1
The value of the determinant \(\begin{vmatrix} \cos 75^\circ & \sin 75^\circ \\ \sin 15^\circ & \cos 15^\circ \end{vmatrix}\) is :
(A) 1 (B) zero (C) \(\frac{1}{2}\) (D) \(\frac{\sqrt{3}}{2}\)
📌 Step-by-Step Answer
📖 Concept
For a 2×2 determinant \(\begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad – bc\). Also use \(\cos A \cos B – \sin A \sin B = \cos(A+B)\).
Step 1
Expand:
\[
\begin{vmatrix}
\cos 75^\circ & \sin 75^\circ\\
\sin 15^\circ & \cos 15^\circ
\end{vmatrix}
=
\cos75^\circ\cos15^\circ-\sin75^\circ\sin15^\circ
\]
\[
\begin{vmatrix}
\cos 75^\circ & \sin 75^\circ\\
\sin 15^\circ & \cos 15^\circ
\end{vmatrix}
=
\cos75^\circ\cos15^\circ-\sin75^\circ\sin15^\circ
\]
Step 2
Apply \(\cos A \cos B – \sin A \sin B = \cos(A+B)\):
\[
= \cos(75^\circ + 15^\circ) = \cos 90^\circ.
\]
\[
= \cos(75^\circ + 15^\circ) = \cos 90^\circ.
\]
Step 3
\(\cos 90^\circ = 0\). Hence the determinant is zero.
✅ Final Answer
(B) zero
🔍 Verification
\(\cos 90^\circ = 0\), so determinant = 0. ✓
2
If A = \(\begin{bmatrix} 1 & 2 \\ 4 & 2 \end{bmatrix}\) and |3A| = k|A|, then the value of k is :
(A) 3 (B) 9 (C) 6 (D) 27
📌 Step-by-Step Answer
📖 Concept
For a square matrix A of order n, \(|kA| = k^n |A|\). Here A is 2×2 ⇒ n = 2.
Step 1
\[
|A| = 1\cdot2 – 2\cdot4 = 2 – 8 = -6.
\]
|A| = 1\cdot2 – 2\cdot4 = 2 – 8 = -6.
\]
Step 2
Since order = 2, \(|3A| = 3^2 |A| = 9|A|\).
Step 3
Given \(|3A| = k|A|\) ⇒ \(9|A| = k|A|\) ⇒ \(k = 9\) (since |A| ≠ 0).
✅ Final Answer
(B) 9
🔍 Verification
\(|3A| = 9(-6) = -54\), \(k|A| = 9(-6) = -54\). ✓
3
If A and B are square matrices both of order 3, such that |A| = -3 and |B| = 2, then |2AB| is equal to :
(A) 48 (B) -48 (C) -24 (D) -12
📌 Step-by-Step Answer
📖 Concept
For n×n matrices, \(|kA| = k^n |A|\) and \(|AB| = |A||B|\). Here n = 3.
Step 1
\[
|2AB| = |2A| \cdot |B| = 2^3 |A| \cdot |B|.
\]
|2AB| = |2A| \cdot |B| = 2^3 |A| \cdot |B|.
\]
Step 2
Substitute: \(8 \cdot (-3) \cdot 2 = 8 \cdot (-6) = -48\).
✅ Final Answer
(B) -48
🔍 Verification
\(8\cdot(-3)\cdot2 = -48\). ✓
4
If \(\left| \begin{array}{cc} x+1 & x-1 \\ x-3 & x+2 \end{array} \right| = \left| \begin{array}{cc} 5 & 5 \\ 3 & 7 \end{array} \right|\), then the value of x is :
(A) 4 (B) 3 (C) 6 (D) 2
📌 Step-by-Step Answer
📖 Concept
Evaluate both determinants using \(ad-bc\) and equate.
Step 1
LHS:
\[
(x+1)(x+2) – (x-1)(x-3)
\]
\[
= (x^2+3x+2) – (x^2-4x+3) = 7x-1.
\]
\[
(x+1)(x+2) – (x-1)(x-3)
\]
\[
= (x^2+3x+2) – (x^2-4x+3) = 7x-1.
\]
Step 2
RHS:
\[
5\cdot7 – 5\cdot3 = 35 – 15 = 20.
\]
\[
5\cdot7 – 5\cdot3 = 35 – 15 = 20.
\]
Step 3
Equate: \(7x – 1 = 20\) ⇒ \(7x = 21\) ⇒ \(x = 3\).
