If the matrix A = [ \beginmatrix 1 & 0 & 0 0 & 2 & 0 3 & 0 & - 1 \endmatrix ] satisfies the equation A^20 + \alpha A^19 + \beta A = [ \beginmatrix 1 & 0 & 0 0 & 4 & 0 0 & 0 & 1 \endmatrix ] for some real numbers \alpha and \beta , then \beta-\alpha is equal to \underline\hspace2cm .

Matrices and Determinants PYQ — JEE Main 2021 · 26 Feb · Shift 2 · Q41 with Solution

View written solutionFree

Correct answer: 4

Write the matrix and observe its structure

We have

A=\begin{bmatrix} 1&0&0\\ 0&2&0\\ 3&0&-1 \end{bmatrix}.

Notice that $A$ is lower triangular, so powers of $A$ are also easy to analyze through its action on basis vectors or via eigenvalues on the diagonal.

We are given

A^{20}+\alpha A^{19}+\beta A= \begin{bmatrix} 1&0&0\\ 0&4&0\\ 0&0&1 \end{bmatrix}.

We need to find $\beta-\alpha$ .

Use diagonal entries

Since $A$ is lower triangular, the diagonal entries of $A^n$ are just the corresponding powers of diagonal entries of $A$ :

\operatorname{diag}(A^n)=(1^n,2^n,(-1)^n).

So,

\operatorname{diag}(A^{20})=(1,2^{20},1),

\operatorname{diag}(A^{19})=(1,2^{19},-1),

\operatorname{diag}(A)=(1,2,-1).

Now compare diagonal entries in

A^{20}+\alpha A^{19}+\beta A= \begin{bmatrix} 1&0&0\\ 0&4&0\\ 0&0&1 \end{bmatrix}.

From the $(1,1)$ entry:

1+\alpha+\beta=1 \implies \alpha+\beta=0.

So,

\beta=-\alpha.

From the $(2,2)$ entry:

2^{20}+\alpha 2^{19}+2\beta=4.

Substitute $\beta=-\alpha$ :

2^{20}+\alpha 2^{19}-2\alpha=4.

But solving this is not even necessary if we use another simpler entry.

From the $(3,3)$ entry:

1-\alpha-\beta=1 \implies \alpha+\beta=0,

which is the same condition.

Use the $(3,1)$ entry

This is the key extra condition.

Let

A^n=\begin{bmatrix} 1&0&0\\ 0&2^n&0\\ x_n&0&(-1)^n \end{bmatrix}.

We determine $x_n$ .

Since

A^{n+1}=A^nA,

we get for the $(3,1)$ entry:

x_{n+1}=x_n\cdot 1+(-1)^n\cdot 3=x_n+3(-1)^n.

Also from $A^1=A$ ,

x_1=3.

Compute a few terms:

$x_1=3$
$x_2=3+3(-1)=0$
$x_3=0+3(1)=3$
$x_4=3+3(-1)=0$

So the pattern is:

x_n= \begin{cases} 3,& n \text{ odd},\\ 0,& n \text{ even}. \end{cases}

Therefore,

(A^{20})_{31}=0,\qquad (A^{19})_{31}=3,\qquad A_{31}=3.

Now compare the $(3,1)$ entry of

A^{20}+\alpha A^{19}+\beta A= \begin{bmatrix} 1&0&0\\ 0&4&0\\ 0&0&1 \end{bmatrix}.

The RHS has $(3,1)$ entry $0$ , so

0+3\alpha+3\beta=0 \implies \alpha+\beta=0,

again same condition.

So we need one equation that actually determines $\alpha$ .

Use the $(2,2)$ entry to solve for $\alpha$ and $\beta$

We already have

\beta=-\alpha.

Now from the $(2,2)$ entry:

2^{20}+\alpha 2^{19}+2\beta=4.

Substitute $\beta=-\alpha$ :

2^{20}+\alpha 2^{19}-2\alpha=4.

Factor out $\alpha$ :

2^{20}+\alpha(2^{19}-2)=4.

Thus,

\alpha(2^{19}-2)=4-2^{20}.

Now,

4-2^{20}=-(2^{20}-4)=-(2)(2^{19}-2).

