Modern Math - Determinants & Metrices - Previous Year IPM/BBA Questions

Answer: Option C

Text Explanation :

Answer: 32

Text Explanation :

Given, A = $\left[\begin{array}{ccc}1& 0& 0\\ 0& 0& 1\\ 0& 1& 0\end{array}\right]$ 

Now, A × A = $\left[\begin{array}{ccc}1& 0& 0\\ 0& 0& 1\\ 0& 1& 0\end{array}\right]$ × $\left[\begin{array}{ccc}1& 0& 0\\ 0& 0& 1\\ 0& 1& 0\end{array}\right]$ = $\left[\begin{array}{ccc}1\times 1+0\times 0+0\times 0& 1\times 0+0\times 0+0\times 1& 1\times 0+0\times 1+0\times 0\\ 0\times 1+0\times 0+1\times 0& 0\times 0+0\times 0+1\times 1& 0\times 0+0\times 1+1\times 0\\ 0\times 1+1\times 0+0\times 0& 0\times 0+1\times 0+0\times 1& 0\times 0+1\times 1+0\times 0\end{array}\right]$ = $\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$ = I

⇒ A³ = A² × A = I × A = A

⇒ A⁶ = A³ × A³ = A × A = I

⇒ A⁹ = A³ × A³ × A³ = A × A × A = A² × A = I × A = A

Now, (A⁹ + A⁶ + A³ + A)

= A + I + A + A + A

= 3A + I

= $\left[\begin{array}{ccc}3& 0& 0\\ 0& 0& 3\\ 0& 3& 0\end{array}\right]$ + $\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$

= $\left[\begin{array}{ccc}4& 0& 0\\ 0& 1& 3\\ 0& 3& 1\end{array}\right]$

∴ The absolute value of the determinant of (A⁹ + A⁶ + A³ + A) = absolute value of $\left[\begin{array}{ccc}4& 0& 0\\ 0& 1& 3\\ 0& 3& 1\end{array}\right]$

= |4 × (1 × 1 - 3 × 3)| = |4 × (1 - 9)| = 32.

Hence, 32.

Answer: Option C

Text Explanation :

A = $\left[\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right]$.

Determinant of A = a(cb - a²) + b(ac - b²) + c(ba - c²)

= 3abc - a³ - b³ - c³

= - (a³ + b³ + c³ - 3abc)

= - (a + b + c)(a² + b² + c² - ab - bc - ca)

(a + b + c)(a² + b² + c² - ab - bc - ca) > 0 for a, b, c > 0

∴ Determinant of A = - (a + b + c)(a² + b² + c² - ab - bc - ca) is always negative.

Hence, option (c).

Answer: Option B

Text Explanation :

Answer: Option A

Text Explanation :

Given,  2A - B - A(A + B)^-1A + B(A + B)^-1B

= 2A - B - (A + B)^-1[A² - B²]

= 2A - B - (A + B)^-1(A - B)(A + B)

= 2A - B - (A + B)^-1(A + B)(A - B)

= 2A - B - I(A - B)

= 2A - B - (A - B)

= 2A - B - A + B

= A 

Hence, option (a).

Answer: 1

Text Explanation :

Given, $\left|\begin{array}{ccc}\mathrm{a}& {\mathrm{a}}^2& {\mathrm{a}}^3-1\\ \mathrm{b}& {\mathrm{b}}^2& {\mathrm{b}}^3-1\\ \mathrm{c}& {\mathrm{c}}^2& {\mathrm{c}}^3-1\end{array}\right|$ = 0

⇒ $\left|\begin{array}{ccc}\mathrm{a}& {\mathrm{a}}^2& {\mathrm{a}}^3\\ \mathrm{b}& {\mathrm{b}}^2& {\mathrm{b}}^3\\ \mathrm{c}& {\mathrm{c}}^2& {\mathrm{c}}^3\end{array}\right|$ + $\left|\begin{array}{ccc}\mathrm{a}& {\mathrm{a}}^2& -1\\ \mathrm{b}& {\mathrm{b}}^2& -1\\ \mathrm{c}& {\mathrm{c}}^2& -1\end{array}\right|$ = 0

