Let $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ and $B = \begin{bmatrix} \alpha \\ \beta \end{bmatrix} \neq \begin{bmatrix} 0 \\ 0 \end{bmatrix}$ such that $AB = B$ and $a + d = 2021$, then the value of $ad - bc$ is equal to ____.

### Analyzing the Setup We are given a matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ and a non-zero vector $B = \begin{bmatrix} \alpha \\ \beta \end{bmatrix}$ satisfying the equation $AB = B$. We are tasked with finding the value of $ad - bc$, given the constraint $a + d = 2021$. ### The Eigenvalue Insight The equation $AB = B$ implies that $B$ is an **eigenvector** of $A$ corresponding to the **eigenvalue** $\lambda = 1$. This geometric interpretation reveals that the transformation $A$ leaves the vector $B$ unchanged. ### The Homogeneous Trap To solve for the determinant, we rearrange the equation: $$AB - B = O$$ Since $B = IB$ (where $I$ is the identity matrix), we can rewrite the expression as: $$(A - I)B = O$$ Because $B \neq \begin{bmatrix} 0 \\ 0 \end{bmatrix}$, the system $(A - I)B = O$ must possess a **non-trivial solution**. This condition implies that the matrix $(A - I)$ is singular. ### The Determinant Constraint For the system to have a non-trivial solution, the determinant of $(A - I)$ must be zero. We construct the matrix $A - I$ as follows: $$A - I = \begin{bmatrix} a-1 & b \\ c & d-1 \end{bmatrix}$$ Setting the determinant to zero, we obtain: $$\det(A - I) = (a-1)(d-1) - bc = 0$$ ### The Algebraic Finale Expanding the determinant expression, we get: $$ad - a - d + 1 - bc = 0$$ Rearranging the terms to isolate the determinant of $A$, which is $ad - bc$, we find: $$ad - bc = a + d - 1$$ Given that $a + d = 2021$, we substitute this value into the equation: $$ad - bc = 2021 - 1$$ The final result is: $$\mathbf{ad - bc = 2020}$$

JEE Main 2021 (17 March Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Matrices and Determinants: Let $A = [a c b d]$ and $B = [α β] \neq = [00]$ such that $A B = B$ and $a + d = 2021$ , then the value of $a d - b c$ is equal to ____.

Q: Why does the existence of a non-trivial solution for $(A - I)B = O$ mean that $\det(A - I) = 0$?

Think of the system $(A - I)B = O$ as a set of linear equations. If the determinant of $(A - I)$ were non-zero, the matrix $(A - I)$ would be invertible. If it were invertible, we could multiply both sides by its inverse to get $B = (A - I)^{-1}O$, which would force $B$ to be the zero vector. Since the problem explicitly tells us $B$ is not the zero vector, the matrix $(A - I)$ cannot be invertible, which means its determinant must be zero!

Q: Is $ad - bc$ always the determinant of matrix $A$?

Yes, exactly! For any $2 \times 2$ matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, the determinant is defined as $\det(A) = ad - bc$. By solving the equation $\det(A - I) = 0$, we are essentially finding a relationship between the trace of the matrix ($a + d$) and its determinant.

Enter Numerical Value:

Visualized Solution

Analyze the Given Matrix Equation $A B = B$

Given matrix $A = [a c b d]$ and non-zero vector $B = [α β]$ .
The fundamental relation is $A B = B$ .
This implies that the vector $B$ is an eigenvector of $A$ corresponding to the eigenvalue $λ = 1$ .

Transform into a Homogeneous System $(A - I) B = O$

Subtract $B$ from both sides: $A B - B = O$ .
Express $B$ as $I B$ , where $I$ is the $2 \times 2$ identity matrix: $A B - I B = O$ .
Factor out the vector $B$ : $(A - I) B = O$ .

Condition for Non-Trivial Solutions

The system $(A - I) B = O$ is a homogeneous system of linear equations.
Since $B \neq = [00]$ , the system has a non-trivial solution.
The condition for a non-trivial solution is $det (A - I) = 0$ .

Construct the Matrix $A - I$

Matrix $A = [a c b d]$ and $I = [1001]$ .
Subtracting gives $A - I = [a - 1 c b d - 1]$ .

