Let the determinant of a square matrix $A$ of order $m$ be $m-n$, where $m$ and $n$ satisfy $4m + n = 22$ and $17m + 4n = 93$. If $det(n \text{ adj}(adj(mA))) = 3^a 5^b 6^c$, then $a + b + c$ is equal to Options: (A) 84 (B) 96 (C) 101 (D) 109

### Analyzing the Setup To begin, we must solve the system of linear equations provided to determine the values of $m$ and $n$: $4m + n = 22$ $17m + 4n = 93$ Using the method of elimination, we multiply the first equation by $4$ to obtain: $16m + 4n = 88$ Subtracting this from the second equation, $17m + 4n = 93$, we find: $(17m - 16m) + (4n - 4n) = 93 - 88$ $m = 5$ Substituting $m = 5$ into the first equation, $4(5) + n = 22$, we solve for $n$: $20 + n = 22 \Rightarrow n = 2$ The order of matrix $A$ is $m = 5$, and its determinant is given by $|A| = m - n = 5 - 2 = 3$. ### The Toolkit of Laws We now address the expression $|n \cdot \text{adj}(\text{adj}(mA))|$. We utilize two fundamental properties of determinants for an $m \times m$ matrix $X$: 1. The scalar property: $|kX| = k^m |X|$ 2. The adjoint property: $|\text{adj}(\text{adj}(X))| = |X|^{(m-1)^2}$ Applying the scalar property to pull $n$ out of the determinant, where the order of the matrix is $m=5$: $$|n \cdot \text{adj}(\text{adj}(mA))| = n^5 \cdot |\text{adj}(\text{adj}(mA))|$$ ### The Transformation Next, we apply the adjoint property to the term $|\text{adj}(\text{adj}(mA))|$, treating $mA$ as the matrix $X$: $$|\text{adj}(\text{adj}(mA))| = |mA|^{(5-1)^2} = |mA|^{16}$$ Substituting this back into our expression, we have: $$n^5 \cdot |mA|^{16} = 2^5 \cdot |5A|^{16}$$ Using the scalar property again for $|5A|$, where $|5A| = 5^5 |A|$ and $|A| = 3$: $$|5A| = 5^5 \cdot 3$$ Substituting this into our expression: $$2^5 \cdot (5^5 \cdot 3)^{16} = 2^5 \cdot 5^{80} \cdot 3^{16}$$ ### Final Calculation We must express the result in the form $3^a 5^b 6^c$. Since $6 = 2 \cdot 3$, we rewrite the expression by grouping the powers of $2$ and $3$: $$2^5 \cdot 3^{16} \cdot 5^{80} = 2^5 \cdot 3^5 \cdot 3^{11} \cdot 5^{80}$$ $$(2 \cdot 3)^5 \cdot 3^{11} \cdot 5^{80} = 6^5 \cdot 3^{11} \cdot 5^{80}$$ Comparing this to $3^a 5^b 6^c$, we identify the exponents: $a = 11, b = 80, c = 5$ The final sum is: **$a + b + c = 11 + 80 + 5 = 96$**

JEE Main 2023 (15 April Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Matrices and Determinants: Let the determinant of a square matrix $A$ of order $m$ be $m - n$ , where $m$ and $n$ satisfy $4 m + n = 22$ and $17 m + 4 n = 93$ . If $d e t (n adj (a d j (m A))) = 3^{a} 5^{b} 6^{c}$ , then $a + b + c$ is equal to

Q: Wait, why does the scalar $n$ come out as $n^m$ and not just $n$?

That is a fantastic question! Remember that for any $m \times m$ matrix $X$, the determinant of $kX$ is $k^m|X|$. This is because when you multiply a matrix by a scalar $k$, you are multiplying every single row by $k$. Since there are $m$ rows, you factor out $k$ exactly $m$ times. Thus, $k$ comes out as $k^m$.

Q: Why is the exponent $(m-1)^2$ in the adjoint of an adjoint property?

Think of it as a nested operation. We know that $|adj(X)| = |X|^{m-1}$. If we take the adjoint again, we are essentially applying this rule twice. So, $|adj(adj(X))| = |adj(X)|^{m-1} = (|X|^{m-1})^{m-1} = |X|^{(m-1)^2}$. It is just a beautiful consequence of the definition of the adjoint!