✅ Final Answer
(B) 3
🔍 Verification
For \(x=3\), LHS = \(7(3)-1=20\), RHS = 20. ✓
5
If A is a square matrix of order 3 such that A(adj A) = 7 I, then | adj A | is equal to :
(A) 1 (C) 7 (D) 49
📌 Step-by-Step Answer
📖 Concept
For a square matrix A of order n: \(A(\text{adj }A) = |A| I\) and \(|\text{adj }A| = |A|^{n-1}\).
Step 1
Given \(A(\text{adj }A) = 7I\) ⇒ \(|A| = 7\).
Step 2
For n=3, \(|\text{adj }A| = |A|^{3-1} = |A|^2 = 7^2 = 49\).
✅ Final Answer
(D) 49
🔍 Verification
\(|A(\text{adj }A)| = |7I| = 7^3 = 343\) and \(|A|\cdot|\text{adj }A| = 7\cdot49 = 343\). ✓
6
The area of a triangle with vertices (3, 0), (0, k) and (-3, 0) is 9 sq units. The value of k is :
(A) 9 (B) -9 (C) 3 (D) 6
📌 Step-by-Step Answer
📖 Concept
Area = \(\frac12 | x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) |\).
Step 1
Substitute:
\[
\text{Area} = \frac12 | 3(k-0) + 0(0-0) + (-3)(0-k) |
\]
\[
= \frac12 | 3k + 3k | = 3|k|.
\]
\[
\text{Area} = \frac12 | 3(k-0) + 0(0-0) + (-3)(0-k) |
\]
\[
= \frac12 | 3k + 3k | = 3|k|.
\]
Step 2
Given \(3|k| = 9\) ⇒ \(|k| = 3\) ⇒ \(k = \pm 3\). From options, \(k=3\).
✅ Final Answer
(C) 3
🔍 Verification
For \(k=3\), Area = \(3\cdot3 = 9\). ✓
7
For a non-singular matrix X, if X² = I, then X⁻¹ is equal to :
(A) X (C) I
📌 Step-by-Step Answer
📖 Concept
If \(X^2 = I\), then \(X \cdot X = I\), so X is its own inverse.
Step 1
Given \(X^2 = I\) ⇒ \(X \cdot X = I\).
Step 2
By definition of inverse, \(X^{-1} = X\).
✅ Final Answer
(A) X
🔍 Verification
\(X \cdot X = I\) ⇒ \(X^{-1} = X\). ✓
8
The cofactor of the element a₃₂ in the determinant \(\Delta = \begin{vmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 3 & 1 & 2 \end{vmatrix}\) is :
(A) ±5 (C) 5
📌 Step-by-Step Answer
📖 Concept
Cofactor \(C_{ij} = (-1)^{i+j} M_{ij}\), where \(M_{ij}\) is the minor.
Step 1
For a₃₂, remove row 3, column 2:
\[
M_{32} = \begin{vmatrix} 1 & 3 \\ 2 & 1 \end{vmatrix} = 1\cdot1 – 3\cdot2 = 1-6 = -5.
\]
\[
M_{32} = \begin{vmatrix} 1 & 3 \\ 2 & 1 \end{vmatrix} = 1\cdot1 – 3\cdot2 = 1-6 = -5.
\]
Step 2
Sign: \((-1)^{3+2} = (-1)^5 = -1\).
Step 3
\[
C_{32} = (-1)^{3+2} \cdot M_{32} = -1 \cdot (-5) = 5.
\]
C_{32} = (-1)^{3+2} \cdot M_{32} = -1 \cdot (-5) = 5.
\]
✅ Final Answer
(C) 5
🔍 Verification
\(C_{32} = (-1)^5 \cdot (-5) = 5\). ✓
9
Assertion (A): If A is a skew-symmetric matrix of order 3, then |A| = 0.
Reason (R): If A is a square matrix of order 3, then |A| = |A′|.
📌 Step-by-Step Answer
📖 Concept
Skew-symmetric of odd order has determinant zero. Also \(|A| = |A’|\) for any square matrix.
Step 1
For skew-symmetric A of order 3: \(A’ = -A\).
Step 2
Taking determinant: \(|A’| = |-A| = (-1)^3 |A| = -|A|\).
Step 3
But \(|A’| = |A|\) (property). So \(|A| = -|A|\) ⇒ \(2|A| = 0\) ⇒ \(|A| = 0\). Assertion (A) is true.
Step 4
Reason (R): \(|A| = |A’|\) is always true for any square matrix. So R is true, but it does not specifically explain why a skew-symmetric matrix of order 3 has determinant 0.
✅ Final Answer
Both A and R are true, but R is not the correct explanation of A.
🔍 Verification
A is true (odd-order skew-symmetric determinant is 0). R is true. R does not directly explain A.
10
If \(A = \begin{bmatrix} 3 & 2 & 1 \\ 4 & -1 & 2 \\ 7 & 3 & -3 \end{bmatrix}\), find \(A^{-1}\). Using \(A^{-1}\), solve the given system of equations \(3x + 4y + 7z = 14;\; 2x – y + 3z = 4;\; x + 2y – 3z = 0\).
📌 Step-by-Step Answer
📖 Concept
\(A^{-1} = \frac{1}{|A|}\text{adj}(A)\). The system \(A’X = B\) has solution \(X = (A^{-1})’B\).
Step 1
Compute \(|A|\):
\[
|A| = 3((-1)(-3)-2\cdot3) – 2(4(-3)-2\cdot7) + 1(4\cdot3 – (-1)\cdot7)
\]
\[
= 3(3-6) – 2(-12-14) + 1(12+7)
\]
\[
= -9 + 52 + 19 = 62 \neq 0.
\]
\[
|A| = 3((-1)(-3)-2\cdot3) – 2(4(-3)-2\cdot7) + 1(4\cdot3 – (-1)\cdot7)
\]
\[
= 3(3-6) – 2(-12-14) + 1(12+7)
\]
\[
= -9 + 52 + 19 = 62 \neq 0.
\]
Step 2
Cofactor matrix:
\[
\begin{bmatrix}
-3 & 26 & 19 \\
9 & -16 & 5 \\
5 & -2 & -11
\end{bmatrix}.
\]
\[
\begin{bmatrix}
-3 & 26 & 19 \\
9 & -16 & 5 \\
5 & -2 & -11
\end{bmatrix}.
\]
Step 3
\(\text{adj}(A) =
\begin{bmatrix}
-3 & 9 & 5 \\
26 & -16 & -2 \\
19 & 5 & -11
\end{bmatrix}\).
\begin{bmatrix}
-3 & 9 & 5 \\
26 & -16 & -2 \\
19 & 5 & -11
\end{bmatrix}\).
Step 4
\[
A^{-1} = \frac{1}{62} \begin{bmatrix}
-3 & 9 & 5 \\
26 & -16 & -2 \\
19 & 5 & -11
\end{bmatrix}.
\]
A^{-1} = \frac{1}{62} \begin{bmatrix}
-3 & 9 & 5 \\
26 & -16 & -2 \\
19 & 5 & -11
\end{bmatrix}.
\]
Step 5
System as \(A’X = B\) with \(X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}\), \(B = \begin{bmatrix} 14 \\ 4 \\ 0 \end{bmatrix}\).
Step 6
\[
X = (A^{-1})’B = \frac{1}{62} \begin{bmatrix}
-3 & 26 & 19 \\
9 & -16 & 5 \\
5 & -2 & -11
\end{bmatrix} \begin{bmatrix} 14 \\ 4 \\ 0 \end{bmatrix}.
\]
X = (A^{-1})’B = \frac{1}{62} \begin{bmatrix}
-3 & 26 & 19 \\
9 & -16 & 5 \\
5 & -2 & -11
\end{bmatrix} \begin{bmatrix} 14 \\ 4 \\ 0 \end{bmatrix}.
\]
Step 7
Compute \(x, y, z\):
\[
x = \frac{1}{62}[(-3)(14)+26(4)+19(0)] = \frac{-42+104}{62} = 1,
\]
\[
y = \frac{1}{62}[9(14)+(-16)(4)+5(0)] = \frac{126-64}{62} = 1,
\]
\[
z = \frac{1}{62}[5(14)+(-2)(4)+(-11)(0)] = \frac{70-8}{62} = 1.
\]
\[
x = \frac{1}{62}[(-3)(14)+26(4)+19(0)] = \frac{-42+104}{62} = 1,
\]
\[
y = \frac{1}{62}[9(14)+(-16)(4)+5(0)] = \frac{126-64}{62} = 1,
\]
\[
z = \frac{1}{62}[5(14)+(-2)(4)+(-11)(0)] = \frac{70-8}{62} = 1.
\]
✅ Final Answer
\(A^{-1} = \frac{1}{62} \begin{bmatrix} -3 & 9 & 5 \\ 26 & -16 & -2 \\ 19 & 5 & -11 \end{bmatrix}\)\(x = 1,\; y = 1,\; z = 1\)
🔍 Verification
\(3(1)+4(1)+7(1)=14\), \(2(1)-1(1)+3(1)=4\), \(1(1)+2(1)-3(1)=0\). ✓