Set the Determinant to Zero

The condition $det (A - I) = 0$ becomes:
$a - 1 c b d - 1 = 0$
Expanding the determinant: $(a - 1) (d - 1) - (b) (c) = 0$ .

Expand the Algebraic Expression

Expand $(a - 1) (d - 1)$ : $a d - a - d + 1$ .
Substitute back into the equation: $a d - a - d + 1 - b c = 0$ .

Group the Terms for $a d - b c$

Rearrange the terms: $(a d - b c) - a - d + 1 = 0$ .
Isolate $a d - b c$ : $a d - b c = a + d - 1$ .

Substitute the Given Value $a + d = 2021$

We are given $a + d = 2021$ .
Substitute this into the equation: $a d - b c = 2021 - 1$ .

Final Calculation and Conclusion

Calculation: $2021 - 1 = 2020$ .
Final Result: $a d - b c = 2020$ .
The value of the determinant of matrix $A$ is $2020$ .

00:00 / 00:00

The Sigma Insight: Solution of System of Linear Equations (Matrix Method and Cramer's Rule)

Analyzing the Setup

We are given a matrix

A = [a c b d]

and a non-zero vector

B = [α β]

satisfying the equation

A B = B

. We are tasked with finding the value of

a d - b c

, given the constraint

a + d = 2021

The Eigenvalue Insight

The equation

A B = B

implies that

B

is an eigenvector of

A

corresponding to the eigenvalue

λ = 1

. This geometric interpretation reveals that the transformation

A

leaves the vector

B

unchanged.

The Homogeneous Trap

To solve for the determinant, we rearrange the equation:

A B - B = O

Since

B = I B

(where

I

is the identity matrix), we can rewrite the expression as:

(A - I) B = O

Because $B eq \begin{bmatrix} 0 \\ 0 \end{bmatrix}$, the system

(A - I) B = O

must possess a non-trivial solution. This condition implies that the matrix

(A - I)

is singular.

The Determinant Constraint

For the system to have a non-trivial solution, the determinant of

(A - I)

must be zero. We construct the matrix

A - I

as follows:

A - I = [a - 1 c b d - 1]

Setting the determinant to zero, we obtain:

det (A - I) = (a - 1) (d - 1) - b c = 0

The Algebraic Finale

Expanding the determinant expression, we get:

a d - a - d + 1 - b c = 0

Rearranging the terms to isolate the determinant of

A

, which is

a d - b c

, we find:

a d - b c = a + d - 1

Given that

a + d = 2021

, we substitute this value into the equation:

a d - b c = 2021 - 1

The final result is:

ad - bc = 2020

Common Traps & FAQs

Wait, why do we reject the possibility that

A

is the identity matrix just because

A B = B

That is a fantastic question! It is a common trap to assume

A = I

because

A B = B

. However,

A B = B

only implies that

B

is an eigenvector of

A

with eigenvalue

λ = 1

. There are infinitely many matrices

A

that satisfy this for a specific vector

B

. We don't need to find

A

itself; we only need to find the value of the expression

a d - b c

Why does the existence of a non-trivial solution for

(A - I) B = O

mean that

det (A - I) = 0

Think of the system

(A - I) B = O

as a set of linear equations. If the determinant of

(A - I)

were non-zero, the matrix

(A - I)

would be invertible. If it were invertible, we could multiply both sides by its inverse to get

B = (A - I)^{- 1} O

, which would force

B

to be the zero vector. Since the problem explicitly tells us

B

is not the zero vector, the matrix

(A - I)

cannot be invertible, which means its determinant must be zero!

a d - b c

always the determinant of matrix

A

Yes, exactly! For any

2 \times 2

matrix

A = [a c b d]

, the determinant is defined as

det (A) = a d - b c

. By solving the equation

det (A - I) = 0

, we are essentially finding a relationship between the trace of the matrix (

a + d

) and its determinant.

Animated Solution for Mathematics - Matrices and Determinants: Let $A = [a c b d]$ and $B = [α β] \neq = [00]$ such that $A B = B$ and $a + d = 2021$ , then the value of $a d - b c$ is equal to ____.

