)
(d) \( A - 4I \) is a skew symmetric matrix
Answer: (b), (c)
Solution:
\( A = \begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix} \)
\( A^2 - 4A + 5I_2 = 0 \) ....(1)
LHS \( A^2 = \begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix} \begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix} = \begin{bmatrix} -1 & -4 \\ 8 & 7 \end{bmatrix} \)
\( 4A = \begin{bmatrix} 4 & -4 \\ 8 & 12 \end{bmatrix} \), \( 5I_2 = \begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix} \)
Put in (1), LHS = RHS
Now, \( A - B = \begin{bmatrix} 1 - \alpha & 0 \\ 0 & -2 \end{bmatrix} = \) Diagonal matrix

Question. Which of the following statement is always true
(a) Adjoint of a symmetric matrix is symmetric matrix
(b) Adjoint of a unit matrix is unit matrix
(c) \( A (\text{adj } A) = (\text{adj } A) A \)
(d) Adjoint of a diagonal matrix is diagonal matrix
Answer: (a), (b), (c), (d)
Solution:
Let \( A = \begin{bmatrix} 1 & 3 \\ 3 & 4 \end{bmatrix} \) Clearly its a symmetric matrix
Its adjoint matrix = \( \begin{bmatrix} 4 & -3 \\ -3 & 1 \end{bmatrix} \) which is also a symmetric matrix.
So, (a) is correct.
Now let \( A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \) which is a unit matrix
Its adjoint matrix = \( \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \) which is also a unit matrix.
So, (b) is correct.
also, \( A (\text{adj } A) = (\text{adj } A) A \). Its a property.
Now let \( A = \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix} \) which is a diagonal matrix
Adjoint of \( A = \begin{bmatrix} 2 & 0 \\ 0 & 1 \end{bmatrix} \) which is a diagonal matrix.
So, (d) is also correct.

Question. If \( A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \) (where \( bc \neq 0 \)) satisfies the equations \( x^2 + k = 0 \), then
(a) \( a + d = 0 \)
(b) \( k = -|A| \)
(c) \( k = |A| \)
(d) None of the options
Answer: (a), (c)
Solution:
Given that \( A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \) \( (bc \neq 0) \)
satisfies \( x^2 + k = 0 \)
\( A^2 + kI = 0 \)
\( A^2 = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} a^2 + bc & ab + bd \\ ac + cd & bc + d^2 \end{bmatrix} \)
\( |A| = ad - bc \)
from relation, \( \begin{bmatrix} a^2 + bc & ab + bd \\ ac + cd & bc + d^2 \end{bmatrix} + \begin{bmatrix} k & 0 \\ 0 & k \end{bmatrix} = 0 \)
\( a^2 + bc + k = 0 \) & \( b(a + d) = 0 \)
\( bc + d^2 + k = 0 \) & \( c(a + d) = 0 \)
\( bc (a + d)^2 = 0 \)
\( (a + d)^2 = 0 \quad (\because bc \neq 0) \)
\( a + d = 0 \)
Now from \( d^2 + bc + k = 0 \)
\( k = -(d^2 + bc) \)
\( k = -(d \cdot d + bc) = -(-ad + bc) \)
\( k = ad - bc = |A| \)

Question. Let \( \phi_1(x) = x + a_1 \), \( \phi_2(x) = x^2 + b_1x + b_2 \) and \( \Delta = \begin{vmatrix} 1 & 1 & 1 \\ \phi_1(x_1) & \phi_1(x_2) & \phi_1(x_3) \\ \phi_2(x_1) & \phi_2(x_2) & \phi_2(x_3) \end{vmatrix} \), then
(a) \( \Delta \) is independent of \( a_1 \)
(b) \( \Delta \) is independent of \( b_1 \) and \( b_2 \)
(c) \( \Delta \) is independent of \( x_1 \), \( x_2 \) and \( x_3 \)
(d) None of the options
Answer: (a), (b)
Solution:
\( \phi_1(x) = x + a_1 \)
\( \phi_2(x) = x^2 + b_1x + b_2 \)
\( \Delta = \begin{vmatrix} 1 & 1 & 1 \\ \phi_1(x_1) & \phi_1(x_2) & \phi_1(x_3) \\ \phi_2(x_1) & \phi_2(x_2) & \phi_2(x_3) \end{vmatrix} \)
\( C_2 \rightarrow C_2 - C_1 \) & \( C_3 \rightarrow C_3 - C_1 \)
\( = \begin{vmatrix} 1 & 0 & 0 \\ x_1 + a_1 & x_2 - x_1 & x_3 - x_1 \\ x_1^2 + b_1x_1 + b_2 & x_2^2 - x_1^2 + b_1(x_2 - x_1) & x_3^2 - x_1^2 + b_1(x_3 - x_1) \end{vmatrix} \)
\( = (x_2 - x_1)(x_3 - x_1) \begin{vmatrix} 1 & 0 & 0 \\ x_1 + a_1 & 1 & 1 \\ x_1^2 + b_1x_1 + b_2 & x_2 + x_1 + b_1 & x_3 + x_1 + b_1 \end{vmatrix} \)
\( = (x_2 - x_1)(x_3 - x_1) \{x_3 + x_1 + b_1 - x_2 - x_1 - b_1\} \)
\( = (x_1 - x_2)(x_2 - x_3)(x_3 - x_1) \)