Select Answer:

Progressive Hints

Hint 1

Hint 1: The Spark. To unlock the secrets of this matrix, you must first solve the system of linear equations to find the values of

m

and

n

. Once you have these, the order of the matrix

A

and its determinant

∣ A ∣

become clear. Visualize the matrix

A

as an

m \times m

square matrix, and remember that the determinant of a scalar multiple

k A

k^{m} ∣ A ∣

Hint 2

Hint 2: The Arsenal. You are dealing with the adjoint of an adjoint. Recall the fundamental property:

∣ a d j (a d j (X)) ∣ = ∣ X ∣^{(m - 1)^{2}}

. This is your primary weapon to simplify the expression

∣ n \cdot a d j (a d j (m A)) ∣

. Combine this with the scalar property

∣ k X ∣ = k^{m} ∣ X ∣

to pull the scalar

n

out of the determinant.

Hint 3

Hint 3: The Execution. Start by expanding the expression:

∣ n \cdot a d j (a d j (m A)) ∣ = n^{m} \cdot ∣ a d j (a d j (m A)) ∣

. Then, substitute

X = m A

into the adjoint property to get

n^{m} \cdot ∣ m A ∣^{(m - 1)^{2}}

. Now, substitute

m = 5

n = 2

, and

∣ A ∣ = 3

to calculate the final numerical value. Finally, express your result in the form

3^{a} 5^{b} 6^{c}

to find

a, b,

and

c

Visualized Solution

The System of Equations

Given equations for matrix order $m$ and parameter $n$ :
$4 m + n = 22$
$17 m + 4 n = 93$

Solving for $m$ and $n$

Multiply the first equation by $4$ :
$16 m + 4 n = 88$
Subtract from the second equation:
$(17 m - 16 m) + (4 n - 4 n) = 93 - 88$
$m = 5$
Substitute $m = 5$ into the first equation:
$4 (5) + n = 22 ⟹ 20 + n = 22 ⟹ n = 2$

Matrix Order and Determinant

Order of square matrix $A$ :
$m = 5$
Determinant of $A$ :
$∣ A ∣ = m - n$
Substitute values:
$∣ A ∣ = 5 - 2 = 3$

Properties of Determinants and Adjoints

Property 1 (Scalar Multiple):
$∣ k X ∣ = k^{m} ∣ X ∣$ (for a matrix $X$ of order $m$ )
Property 2 (Adjoint of Adjoint):
$∣ a d j (a d j (X)) ∣ = ∣ X ∣^{(m - 1)^{2}}$

Extracting the Outer Scalar

Expression to evaluate:
$∣ n \cdot a d j (a d j (m A)) ∣$
The matrix $a d j (a d j (m A))$ has the same order as $A$ , which is $m = 5$ .
Using Property 1:
$∣ n \cdot a d j (a d j (m A)) ∣ = n^{m} ∣ a d j (a d j (m A)) ∣$

Applying the Adjoint Property

Let $X = m A$ .
Using Property 2:
$∣ a d j (a d j (m A)) ∣ = ∣ m A ∣^{(m - 1)^{2}}$
Combining with previous step:
$n^{m} (∣ m A ∣)^{(m - 1)^{2}}$

Substituting $m$ and $n$

Expression:
$n^{m} (∣ m A ∣)^{(m - 1)^{2}}$
Substitute $m = 5$ and $n = 2$ :
$= 2^{5} (∣5 A ∣)^{(5 - 1)^{2}}$
$= 2^{5} (∣5 A ∣)^{16}$

Expanding $∣5 A ∣$

Using Property 1 again on $∣5 A ∣$ :
$∣5 A ∣ = 5^{m} ∣ A ∣$
Substitute $m = 5$ and $∣ A ∣ = 3$ :
$∣5 A ∣ = 5^{5} \cdot 3$

Combining the Powers

Substitute $∣5 A ∣$ back into the main expression:
$= 2^{5} (5^{5} \cdot 3)^{16}$
Distribute the power of $16$ :
$= 2^{5} \cdot (5^{5})^{16} \cdot 3^{16}$
$= 2^{5} \cdot 5^{80} \cdot 3^{16}$

