JEE Advanced 2020

LEVELJEE Main

Animated Solution for Mathematics - Matrices and Determinants: Let $M$ be a $3 \times 3$ invertible matrix with real entries and let $I$ denote the $3 \times 3$ identity matrix. If $M^{- 1} = adj (adj M)$ , then which of the following statement is/are ALWAYS TRUE?

Q: Wait, why can't we just assume $M = I$ if $M^2 = I$?

That is a classic trap! While it is true that $I^2 = I$, the condition $M^2 = I$ defines an involutory matrix, not just the identity matrix. Think of diagonal matrices with entries like $-1, -1, 1$. Their square is $I$, but they are clearly not $I$. Always look for the most general case!

Q: Why did we take the determinant of both sides in the middle of the problem?

Because we had a matrix equation $I = (\det M) M^2$ and we needed to find the value of the scalar $\det M$. Taking the determinant converts the matrix equation into a simple algebraic equation involving the determinant, which allows us to solve for its value directly.

Select Answer:

* Multiple Correct

Visualized Solution

The Given Equation

We are given a $3 \times 3$ invertible matrix $M$ with real entries.
This means $det (M) \neq = 0$ .
The core equation provided is: $M^{- 1} = adj (adj M)$

Adjoint of Adjoint Property

Recall the standard property for an $n \times n$ matrix $A$ :
$adj (adj A) = (det A)^{n - 2} A$
For our $3 \times 3$ matrix $M$ , $n = 3$ .
Therefore, $adj (adj M) = (det M)^{3 - 2} M = (det M) M$

Substituting the Property

Substitute this result back into the given equation:
$M^{- 1} = (det M) M$
To eliminate the inverse, multiply both sides by $M$ :
$M M^{- 1} = (det M) M^{2}$
$I = (det M) M^{2}$

Applying Determinant

Take the determinant of both sides of the equation $I = (det M) M^{2}$ .
$det (I) = det ((det M) M^{2})$
We know $det (I) = 1$ .
For a scalar $k$ and an $n \times n$ matrix $A$ , $det (k A) = k^{n} det (A)$ .

Finding $det (M)$

Here, the scalar $k = det M$ and $n = 3$ .
So, $det ((det M) M^{2}) = (det M)^{3} \cdot det (M^{2})$
We also know $det (M^{2}) = (det M)^{2}$ .
Thus, $1 = (det M)^{3} \cdot (det M)^{2} = (det M)^{5}$ .
Since $M$ has real entries, $det M$ must be real.
Therefore, $det M = 1$ .

Evaluating $M^{2}$

Now, substitute $det M = 1$ back into our earlier equation: $I = (det M) M^{2}$ .
$I = (1) M^{2}$
$M^{2} = I$
This proves that $M^{2} = I$ is always true.

Checking $(adj M)^{2}$

We need to check the statement $(adj M)^{2} = I$ .
Using the property $adj (A^{k}) = (adj A)^{k}$ , we can write $(adj M)^{2} = adj (M^{2})$ .
Since we already proved $M^{2} = I$ , we get:
$adj (M^{2}) = adj (I)$
The adjoint of an identity matrix is the identity matrix itself, so $adj (I) = I$ .
Thus, $(adj M)^{2} = I$ is always true.

Counterexample for $M = I$

We found $M^{2} = I$ , which means $M$ is an involutory matrix.
Does $M^{2} = I$ imply $M = I$ ? Not necessarily!
Consider a diagonal matrix with entries $- 1, - 1, 1$ .
Its determinant is $1$ , and its square is $I$ , but it is not equal to $I$ .
Hence, $M = I$ is not always true.