Hence,

\alpha=-2.

Therefore,

\beta=2.

So,

\beta-\alpha=2-(-2)=4.

Final answer

\boxed{4}

This matches the stored correct answer.

More from Matrices and Determinants

View written solutionFree

Correct answer: 4

Write the matrix and observe its structure

We have

A=\begin{bmatrix} 1&0&0\\ 0&2&0\\ 3&0&-1 \end{bmatrix}.

Notice that $A$ is lower triangular, so powers of $A$ are also easy to analyze through its action on basis vectors or via eigenvalues on the diagonal.

We are given

A^{20}+\alpha A^{19}+\beta A= \begin{bmatrix} 1&0&0\\ 0&4&0\\ 0&0&1 \end{bmatrix}.

We need to find $\beta-\alpha$ .

Use diagonal entries

Since $A$ is lower triangular, the diagonal entries of $A^n$ are just the corresponding powers of diagonal entries of $A$ :

\operatorname{diag}(A^n)=(1^n,2^n,(-1)^n).

So,

\operatorname{diag}(A^{20})=(1,2^{20},1),

\operatorname{diag}(A^{19})=(1,2^{19},-1),

\operatorname{diag}(A)=(1,2,-1).

Now compare diagonal entries in

A^{20}+\alpha A^{19}+\beta A= \begin{bmatrix} 1&0&0\\ 0&4&0\\ 0&0&1 \end{bmatrix}.

From the $(1,1)$ entry:

1+\alpha+\beta=1 \implies \alpha+\beta=0.

So,

\beta=-\alpha.

From the $(2,2)$ entry:

2^{20}+\alpha 2^{19}+2\beta=4.

Substitute $\beta=-\alpha$ :

2^{20}+\alpha 2^{19}-2\alpha=4.

But solving this is not even necessary if we use another simpler entry.

From the $(3,3)$ entry:

1-\alpha-\beta=1 \implies \alpha+\beta=0,

which is the same condition.

Use the $(3,1)$ entry

This is the key extra condition.

Let

A^n=\begin{bmatrix} 1&0&0\\ 0&2^n&0\\ x_n&0&(-1)^n \end{bmatrix}.

We determine $x_n$ .

Since

A^{n+1}=A^nA,

we get for the $(3,1)$ entry:

x_{n+1}=x_n\cdot 1+(-1)^n\cdot 3=x_n+3(-1)^n.

Also from $A^1=A$ ,

x_1=3.

Compute a few terms:

$x_1=3$
$x_2=3+3(-1)=0$
$x_3=0+3(1)=3$
$x_4=3+3(-1)=0$

So the pattern is:

x_n= \begin{cases} 3,& n \text{ odd},\\ 0,& n \text{ even}. \end{cases}

Therefore,

(A^{20})_{31}=0,\qquad (A^{19})_{31}=3,\qquad A_{31}=3.

Now compare the $(3,1)$ entry of

A^{20}+\alpha A^{19}+\beta A= \begin{bmatrix} 1&0&0\\ 0&4&0\\ 0&0&1 \end{bmatrix}.

The RHS has $(3,1)$ entry $0$ , so

0+3\alpha+3\beta=0 \implies \alpha+\beta=0,

again same condition.

So we need one equation that actually determines $\alpha$ .

Use the $(2,2)$ entry to solve for $\alpha$ and $\beta$

We already have

\beta=-\alpha.

Now from the $(2,2)$ entry:

2^{20}+\alpha 2^{19}+2\beta=4.

Substitute $\beta=-\alpha$ :

2^{20}+\alpha 2^{19}-2\alpha=4.

Factor out $\alpha$ :

2^{20}+\alpha(2^{19}-2)=4.

Thus,

\alpha(2^{19}-2)=4-2^{20}.

Now,

4-2^{20}=-(2^{20}-4)=-(2)(2^{19}-2).

Hence,

\alpha=-2.

Therefore,

\beta=2.

So,

\beta-\alpha=2-(-2)=4.

Final answer

\boxed{4}

This matches the stored correct answer.

Matrices and Determinants question

Matrices and Determinants question

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From the $(2,2)$ entry:

From the $(3,3)$ entry:

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