Matching the Target Form

Target form: $3^{a} 5^{b} 6^{c}$
Current expression: $2^{5} \cdot 5^{80} \cdot 3^{16}$
We need a base of $6$ . Since $6 = 2 \cdot 3$ , we pair the $2$ s and $3$ s.
Rewrite $3^{16}$ as $3^{11} \cdot 3^{5}$ :
$= 3^{11} \cdot 5^{80} \cdot (2^{5} \cdot 3^{5})$
$= 3^{11} \cdot 5^{80} \cdot 6^{5}$

Final Calculation of $a + b + c$

Compare $3^{11} \cdot 5^{80} \cdot 6^{5}$ with $3^{a} 5^{b} 6^{c}$ :
$a = 11$ , $b = 80$ , $c = 5$
Calculate the sum:
$a + b + c = 11 + 80 + 5 = 96$
Final Answer: 96

00:00 / 00:00

The Sigma Insight: Adjoint and Inverse of a Matrix

Analyzing the Setup

To begin, we must solve the system of linear equations provided to determine the values of

m

and

n

4 m + n = 22

17 m + 4 n = 93

Using the method of elimination, we multiply the first equation by

4

to obtain:

16 m + 4 n = 88

Subtracting this from the second equation,

17 m + 4 n = 93

, we find:

(17 m - 16 m) + (4 n - 4 n) = 93 - 88

m = 5

Substituting

m = 5

into the first equation,

4 (5) + n = 22

, we solve for

n

20 + n = 22 \Rightarrow n = 2

The order of matrix

A

m = 5

, and its determinant is given by

∣ A ∣ = m - n = 5 - 2 = 3

The Toolkit of Laws

We now address the expression

∣ n \cdot adj (adj (m A)) ∣

. We utilize two fundamental properties of determinants for an

m \times m

matrix

X

1. The scalar property:

∣ k X ∣ = k^{m} ∣ X ∣

2. The adjoint property:

∣ adj (adj (X)) ∣ = ∣ X ∣^{(m - 1)^{2}}

Applying the scalar property to pull

n

out of the determinant, where the order of the matrix is

m = 5

∣ n \cdot adj (adj (m A)) ∣ = n^{5} \cdot ∣ adj (adj (m A)) ∣

The Transformation

Next, we apply the adjoint property to the term

∣ adj (adj (m A)) ∣

, treating

m A

as the matrix

X

∣ adj (adj (m A)) ∣ = ∣ m A ∣^{(5 - 1)^{2}} = ∣ m A ∣^{16}

Substituting this back into our expression, we have:

n^{5} \cdot ∣ m A ∣^{16} = 2^{5} \cdot ∣5 A ∣^{16}

Using the scalar property again for

∣5 A ∣

, where

∣5 A ∣ = 5^{5} ∣ A ∣

and

∣ A ∣ = 3

∣5 A ∣ = 5^{5} \cdot 3

Substituting this into our expression:

2^{5} \cdot (5^{5} \cdot 3)^{16} = 2^{5} \cdot 5^{80} \cdot 3^{16}

Final Calculation

We must express the result in the form

3^{a} 5^{b} 6^{c}

. Since

6 = 2 \cdot 3

, we rewrite the expression by grouping the powers of

2

and

3

2^{5} \cdot 3^{16} \cdot 5^{80} = 2^{5} \cdot 3^{5} \cdot 3^{11} \cdot 5^{80}

(2 \cdot 3)^{5} \cdot 3^{11} \cdot 5^{80} = 6^{5} \cdot 3^{11} \cdot 5^{80}

Comparing this to

3^{a} 5^{b} 6^{c}

, we identify the exponents:

a = 11, b = 80, c = 5

The final sum is: $a + b + c = 11 + 80 + 5 = 96$

Common Traps & FAQs

Wait, why does the scalar

n

come out as

n^{m}

and not just

n

That is a fantastic question! Remember that for any

m \times m

matrix

X

, the determinant of

k X

k^{m} ∣ X ∣

. This is because when you multiply a matrix by a scalar

k

, you are multiplying every single row by

k

. Since there are

m

rows, you factor out

k

exactly

m

times. Thus,

k

comes out as

k^{m}

Why is the exponent

(m - 1)^{2}

in the adjoint of an adjoint property?