Final Conclusion

The statements that are ALWAYS TRUE are:
$det M = 1$
$M^{2} = I$
$(adj M)^{2} = I$

00:00 / 00:00

The Sigma Insight: Adjoint and Inverse of a Matrix

Analyzing the Setup

We are given a

3 \times 3

invertible matrix

M

satisfying the relationship:

M^{- 1} = adj (adj M)

To solve this, we must utilize the fundamental property of the adjoint of an adjoint. For any

n \times n

matrix

A

, the identity is:

adj (adj A) = (det A)^{n - 2} A

The Master Equation

Given that

M

is a

3 \times 3

matrix, we set

n = 3

. Substituting this into our property, we obtain:

adj (adj M) = (det M)^{3 - 2} M = (det M) M

Equating this to the given expression

M^{- 1} = adj (adj M)

, we arrive at the simplified matrix equation:

M^{- 1} = (det M) M

The Determinant Dance

To isolate the scalar, we multiply both sides of the equation by

M

M M^{- 1} = (det M) M^{2}

I = (det M) M^{2}

To find the value of

det M

, we take the determinant of both sides. Using the property

det (k A) = k^{n} det (A)

for an

n \times n

matrix, where

k = det M

and

n = 3

det (I) = det ((det M) M^{2})

1 = (det M)^{3} \cdot det (M^{2})

Since

det (M^{2}) = (det M)^{2}

, the equation becomes:

1 = (det M)^{3} \cdot (det M)^{2} = (det M)^{5}

Given that

M

has real entries, the only real solution is:

det M = 1

Final Verification

Substituting

det M = 1

back into our equation

I = (det M) M^{2}

, we find:

M^{2} = I

This confirms that

M

is an involutory matrix. Note that

M^{2} = I

does not necessarily imply

M = I

; for example, a diagonal matrix

M = diag (- 1, - 1, 1)

satisfies both

det M = 1

and

M^{2} = I

without being the identity matrix.

Finally, we evaluate

(adj M)^{2}

. Using the property

adj (A^{k}) = (adj A)^{k}

(adj M)^{2} = adj (M^{2})

Since

M^{2} = I

, we have:

(adj M)^{2} = adj (I) = I

Thus, the relations $det M = 1$ , $M^{2} = I$ , and $(adj M)^{2} = I$ are always true for the given matrix

M

Common Traps & FAQs

Wait, why can't we just assume

M = I

M^{2} = I

That is a classic trap! While it is true that

I^{2} = I

, the condition

M^{2} = I

defines an involutory matrix, not just the identity matrix. Think of diagonal matrices with entries like

- 1, - 1, 1

. Their square is

I

, but they are clearly not

I

. Always look for the most general case!

How do I remember the

adj (adj M)

formula?

Think of it as a scaling factor. The adjoint operation essentially 'flips' the matrix and scales it by the determinant. Doing it twice brings you back to the original matrix, but scaled by the determinant raised to the power of

(n - 2)

. For

n = 3

, the power is

3 - 2 = 1

, so it is just

(det M) M

Why did we take the determinant of both sides in the middle of the problem?

Because we had a matrix equation

I = (det M) M^{2}

and we needed to find the value of the scalar

det M

. Taking the determinant converts the matrix equation into a simple algebraic equation involving the determinant, which allows us to solve for its value directly.

Animated Solution for Mathematics - Matrices and Determinants: Let $M$ be a $3 \times 3$ invertible matrix with real entries and let $I$ denote the $3 \times 3$ identity matrix. If $M^{- 1} = adj (adj M)$ , then which of the following statement is/are ALWAYS TRUE?

Select Answer:

Visualized Solution

The Given Equation

Adjoint of Adjoint Property

Substituting the Property

Applying Determinant

Finding $det (M)$

Evaluating $M^{2}$

Checking $(adj M)^{2}$

Counterexample for $M = I$

Final Conclusion

The Sigma Insight: Adjoint and Inverse of a Matrix

Analyzing the Setup

The Master Equation

The Determinant Dance

Final Verification

Similar Questions

For $3 \times 3$ matrices $M$ and $N$ , which of the following statement(s) is (are) NOT correct?

Let $A$ be a $2 \times 2$ matrix. \\ Statement-1: $a d j (a d j A) = A$ \\ Statement-2: $∣ a d j A ∣ = ∣ A ∣$

Let $I$ be the identity matrix of order $3 \times 3$ and for the matrix $A = λ 47 25 - 1 362$ , $∣ A ∣ = - 1$ . Let $B$ be the inverse of the matrix $adj (A adj (A^{2}))$ . Then $∣ (λ B + I) ∣$ is equal to _____

Let $A$ be a square matrix all of whose entries are integers. Then which one of the following is true?

Let $A$ be a $3 \times 3$ matrix such that $∣ adj (adj (adj A)) ∣ = 1 2^{4}$ . Then $∣ A^{- 1} adj A ∣$ is equal to

If A, B, and $(a d j (A^{- 1}) + a d j (B^{- 1}))$ are non-singular matrices of same order, then the inverse of $A (a d j (A^{- 1}) + a d j (B^{- 1}))^{- 1} B$ , is equal to

Let $A = [1 - 1 24]$ . If $A^{- 1} = α I + β A, α, β \in R, I$ is a $2 \times 2$ identity matrix, then $4 (α - β)$ 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 $M$ be a $2 \times 2$ symmetric matrix with integer entries. Then $M$ is invertible if

Let $A$ be a $2 \times 2$ symmetric matrix such that $A [11] = [37]$ and the determinant of $A$ be 1. If $A^{- 1} = α A + β I$ , where $I$ is an identity matrix of order $2 \times 2$ , then $α + β$ equals _______

Select Answer:

Visualized Solution

The Given Equation

Adjoint of Adjoint Property

Substituting the Property

Applying Determinant

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } det(M)

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } M2

Checking .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (adj M)2

Counterexample for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } M=I

Final Conclusion

The Sigma Insight: Adjoint and Inverse of a Matrix

Analyzing the Setup

The Master Equation

The Determinant Dance

Final Verification

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For 3×3 matrices M and N, which of the following statement(s) is (are) NOT correct?

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let A be a 2×2 matrix. \\ Statement-1: adj(adjA)=A \\ Statement-2: ∣adjA∣=∣A∣

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let A be a square matrix all of whose entries are integers. Then which one of the following is true?

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let A be a 3×3 matrix such that ∣adj(adj(adjA))∣=124. Then ∣A−1adjA∣ is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If A, B, and (adj(A−1)+adj(B−1)) are non-singular matrices of same order, then the inverse of A(adj(A−1)+adj(B−1))−1B, is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let A=[1−1​24​]. If A−1=αI+βA,α,β∈R,I is a 2×2 identity matrix, then 4(α−β) 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 M be a 2×2 symmetric matrix with integer entries. Then M is invertible if

Finding $det (M)$

Evaluating $M^{2}$

Checking $(adj M)^{2}$

Counterexample for $M = I$

For $3 \times 3$ matrices $M$ and $N$ , which of the following statement(s) is (are) NOT correct?

Let $A$ be a $2 \times 2$ matrix. \\ Statement-1: $a d j (a d j A) = A$ \\ Statement-2: $∣ a d j A ∣ = ∣ A ∣$

Let $A$ be a square matrix all of whose entries are integers. Then which one of the following is true?

Let $A$ be a $3 \times 3$ matrix such that $∣ adj (adj (adj A)) ∣ = 1 2^{4}$ . Then $∣ A^{- 1} adj A ∣$ is equal to

If A, B, and $(a d j (A^{- 1}) + a d j (B^{- 1}))$ are non-singular matrices of same order, then the inverse of $A (a d j (A^{- 1}) + a d j (B^{- 1}))^{- 1} B$ , is equal to

Let $A = [1 - 1 24]$ . If $A^{- 1} = α I + β A, α, β \in R, I$ is a $2 \times 2$ identity matrix, then $4 (α - β)$ 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 $M$ be a $2 \times 2$ symmetric matrix with integer entries. Then $M$ is invertible if