Question. Suppose \( a_1, a_2, a_3 \) are in A.P. and \( b_1, b_2, b_3 \) are in H.P. and let \( \Delta = \begin{vmatrix} a_1 - b_1 & a_1 - b_2 & a_1 - b_3 \\ a_2 - b_1 & a_2 - b_2 & a_2 - b_3 \\ a_3 - b_1 & a_3 - b_2 & a_3 - b_3 \end{vmatrix} \), then prove that
(a) \( \Delta \) is independent of \( a_1, a_2, a_3 \)
(b) \( a_1 - \Delta \), \( a_2 - 2\Delta \), \( a_3 - 3\Delta \) are in A.P.
(c) \( b_1 + \Delta \), \( b_2 + \Delta^2 \), \( b_3 + \Delta \) are in H.P.
(d) \( \Delta \) is independent of \( b_1, b_2, b_3 \)
Answer: (a), (b), (c), (d)
Solution:
Given \( a_1, a_2, a_3 \) in A.P. & \( b_1, b_2, b_3 \) in H.P.
\( \Delta = \begin{vmatrix} a_1 - b_1 & a_1 - b_2 & a_1 - b_3 \\ a_2 - b_1 & a_2 - b_2 & a_2 - b_3 \\ a_3 - b_1 & a_3 - b_2 & a_3 - b_3 \end{vmatrix} \)
\( R_2 \rightarrow R_2 - R_1 \) & \( R_3 \rightarrow R_3 - R_1 \)
\( \Delta = \begin{vmatrix} a_1 - b_1 & a_1 - b_2 & a_1 - b_3 \\ a_2 - a_1 & a_2 - a_1 & a_2 - a_1 \\ a_3 - a_1 & a_3 - a_1 & a_3 - a_1 \end{vmatrix} \)
\( \Delta = 0 \)

Question. If \( \Delta = \begin{vmatrix} x & 2y - z & -z \\ y & 2x - z & -z \\ y & 2y - z & 2x - 2y - z \end{vmatrix} \), then
(a) \( x - y \) is a factor of \( \Delta \)
(b) \( (x - y)^2 \) is a factor of \( \Delta \)
(c) \( (x - y)^3 \) is a factor of \( \Delta \)
(d) \( \Delta \) is independent of \( z \)
Answer: (a), (b)
Solution:
Given \( \Delta = \begin{vmatrix} x & 2y - z & -z \\ y & 2x - z & -z \\ y & 2y - z & 2x - 2y - z \end{vmatrix} \)
Applying \( R_2 \rightarrow R_2 - R_1 \), \( R_3 \rightarrow R_3 - R_1 \)
\( = \begin{vmatrix} x & 2y - z & -z \\ -(x - y) & 2(x - y) & 0 \\ -(x - y) & 0 & 2(x - y) \end{vmatrix} \)
\( = (x - y)^2 \begin{vmatrix} x & 2y - z & -z \\ -1 & 2 & 0 \\ -1 & 0 & 2 \end{vmatrix} \)
\( = (x - y)^2 \cdot 4(x + y - z) \)
\( = 4(x - y)^2 (x + y - z) \)

Question. Let \( \Delta = \begin{vmatrix} a & a^2 & 0 \\ 1 & 2a + b & (a + b)^2 \\ 0 & 1 & 2a + 3b \end{vmatrix} \) then
(a) \( a + b \) is a factor of \( \Delta \)
(b) \( a + 2b \) is a factor of \( \Delta \)
(c) \( 2a + 3b \) is a factor of \( \Delta \)
(d) \( a^2 \) is a factor of \( \Delta \)
Answer: (a), (b)
Solution:
Given \( \Delta = \begin{vmatrix} a & a^2 & 0 \\ 1 & 2a + b & (a + b)^2 \\ 0 & 1 & 2a + 3b \end{vmatrix} \)
Applying \( C_2 \rightarrow C_2 - aC_1 \)
\( \Delta = \begin{vmatrix} a & 0 & 0 \\ 1 & a + b & (a + b)^2 \\ 0 & 1 & 2a + 3b \end{vmatrix} \)
\( \Delta = a [(a + b) (2a + 3b) - (a + b)^2] \)
\( \Delta = a(a + b) (2a + 3b - a - b) \)
\( \Delta = a(a + b) (a + 2b) \)

Question. Let a, b > 0 and \( \Delta = \begin{vmatrix} -x & a & b \\ b & -x & a \\ a & b & -x \end{vmatrix} \), then
(a) \( a + b - x \) is a factor of \( \Delta \)
(b) \( x^2 + (a + b)x + a^2 + b^2 - ab \) is a factor of \( \Delta \)
(c) \( \Delta = 0 \) has three real roots if \( a = b \)
(d) None of the options
Answer: (a), (b), (c)
Solution:
Given \( a, b > 0 \) and \( \Delta = \begin{vmatrix} -x & a & b \\ b & -x & a \\ a & b & -x \end{vmatrix} \)
Applying \( R_1 \rightarrow R_1 + R_2 + R_3 \)
\( \Delta = \begin{vmatrix} a + b - x & a + b - x & a + b - x \\ b & -x & a \\ a & b & -x \end{vmatrix} \)
\( \Delta = (a + b - x) \begin{vmatrix} 1 & 1 & 1 \\ b & -x & a \\ a & b & -x \end{vmatrix} \)
\( \Delta = (a + b - x) [1(x^2 - ab) - 1(-bx - a^2) + 1(b^2 + ax)] \)
\( \Delta = (a + b - x) [x^2 - ab + bx + a^2 + b^2 + ax] \)
\( \Delta = (a + b - x) [x^2 + (a + b)x + a^2 + b^2 - ab] \)
If \( a = b \) then \( \Delta = (2b - x) (x + b)^2 \)
If \( \Delta = 0 \) which gives three real roots.