Think of it as a nested operation. We know that

∣ a d j (X) ∣ = ∣ X ∣^{m - 1}

. If we take the adjoint again, we are essentially applying this rule twice. So,

∣ a d j (a d j (X)) ∣ = ∣ a d j (X) ∣^{m - 1} = (∣ X ∣^{m - 1})^{m - 1} = ∣ X ∣^{(m - 1)^{2}}

. It is just a beautiful consequence of the definition of the adjoint!

How do I know when to stop and convert to the

3^{a} 5^{b} 6^{c}

form?

The target form is your compass. Once you have your final expression, like

2^{5} \cdot 5^{80} \cdot 3^{16}

, look at the bases. You have

2, 3,

and

5

, but the target requires

6

. Since

6 = 2 \cdot 3

, you simply need to group the

2

s and

3

s together to form

6

s. It is a puzzle of exponents!

Animated Solution for Mathematics - Matrices and Determinants: Let the determinant of a square matrix $A$ of order $m$ be $m - n$ , where $m$ and $n$ satisfy $4 m + n = 22$ and $17 m + 4 n = 93$ . If $d e t (n adj (a d j (m A))) = 3^{a} 5^{b} 6^{c}$ , then $a + b + c$ is equal to

Select Answer:

Visualized Solution

The System of Equations

Solving for $m$ and $n$

Matrix Order and Determinant

Properties of Determinants and Adjoints

Extracting the Outer Scalar

Applying the Adjoint Property

Substituting $m$ and $n$

Expanding $∣5 A ∣$

Combining the Powers

Matching the Target Form

Final Calculation of $a + b + c$

The Sigma Insight: Adjoint and Inverse of a Matrix

Analyzing the Setup

The Toolkit of Laws

The Transformation

Final Calculation

Similar Questions

Let $A = 246 2 + p 6 + 2 p 12 + 3 p 2 + p + q 8 + 3 p + 2 q 20 + 6 p + 3 q$ . If $det (adj (adj (3 A))) = 2^{m} \cdot 3^{n}, m, n \in N$ , then $m + n$ is equal to

Let $a \in R$ and $A$ be a matrix of order $3 \times 3$ such that $det (A) = - 4$ and $A + I = 12 a a 11 102$ , where $I$ is the identity matrix of order $3 \times 3$ . If $det ((a + 1) adj ((a - 1) A))$ is $2^{m} 3^{n}$ , $m, n \in {0, 1, 2, \dots, 20}$ , then $m + n$ is equal to :

Let $M = 013 12 b a 31$ and $adj M = - 1 8 - 5 1 - 6 3 - 1 2 - 1$ where $a$ and $b$ are real numbers. Which of the following options is/are correct?

Let $A$ be a matrix of order $3 \times 3$ and $∣ A ∣ = 5$ . If $∣2 adj (3 A adj (2 A)) ∣ = 2^{α} \cdot 3^{β} \cdot 5^{γ}, α, β, γ \in N$ then $α + β + γ$ is equal to

Let $A = {a_{ij}}$ be a $3 \times 3$ matrix, where $a_{ij} = ⎩ ⎨ ⎧ (- 1)^{j - i} 2 (- 1)^{i + j} if i < j if i = j if i > j$ then $det (3 Adj (2 A^{- 1}))$ is equal to

Let $A$ be a $3 \times 3$ real matrix. If $det (2 Adj (2 Adj (Adj (2 A)))) = 2^{41}$ , then the value of $det (A^{2})$ equal .

Let A be a $3 \times 3$ matrix such that $X^{T} A X = 0$ for all nonzero $3 \times 1$ matrices $X = x y z$ . If $A 111 = 14 - 5$ and $A 121 = 04 - 8$ , and $det (a d j (2 (A + I))) = 2^{α} 3^{β} 5^{γ}$ , $α, β, γ \in N,$ then $α^{2} + β^{2} + γ^{2}$ is

Let $P = [p_{ij}]$ and $Q = [q_{ij}]$ be two square matrices of order 3 such that $q_{ij} = 2^{(i + j - 1)} p_{ij}$ and $det (Q) = 2^{10}$ . Then the value of $det (adj (adj P))$ is:

Let A be a $3 \times 3$ matrix such that $A + A^{T} = O$ . If $A 1 - 1 0 = 332, A^{2} 1 - 1 0 = - 3 19 - 24$ and $det (adj (2 adj (A + I))) = (2)^{α} \cdot (3)^{β} \cdot (11)^{γ}, α, β, γ$ are non-negative integers, then $α + β + γ$ is equal to .........

Let the matrix $A = 001100010$ and the matrix $B_{0} = A^{49} + 2 A^{98}$ . If $B_{n} = adj (B_{n - 1})$ for all $n \geq 1$ , then $det (B_{4})$ is equal to :

Select Answer:

Visualized Solution

The System of Equations

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n

Matrix Order and Determinant

Properties of Determinants and Adjoints

Extracting the Outer Scalar

Applying the Adjoint Property

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n

Expanding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣5A∣

Combining the Powers

Matching the Target Form

Final Calculation of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a+b+c

The Sigma Insight: Adjoint and Inverse of a Matrix

Analyzing the Setup

The Toolkit of Laws

The Transformation

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let A=​246​2+p6+2p12+3p​2+p+q8+3p+2q20+6p+3q​​. If det(adj(adj(3A)))=2m⋅3n,m,n∈N, then m+n is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let M=​013​12b​a31​​ and adj M=​−18−5​1−63​−12−1​​ where a and b are real numbers. Which of the following options is/are correct?

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let A be a matrix of order 3×3 and ∣A∣=5. If ∣2 adj(3A adj(2A))∣=2α⋅3β⋅5γ,α,β,γ∈N then α+β+γ is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let A={aij​} be a 3×3 matrix, where aij​=⎩⎨⎧​(−1)j−i2(−1)i+j​if i<jif i=jif i>j​ then det(3Adj(2A−1)) is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let A be a 3×3 real matrix. If det(2 Adj(2 Adj(Adj(2A))))=241, then the value of det(A2) equal .

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let P=[pij​] and Q=[qij​] be two square matrices of order 3 such that qij​=2(i+j−1)pij​ and det(Q)=210. Then the value of det(adj(adj P)) is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let the matrix A=​001​100​010​​ and the matrix B0​=A49+2A98. If Bn​=adj(Bn−1​) for all n≥1, then det(B4​) is equal to :

Solving for $m$ and $n$

Substituting $m$ and $n$

Expanding $∣5 A ∣$

Final Calculation of $a + b + c$

Let $A = 246 2 + p 6 + 2 p 12 + 3 p 2 + p + q 8 + 3 p + 2 q 20 + 6 p + 3 q$ . If $det (adj (adj (3 A))) = 2^{m} \cdot 3^{n}, m, n \in N$ , then $m + n$ is equal to

Let $M = 013 12 b a 31$ and $adj M = - 1 8 - 5 1 - 6 3 - 1 2 - 1$ where $a$ and $b$ are real numbers. Which of the following options is/are correct?

Let $A$ be a matrix of order $3 \times 3$ and $∣ A ∣ = 5$ . If $∣2 adj (3 A adj (2 A)) ∣ = 2^{α} \cdot 3^{β} \cdot 5^{γ}, α, β, γ \in N$ then $α + β + γ$ is equal to

Let $A = {a_{ij}}$ be a $3 \times 3$ matrix, where $a_{ij} = ⎩ ⎨ ⎧ (- 1)^{j - i} 2 (- 1)^{i + j} if i < j if i = j if i > j$ then $det (3 Adj (2 A^{- 1}))$ is equal to

Let $A$ be a $3 \times 3$ real matrix. If $det (2 Adj (2 Adj (Adj (2 A)))) = 2^{41}$ , then the value of $det (A^{2})$ equal .

Let $P = [p_{ij}]$ and $Q = [q_{ij}]$ be two square matrices of order 3 such that $q_{ij} = 2^{(i + j - 1)} p_{ij}$ and $det (Q) = 2^{10}$ . Then the value of $det (adj (adj P))$ is:

Let the matrix $A = 001100010$ and the matrix $B_{0} = A^{49} + 2 A^{98}$ . If $B_{n} = adj (B_{n - 1})$ for all $n \geq 1$ , then $det (B_{4})$ is equal to :