⇒ abc × $\left|\begin{array}{ccc}1& \mathrm{a}& {\mathrm{a}}^2\\ 1& \mathrm{b}& {\mathrm{b}}^2\\ 1& \mathrm{c}& {\mathrm{c}}^2\end{array}\right|$ - $\left|\begin{array}{ccc}\mathrm{a}& {\mathrm{a}}^2& 1\\ \mathrm{b}& {\mathrm{b}}^2& 1\\ \mathrm{c}& {\mathrm{c}}^2& 1\end{array}\right|$ = 0

⇒ abc × $\left|\begin{array}{ccc}1& \mathrm{a}& {\mathrm{a}}^2\\ 1& \mathrm{b}& {\mathrm{b}}^2\\ 1& \mathrm{c}& {\mathrm{c}}^2\end{array}\right|$ + $\left|\begin{array}{ccc}1& {\mathrm{a}}^2& \mathrm{a}\\ 1& {\mathrm{b}}^2& \mathrm{b}\\ 1& {\mathrm{c}}^2& \mathrm{c}\end{array}\right|$ = 0

⇒ abc × $\left|\begin{array}{ccc}1& \mathrm{a}& {\mathrm{a}}^2\\ 1& \mathrm{b}& {\mathrm{b}}^2\\ 1& \mathrm{c}& {\mathrm{c}}^2\end{array}\right|$ - $\left|\begin{array}{ccc}1& \mathrm{a}& {\mathrm{a}}^2\\ 1& \mathrm{b}& {\mathrm{b}}^2\\ 1& \mathrm{c}& {\mathrm{c}}^2\end{array}\right|$ = 0

⇒ (abc - 1) × $\left|\begin{array}{ccc}1& \mathrm{a}& {\mathrm{a}}^2\\ 1& \mathrm{b}& {\mathrm{b}}^2\\ 1& \mathrm{c}& {\mathrm{c}}^2\end{array}\right|$ = 0

$\left|\begin{array}{ccc}1& \mathrm{a}& {\mathrm{a}}^2\\ 1& \mathrm{b}& {\mathrm{b}}^2\\ 1& \mathrm{c}& {\mathrm{c}}^2\end{array}\right|$ ≠ 0

∴ abc - 1 = 0

⇒ abc = 1

Hence, 1.

Answer: Option A

Text Explanation :

Total number of matrices formed = Number of ways of selecting 4 distinct integer × Arranging these 4 integers in the matrix.
= ⁶C₄ × 4! = 15 × 24 = 360

For a singular matrix $\left|\begin{array}{cc}\mathrm{a}& \mathrm{b}\\ \mathrm{c}& \mathrm{d}\end{array}\right|$, its determinant = 0.
∴ ad - bc = 0
⇒ ad = bc

Now, possible pairs of a and d can be

Case 1: a and b are 2 or 3, 
∴ a × d = b × c = 6
⇒ b and c are 1 or 6.
∴ Possible arrangements for a and d = 2! and that for b and c = 2!
∴ Total possible arrangements for a, b, c and d = 2! × 2! = 4.

Case 2: a and b are 1 or 6,
∴ a × d = b × c = 6
⇒ b and c are 2 or 3.
∴ Possible arrangements for a and d = 2! and that for b and c = 2!
∴ Total possible arrangements for a, b, c and d = 2! × 2! = 4.

Case 3: a and b are 3 or 4, 
∴ a × d = b × c = 12
⇒ b and c are 2 or 6.
∴ Possible arrangements for a and d = 2! and that for b and c = 2!
∴ Total possible arrangements for a, b, c and d = 2! × 2! = 4.

Case 4: a and b are 2 or 6, 
∴ a × d = b × c = 12
⇒ b and c are 3 or 4.
∴ Possible arrangements for a and d = 2! and that for b and c = 2!
∴ Total possible arrangements for a, b, c and d = 2! × 2! = 4.

⇒ Total number of required matrices = 4 + 4 + 4 + 4 = 16

Required probability = 16/360 = 2/45.

Hence, option (a).

Answer: 10

Text Explanation :

Answer: 3

Text Explanation :

Answer: 4

Text Explanation :

Answer: Option A

Text Explanation :