Enter Numerical Value:

Visualized Solution

Analyze the Given Matrix Equation $A B = B$

Transform into a Homogeneous System $(A - I) B = O$

Condition for Non-Trivial Solutions

Construct the Matrix $A - I$

Set the Determinant to Zero

Expand the Algebraic Expression

Group the Terms for $a d - b c$

Substitute the Given Value $a + d = 2021$

Final Calculation and Conclusion

The Sigma Insight: Solution of System of Linear Equations (Matrix Method and Cramer's Rule)

Analyzing the Setup

The Eigenvalue Insight

The Homogeneous Trap

The Determinant Constraint

The Algebraic Finale

Similar Questions

Let $A = 211010101$ , $B = [B_{1}, B_{2}, B_{3}]$ , where $B_{1}, B_{2}, B_{3}$ are column matrices, and $A B_{1} = 100$ , $A B_{2} = 230$ , $A B_{3} = 321$ . If $α = ∣ B ∣$ and $β$ is the sum of all the diagonal elements of $B$ , then $α^{3} + β^{3}$ is equal to

Comprehension Passage

If the point $P (a, b, c)$ , with reference to (E), lies on the plane $2 x + y + z = 1$ , then the value of $7 a + b + c$ is

Let $ω$ be a solution of $x^{3} - 1 = 0$ with $Im (ω) > 0$ , if $a = 2$ with $b$ and $c$ satisfying (E), then the value of $\frac{3}{ω ^{a}} + \frac{1}{ω ^{b}} + \frac{3}{ω ^{c}}$ is equal to

Let $b = 6$ , with $a$ and $c$ satisfying (E). If $α$ and $β$ are the roots of the quadratic equation $a x^{2} + b x + c = 0$ , then $\sum_{n = 0}^{\infty} (\frac{1}{α} + \frac{1}{β})^{n}$ is

Let $a, b, c$ be any real numbers. Suppose that there are real numbers $x, y, z$ not all zero such that $x = cy + b z, y = a z + c x$ , and $z = b x + a y$ . Then $a^{2} + b^{2} + c^{2} + 2 ab c$ is equal to

If $A = a 11 1 b b 0 d c, B = a 0 f 1 d g 1 c h, U = f g h, V = a^{2} 00, X = x y z$ and $A X = U$ has infinitely many solutions, prove that $B X = V$ has no unique solution. Also show that if $a f d \neq = 0$ , then $B X = V$ has no solution.

Consider the matrices : $A = [23 - 5 m], B = [20 m]$ and $X = [x y]$ . Let the set of all $m$ , for which the system of equations $A X = B$ has a negative solution (i.e., $x < 0$ and $y < 0$ ), be the interval $(a, b)$ . Then $8 \int_{a}^{b} ∣ A ∣ d m$ is equal to_________

If the system of linear equations: $x + y + 2 z = 6$ $2 x + 3 y + a z = a + 1$ $- x - 3 y + b z = 2 b$ where $a, b \in R,$ has infinitely many solutions, then $7 a + 3 b$ is equal to:

If the system of equations $x + y + a z = b$ , $2 x + 5 y + 2 z = 6$ , $x + 2 y + 3 z = 3$ has infinitely many solutions, then $2 a + 3 b$ is equal to

If the system of linear equations $2 x - 3 y = γ + 5$ , $α x + 5 y = β + 1$ , where $α, β, γ \in R$ has infinitely many solutions, then the value of $∣9 α + 3 β + 5 γ ∣$ is equal to

Let $S$ be the set of all values of $θ \in [- π, π]$ for which the system of linear equations $x + y + 3 z = 0$ , $- x + (tan θ) y + 7 z = 0$ , $x + y + (tan θ) z = 0$ has non-trivial solution. Then $120 \sum_{θ \in S} θ$ is equal to

If for some $α$ and $β$ in $R$ , the intersection of the following three planes $x + 4 y - 2 z = 1$ , $x + 7 y - 5 z = β$ , $x + 5 y + α z = 5$ is a line in $R^{3}$ , then $α + β$ is equal to:

Enter Numerical Value:

Visualized Solution

Analyze the Given Matrix Equation .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AB=B

Transform into a Homogeneous System .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (A−I)B=O

Condition for Non-Trivial Solutions

Construct the Matrix .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A−I

Set the Determinant to Zero

Expand the Algebraic Expression

Group the Terms for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ad−bc

Substitute the Given Value .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a+d=2021

Final Calculation and Conclusion

The Sigma Insight: Solution of System of Linear Equations (Matrix Method and Cramer's Rule)

Analyzing the Setup

The Eigenvalue Insight

The Homogeneous Trap

The Determinant Constraint

The Algebraic Finale

Similar Questions

Comprehension Passage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the point P(a,b,c), with reference to (E), lies on the plane 2x+y+z=1, then the value of 7a+b+c is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let ω be a solution of x3−1=0 with Im(ω)>0, if a=2 with b and c satisfying (E), then the value of ωa3​+ωb1​+ωc3​ is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let b=6, with a and c satisfying (E). If α and β are the roots of the quadratic equation ax2+bx+c=0, then ∑n=0∞​(α1​+β1​)n is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a,b,c be any real numbers. Suppose that there are real numbers x,y,z not all zero such that x=cy+bz,y=az+cx, and z=bx+ay. Then a2+b2+c2+2abc is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the system of linear equations: x+y+2z=6 2x+3y+az=a+1 −x−3y+bz=2b where a,b∈R, has infinitely many solutions, then 7a+3b is equal to:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the system of equations x+y+az=b, 2x+5y+2z=6, x+2y+3z=3 has infinitely many solutions, then 2a+3b is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the system of linear equations 2x−3y=γ+5, αx+5y=β+1, where α,β,γ∈R has infinitely many solutions, then the value of ∣9α+3β+5γ∣ is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If for some α and β in R, the intersection of the following three planes x+4y−2z=1, x+7y−5z=β, x+5y+αz=5 is a line in R3, then α+β is equal to:

Analyze the Given Matrix Equation $A B = B$

Transform into a Homogeneous System $(A - I) B = O$

Construct the Matrix $A - I$

Group the Terms for $a d - b c$

Substitute the Given Value $a + d = 2021$

If the point $P (a, b, c)$ , with reference to (E), lies on the plane $2 x + y + z = 1$ , then the value of $7 a + b + c$ is

Let $ω$ be a solution of $x^{3} - 1 = 0$ with $Im (ω) > 0$ , if $a = 2$ with $b$ and $c$ satisfying (E), then the value of $\frac{3}{ω ^{a}} + \frac{1}{ω ^{b}} + \frac{3}{ω ^{c}}$ is equal to

Let $b = 6$ , with $a$ and $c$ satisfying (E). If $α$ and $β$ are the roots of the quadratic equation $a x^{2} + b x + c = 0$ , then $\sum_{n = 0}^{\infty} (\frac{1}{α} + \frac{1}{β})^{n}$ is

Let $a, b, c$ be any real numbers. Suppose that there are real numbers $x, y, z$ not all zero such that $x = cy + b z, y = a z + c x$ , and $z = b x + a y$ . Then $a^{2} + b^{2} + c^{2} + 2 ab c$ is equal to

If the system of linear equations: $x + y + 2 z = 6$ $2 x + 3 y + a z = a + 1$ $- x - 3 y + b z = 2 b$ where $a, b \in R,$ has infinitely many solutions, then $7 a + 3 b$ is equal to:

If the system of equations $x + y + a z = b$ , $2 x + 5 y + 2 z = 6$ , $x + 2 y + 3 z = 3$ has infinitely many solutions, then $2 a + 3 b$ is equal to

If the system of linear equations $2 x - 3 y = γ + 5$ , $α x + 5 y = β + 1$ , where $α, β, γ \in R$ has infinitely many solutions, then the value of $∣9 α + 3 β + 5 γ ∣$ is equal to

If for some $α$ and $β$ in $R$ , the intersection of the following three planes $x + 4 y - 2 z = 1$ , $x + 7 y - 5 z = β$ , $x + 5 y + α z = 5$ is a line in $R^{3}$ , then $α + β$ is equal to: