Matrices

Given \(A^3-4A^2+A+21I=O\)

Therefore every eigenvalue of \(A\) satisfies

\(x^3-4x^2+x+21=0\)

\(=(x-3)(x^2-x-7)\)

Since the factors are distinct, the eigenvalues of \(A\) are

\(3,\ \dfrac{1+\sqrt{29}}{2},\ \dfrac{1-\sqrt{29}}{2}\)

The trace of \(A\) is

\(\operatorname{tr}(A)=2+3+b=5+b\)

Also, the sum of eigenvalues is

\(3+\dfrac{1+\sqrt{29}}{2}+\dfrac{1-\sqrt{29}}{2}=4\)

\(5+b=4\)

\(\Rightarrow b=-1\)

\(\det(A)=2(3b-5)-ab\)

Putting \(b=-1\),

\(\det(A)=-16+a\)

The product of eigenvalues is

\(3\cdot\dfrac{1+\sqrt{29}}{2}\cdot\dfrac{1-\sqrt{29}}{2}\)

\(=3\cdot\dfrac{1-29}{4}\)

\(=-21\)

\(-16+a=-21\)

\(\Rightarrow a=-5\)

\(2a+3b=2(-5)+3(-1)\)

\(=-10-3\)

\(=-13\)

\( \Delta=\left|\begin{array}{lll}
1 & 5 & 6 \\
2 & 3 & 4 \\
1 & 6 & a
\end{array}\right|=0 \)
\( \Rightarrow(3 a-24)-5(2 a-4)+6(9)=0 \)
\( \Rightarrow-7 a+50=0 \)
\( a=\frac{50}{7}\)
Now \( \Delta_z=0 \Rightarrow\left|\begin{array}{lll}
1 & 5 & 4 \\
2 & 3 & 7 \\
1 & 6 & b
\end{array}\right|=0 \)
\(\Rightarrow(3 b-42)-5(2 b-7) +4(9)=0 \)
\( \Rightarrow-7 b+29=0\)
\(b=\frac{29}{7} \)
\( a-b=3 \)
\( \therefore x-y=3\)

\(\begin{aligned}
&\text { Given system has infinite solution }\\
&\begin{aligned}
& \Rightarrow\left|\begin{array}{ccc}
\lambda-4 & \lambda-2 & \lambda \\
2 & -3 & 5 \\
1 & 2 & 6
\end{array}\right|=0 \\
& \Rightarrow(\lambda-4)(-28)-(\lambda-2) 7+\lambda \times 7=0 \\
& \Rightarrow-4(\lambda-4)-(\lambda-2)+d=0 \\
& \Rightarrow-4 \lambda+16-\lambda+2+\lambda=0 \\
& =-4 \lambda=-18 \quad\lambda=\frac{9}{2}
\end{aligned}
\end{aligned}\)

For infinitely many solutions, Cramer's rule requires

\(D=0,\quad D_1=0,\quad D_2=0,\quad D_3=0\).

The coefficient determinant is

\(D=\begin{vmatrix}
1&1&1\\
2&5&5\\
1&2&m
\end{vmatrix}\).

Expanding along the first row,

\(D=\begin{vmatrix}
5&5\\
2&m
\end{vmatrix}
-\begin{vmatrix}
2&5\\
1&m
\end{vmatrix}
+\begin{vmatrix}
2&5\\
1&2
\end{vmatrix}\)

\(=(5m-10)-(2m-5)+(4-5)\)

\(=3m-6\).

Since \(D=0\),

\(3m-6=0\),

\(m=2\).

Now

\(D_1=\begin{vmatrix}
4&1&1\\
17&5&5\\
n&2&2
\end{vmatrix}\).

Since the second and third columns are identical,

\(D_1=0\).

Next,

\(D_2=\begin{vmatrix}
1&4&1\\
2&17&5\\
1&n&2
\end{vmatrix}\).

Expanding along the first row,

\(D_2=
\begin{vmatrix}
17&5\\
n&2
\end{vmatrix}
-4\begin{vmatrix}
2&5\\
1&2
\end{vmatrix}
+\begin{vmatrix}
2&17\\
1&n
\end{vmatrix}\)

\(=(34-5n)-4(4-5)+(2n-17)\)

\(=21-3n\).

Since \(D_2=0\),

\(21-3n=0\),

\(n=7\).

Also,

\(D_3=\begin{vmatrix}
1&1&4\\
2&5&17\\
1&2&7
\end{vmatrix}\)

\(=\begin{vmatrix}
1&1&4\\
2&5&17\\
1&2&7
\end{vmatrix}\)

\(=0\)

after substituting \(n=7\), confirming consistency.

Thus

\(m=2,\quad n=7\).

Now,

\(m^2+n^2-mn
=2^2+7^2-(2)(7)\)

\(=4+49-14\)

\(=39\).

$A=\left[\begin{array}{ccc}{\left(\sqrt{2}\right)}^2& {\left(\sqrt{2}\right)}^3& {\left(\sqrt{2}\right)}^4\\ {\left(\sqrt{2}\right)}^3& {\left(\sqrt{2}\right)}^4& {\left(\sqrt{2}\right)}^5\\ {\left(\sqrt{2}\right)}^4& {\left(\sqrt{2}\right)}^5& {\left(\sqrt{2}\right)}^6\end{array}\right]$

$A=\left[\begin{array}{ccc}2& 2\sqrt{2}& 4\\ 2\sqrt{2}& 4& 4\sqrt{2}\\ 4& 4\sqrt{2}& 8\end{array}\right]$

$A^2=2^2\left[\begin{array}{ccc}1& \sqrt{2}& 2\\ \sqrt{2}& 2& 2\sqrt{2}\\ 2& 2\sqrt{2}& 4\end{array}\right]\left[\begin{array}{ccc}1& \sqrt{2}& 2\\ \sqrt{2}& 2& 2\sqrt{2}\\ 2& 2\sqrt{2}& 4\end{array}\right]$

$=4\left[\begin{array}{ccc}-& -& -\\ -& -& -\\ \left(14\right)& \left(14\sqrt{2}\right)& \left(28\right)\end{array}\right]$

Sum of the all elements of third row

$$\begin{gathered} =4(14+14\sqrt{2}+28) \\ =4(42+14\sqrt{2}) \\ =168+56\sqrt{2} \end{gathered}$$

$$\begin{gathered} =\alpha +\beta \sqrt{2} \\ \therefore \alpha +\beta =168+56=224 \end{gathered}$$

$$\begin{gathered} ∆=\left|\begin{array}{ccc}1& 1& 1\\ 1& 2& {\lambda }^2\\ 1& 3& \lambda \end{array}\right|=1\left(3-2\right)-{\lambda }^2\left(3-1\right)+\lambda \left(2-1\right) \\ =-\left(2{\lambda }^2-\lambda -1\right) \\ =-\left(\lambda -1\right)\left(2\lambda +1\right) \end{gathered}$$

For unique solution \(\lambda \neq 1\) and \(\lambda \neq-\frac{1}{2}\)

If \(\lambda = 1\)

${\Delta }_x=\left|\begin{array}{lll}5& 1& 1\\ 9& 2& 1\\ \mu & 3& 1\end{array}\right|=-\mu +13$

${\Delta }_y=\left|\begin{array}{lll}1& 5& 1\\ 1& 9& 1\\ 1& \mu & 1\end{array}\right|=0$

${\Delta }_z=\left|\begin{array}{lll}1& 1& 5\\ 1& 2& 9\\ 1& 3& \mu \end{array}\right|=\mu -13$

Infinite solution $\lambda =1 \& \mu =13$
For no solution$\lambda =1 \& \mu \ne 13$
when $\lambda =-\frac{1}{2}$ and \(\mu \neq 13\) gives no solution.

\(If the system of equations

2x+3y−z=5x+αy+3z=−43x−y+βz=7

has infinitely many solutions, then 13αβ is equal to

\begin{aligned}
&2=\mathrm{k}_1+3 \mathrm{k}_2, 3=\mathrm{k}_1 \alpha-\mathrm{k}_2,-1=3 \mathrm{k}_1+\beta \mathrm{k}_2,-5=4 \mathrm{k}_1-7 \mathrm{k}_2\\
&\text { On solving we get }\\
&\begin{aligned}
& k_2=\frac{13}{19}, k_1=\frac{-1}{19}, \alpha=-70, \beta=\frac{-16}{13} \\
& 13 \alpha \beta=13(-70)\left(\frac{-16}{13}\right) \\
& =1120
\end{aligned}
\end{aligned}\)

$$\begin{gathered} \text{using cramer's rule :} \\ \Delta =\left|\begin{array}{ccc}1& 1& 2\\ 2& 3& \mathrm{a}\\ -1& -3& \mathrm{b}\end{array}\right|=0 \\ \Rightarrow 2\mathrm{a}+\mathrm{b}-6=0 ...\text{(1)} \\ {\Delta }_3=\left|\begin{array}{ccc}1& 1& 6\\ 2& 3& a+1\\ -1& -3& 2b\end{array}\right|=0 \\ \Rightarrow \mathrm{a}+\mathrm{b}-8=0 ...\text{(2)} \\ \text{from eqn} \left(1\right) \text{and} \left(2\right) \\ a=-2, b=10 \\ \Rightarrow 7a+3b=16 \end{gathered}$$

$\begin{array}{rl}& x+2y+3z=5\\ & 2x+3y+z=9\\ & 4x+3y+\lambda z=\mu \end{array}$

for infinite following $\Delta ={\Delta }_1={\Delta }_2={\Delta }_3=0$

$\Delta =\left|\begin{array}{lll}1& 2& 3\\ 2& 3& 1\\ 4& 3& \lambda \end{array}\right|=0\Rightarrow \lambda =-13$

${\Delta }_1=\left|\begin{array}{ccc}5& 2& 3\\ 9& 3& 1\\ \mu & 3& -13\end{array}\right|=0\Rightarrow \mu =15$

${\Delta }_2=\left|\begin{array}{ccc}1& 5& 3\\ 2& 9& 1\\ 4& 15& -13\end{array}\right|=0$

${\Delta }_3=\left|\begin{array}{ccc}1& 2& 5\\ 2& 3& 9\\ 4& 3& 15\end{array}\right|=0$

for $\lambda =-13,\mu =15$ system of equation has infinite solution hence \(\lambda+2 \mu=17\).

We are given:

• $A = \begin{bmatrix} \alpha & -1 \\ 6 & \beta \end{bmatrix}$ with $\alpha > 0$.
• $\det \left(A\right)=0⟹\alpha \beta -(-1)\left(6\right)=0$$⟹\alpha \beta +6=0⟹\alpha \beta =-6$.
• $\alpha + \beta = 1 \implies \beta = 1 - \alpha$.

Substitute $\beta$ in the determinant equation:

$\alpha(1 - \alpha) = -6$ $\alpha - \alpha^2 = -6$ $\alpha^2 - \alpha - 6 = 0$

Factoring the quadratic equation:

$(\alpha - 3)(\alpha + 2) = 0$

Since it's given that $\alpha > 0$, we must have $\alpha = 3$.

Then, $\beta = 1 - 3 = -2$.

Thus, the matrix $A$ is: $A = \begin{bmatrix} 3 & -1 \\ 6 & -2 \end{bmatrix}$

$A^2=\left[\begin{array}{cc}3& -1\\ 6& -2\end{array}\right]\left[\begin{array}{cc}3& -1\\ 6& -2\end{array}\right]$$=\left[\begin{array}{cc}9-6& -3+2\\ 18-12& -6+4\end{array}\right]=\left[\begin{array}{cc}3& -1\\ 6& -2\end{array}\right]=A$

Since $A^2 = A$, matrix $A$ is idempotent. .

Using the binomial expansion for matrices that commute (since $I$ and $A$ always commute):

$(I+A{)}^8={\sum }_{k=0}^8\left(\genfrac{}{}{0px}{}{8}{k}\right)I^{8-k}{A}^k$$=I+{\sum }_{k=1}^8\left(\genfrac{}{}{0px}{}{8}{k}\right)A^k$

Because null for all $k \ge 1$:

$(I + A)^8 = I + \left( \sum_{k=1}^{8} \binom{8}{k} \right) A$

We know that $\sum_{k=0}^{8} \binom{8}{k} = 2^8 = 256$,

so $\sum_{k=1}^{8} \binom{8}{k} = 2^8 - \binom{8}{0} = 256 - 1 = 255$.

$(I + A)^8 = I + 255A$

Substitute $I$ and $A$ back into the formula:

$(I + A)^8 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + 255 \begin{bmatrix} 3 & -1 \\ 6 & -2 \end{bmatrix}$

$(I + A)^8 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + \begin{bmatrix} 765 & -255 \\ 1530 & -510 \end{bmatrix}$

$(I+A{)}^8=\left[\begin{array}{cc}1+765& -255\\ 1530& 1-510\end{array}\right]$$=\left[\begin{array}{cc}766& -255\\ 1530& -509\end{array}\right]$

\(D=\left|\begin{array}{ccc}1 & 2 & t \\ 6 & 1 & 5 t \\ 3 & t^2 & f(t)\end{array}\right|=0\)

\(\Rightarrow 1\left(\mathrm{f}(\mathrm{t})-5 \mathrm{t}^3\right)-2(6 \mathrm{f}(\mathrm{t})-15 \mathrm{t})+\mathrm{t}\left(6 \mathrm{t}^2-3\right)=0\)

\(f(t)=\frac{t^3+27 t}{11}\)

\(f^{\prime}(\mathrm{t})=\frac{1}{11}\left(3 \mathrm{t}^2+27\right)>0 \ \forall \mathrm{t} \in \mathrm{R}\)

\(\Rightarrow f(t)\) is strictly increasing on \(\mathbb{R}\)

1. Find $\alpha$ and $\beta$ :
Given $det\left(A\right)=0⟹\alpha \beta -(-1)\left(6\right)=0⟹\alpha \beta +6=0⟹\alpha \beta =-6$.
Given $\alpha +\beta =1⟹\beta =1-\alpha$.
Substitute $\beta$ into the first equation: $\alpha (1-\alpha )=-6⟹\alpha -{\alpha }^2=-6⟹{\alpha }^2-\alpha -6=0$.
Factorizing the quadratic: $(\alpha -3)(\alpha +2)=0$.
Since $\alpha >0$, we have $\alpha =3$. Consequently, $\beta =1-3=-\mathbf{2}$.
Therefore, matrix $A=\left[\begin{array}{cc}3& -1\\ 6& -2\end{array}\right]$.

2. Check properties of A :
Let's compute $A^2$ :

$$\begin{gathered} A^2=\left[\begin{array}{ll}3& -1\\ 6& -2\end{array}\right]\left[\begin{array}{ll}3& -1\\ 6& -2\end{array}\right] \\ =\left[\begin{array}{cc}9-6& -3+2\\ 18-12& -6+4\end{array}\right]=\left[\begin{array}{ll}3& -1\\ 6& -2\end{array}\right]=A. \end{gathered}$$

Since $A^2=A$, matrix A is an idempotent matrix. This means $A^n=A$ for any positive integer n.
3. Compute $(I+A{)}^8$ :
Using binomial expansion: $(I+A{)}^8=I+\left(\genfrac{}{}{0ex}{}{8}{1}\right)A+\left(\genfrac{}{}{0ex}{}{8}{2}\right)A^2+\cdots +\left(\genfrac{}{}{0ex}{}{8}{8}\right)A^8$.
Since $A^k=A$ for all $k\ge 1$ :

$(I+A{)}^8=I+A\left[\left(\genfrac{}{}{0ex}{}{8}{1}\right)+\left(\genfrac{}{}{0ex}{}{8}{2}\right)+\cdots +\left(\genfrac{}{}{0ex}{}{8}{8}\right)\right].$

The sum of binomial coefficients $\sum _{k=1}^n\left(\genfrac{}{}{0ex}{}{n}{k}\right)=2^n-1$.
So, $(I+A{)}^8=I+\left(2^8-1\right)A=I+(256-1)A=I+255A$.

$$\begin{gathered} (I+A{)}^8=\left[\begin{array}{ll}1& 0\\ 0& 1\end{array}\right]+255\left[\begin{array}{ll}3& -1\\ 6& -2\end{array}\right] \\ =\left[\begin{array}{cc}1+765& 0-255\\ 0+1530& 1-510\end{array}\right]=\left[\begin{array}{cc}766& -255\\ 1530& -509\end{array}\right] \end{gathered}$$

Given

\(Q^{-1}=Q^T\)

So \(Q\) is an orthogonal matrix.

Since \(Q\) has integer entries, each row and each column must have exactly one non-zero entry equal to \(1\) or \(-1\).

Thus \(Q\) is a signed permutation matrix.

Now

\(PQ=QP\)

with

\(P=\operatorname{diag}(2,2,3)\).

Let

\(Q=(q_{ij})\).

Since \(PQ=QP\),

\((p_i-p_j)q_{ij}=0\)

for all \(i,j\),

where

\(p_1=p_2=2,\quad p_3=3\).

Hence

\(q_{13}=q_{31}=q_{23}=q_{32}=0\).

Therefore \(Q\) must have the form

\(\begin{pmatrix}
*&*&0\\
*&*&0\\
0&0&*
\end{pmatrix}\),

where \(Q\) is a signed permutation matrix.

The first two coordinates may be permuted in

\(2!\) ways.

Each non-zero entry can independently be assigned sign \(+1\) or \(-1\).

There are \(3\) non-zero entries, giving

\(2^3\) sign choices.

Therefore total number of such matrices

\(=2!\times2^3\)

\(=2\times8\)

\(=16\).

$$\begin{gathered} \mathrm{Given}: \mathrm{A}=\left[\begin{array}{cc}\frac{1}{\sqrt{2}}& -2\\ 0& 1\end{array}\right] \\ \text{and} P=\left[\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right] \\ \text{Clearly} {\mathrm{P}}^{\mathrm{T}}\mathrm{P}=\text{P}{\mathrm{P}}^{\mathrm{T}}=\text{I} \\ \mathrm{B} = {\mathrm{PAP}}^T \\ \mathrm{Pre} \mathrm{multiply} \mathrm{by} {\mathrm{P}}^{\mathrm{T}} \\ {\mathrm{P}}^{\mathrm{T}}\mathrm{B}={\mathrm{P}}^{\mathrm{T}}{\mathrm{PAP}}^{\mathrm{T}}={\mathrm{AP}}^{\mathrm{T}} \\ \mathrm{Now} \mathrm{post} \mathrm{multiply} \mathrm{by} \mathrm{P} \\ {\mathrm{P}}^{\mathrm{T}}\mathrm{BP}={\mathrm{AP}}^{\mathrm{T}}\mathrm{P}=\mathrm{A} \\ So {\mathrm{A}}^2=\left({\mathrm{P}}^{\mathrm{T}}\mathrm{BP}\right)\left(P^{\mathrm{T}}\mathrm{BP}\right) \\ {\mathrm{A}}^2={\mathrm{P}}^{\mathrm{T}}{\mathrm{B}}^2\mathrm{P} \\ \mathrm{Similarly} {\mathrm{A}}^{10}={\mathrm{P}}^{\mathrm{T}}{\mathrm{B}}^{10}\mathrm{P}=\mathrm{C} \\ \mathrm{A}=\left[\begin{array}{cc}\frac{1}{\sqrt{2}}& -2\\ 0& 1\end{array}\right] \\ \Rightarrow {\mathrm{A}}^2=\left[\begin{array}{cc}\frac{1}{2}& -\sqrt{2}-2\\ 0& 1\end{array}\right] \\ \mathrm{Similarly} \mathrm{check} {\mathrm{A}}^3 \mathrm{and} \mathrm{so} \mathrm{on} \\ \mathrm{since} \mathrm{C} = {\mathrm{A}}^{10} \\ \mathrm{Sum} \mathrm{of} \mathrm{diagonal} \mathrm{elements} \mathrm{of} \mathrm{C} \mathrm{is}{\left(\frac{1}{\sqrt{2}}\right)}^{10}+1 \\ =\frac{1}{32}+1=\frac{33}{32}=\frac{\mathrm{m}}{\mathrm{n}} \\ \mathrm{m} + \mathrm{n} = 65 \end{gathered}$$

$B={(I+A)}^{-1}, A+C=I$

$\Rightarrow B\left(I+A\right)=\left(I+A\right)B=I$

$\Rightarrow B+BA=B+AB$

$\Rightarrow B+B\left(I-C\right)=B+\left(I-C\right)B$

$\Rightarrow 2B-BC=2B-CB$

$\Rightarrow BC=CB$

$\therefore CB\left[\begin{array}{l}{x}_1\\ x_2\end{array}\right]=\left[\begin{array}{cc}1& -5\\ -1& 2\end{array}\right]\left[\begin{array}{l}{x}_1\\ x_2\end{array}\right]=\left[\begin{array}{c}12\\ -6\end{array}\right]$

$\Rightarrow \left[\begin{array}{l}{x}_1\\ x_2\end{array}\right]={\left[\begin{array}{cc}1& -5\\ -1& 2\end{array}\right]}^{-1}\left[\begin{array}{c}12\\ -6\end{array}\right]=-\frac{1}{3}\left[\begin{array}{ll}2& 5\\ 1& 1\end{array}\right]\left[\begin{array}{l}12\\ -6\end{array}\right]$

$\Rightarrow \left[\begin{array}{l}{x}_1\\ x_2\end{array}\right]=\left[\begin{array}{l}2\\ -2\end{array}\right]$

$\therefore x_1+x_2=0$

Ans. (2)
Sol.
\(\begin{aligned}
& \Delta=\left|\begin{array}{lll}
1 & 2 & -3 \\
2 & \lambda & 5 \\
4 & 3 & \mu
\end{array}\right|=0 \\
& 12 \lambda+\lambda \mu-4 \mu+7=0\ldots \ldots (i) \\
& \Delta z=\left|\begin{array}{lll}
1 & 2 & 2 \\
2 & \lambda & 5 \\
4 & 3 & 33
\end{array}\right|=0 \\
& \lambda=\frac{19}{5} \\
& \text { from (i) } \mu=263 \\
& \lambda+\mu=\frac{19}{5}+263=\frac{1334}{5}
\end{aligned}\)

Given

\(
x+2y-3z=2
\)

\(
2x+\lambda y+5z=5
\)

\(
14x+3y+\mu z=33
\)

For infinitely many solutions, the third equation must be a linear combination of the first two:

Checking the standard condition for infinitely many solutions using determinants,

\(
\Delta=
\begin{vmatrix}
1&2&-3\\
2&\lambda&5\\
14&3&\mu
\end{vmatrix}
=0
\)

\(
\Delta_x=\Delta_y=\Delta_z=0
\)

\(
\lambda=4,\qquad \mu=8
\)

\(
\lambda+\mu=12
\)

$$\begin{gathered} \left|\begin{array}{ccc}1& 1& 1\\ 1& 2& {\lambda }^2\\ 1& 3& \lambda \end{array}\right|=0 \\ \begin{array}{rl}& \Rightarrow 2{\lambda }^2-\lambda -1=0\\ & \lambda =1,-\frac{1}{2}\end{array} \\ \left|\begin{array}{ccc}1& 1& 5\\ 2& {\lambda }^2& 9\\ 3& \lambda & \mu \end{array}\right|=0\Rightarrow \mu =13 \end{gathered}$$

Infinite solution $\lambda =1\&\mu =13$
For unique solution $\lambda \ne 1$, $\mu \ne 13$
For no solution$\lambda =1\&\mu \ne 13$
Considering the case when$\lambda =-\frac{1}{2}$ and \(\mu \neq 13\) this will generate no solution case.

$$\begin{gathered} P=\left[\begin{array}{lr}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right] \\ P{P}^T=P^{T}P=I \\ {B}^2=PA{P}^{T}PA{P}^T \\ {B}^2=P{A}^2{P}^T \\ \mathrm{Similarly}, B^{10}=P{A}^{10}{P}^T \\ C = P^T{B}^{10}P \\ =P^{T}P{A}^{10}{P}^{T}P=A^{10} \\ {A}^2=\left[\begin{array}{cc}\frac{1}{\sqrt{2}}& -2\\ 0& 1\end{array}\right]\left[\begin{array}{cc}\frac{1}{\sqrt{2}}& -2\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}{\left(\frac{1}{\sqrt{2}}\right)}^2& -\\ 0& 1\end{array}\right] \\ {A}^3=\left[\begin{array}{cc}{\left(\frac{1}{\sqrt{2}}\right)}^2& -\\ 0& 1\end{array}\right]\left[\begin{array}{cc}\frac{1}{\sqrt{2}}& -2\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}{\left(\frac{1}{\sqrt{2}}\right)}^3& -\\ 0& 1\end{array}\right] \\ . \\ . \\ . \\ . \\ \mathrm{Similarly} \mathrm{for} A^{10}=\left[\begin{array}{cc}{\left(\frac{1}{\sqrt{2}}\right)}^{10}& -\\ 0& 1\end{array}\right] \\ \mathrm{Sum} \mathrm{of} \mathrm{diagonal} \mathrm{elements} \mathrm{of} C = \frac{1}{32}+1=\frac{33}{32}=\frac{m}{n} \\ \mathrm{and} m+n = 33+32 = 65. \end{gathered}$$

\(
\begin{aligned}
&11x+y+\lambda z=-5\\
&2x+3y+5z=3\\
&8x-19y-39z=\mu
\end{aligned}
\)

For infinite solutions.

\(
\begin{aligned}
&D=
\left|\begin{array}{ccc}
11 & 1 & \lambda\\
2 & 3 & 5\\
8 & -19 & -39
\end{array}\right|
=0\\
&\Rightarrow 11(-117+95)-1(-78-40)+\lambda(-38-24)\\
&\Rightarrow 11(-22)+118-\lambda(62)=0\\
&\Rightarrow 62\lambda=118-242\\
&\Rightarrow \lambda=\frac{-124}{62}=-2\\[4pt]
&D_1=
\left|\begin{array}{ccc}
-5 & 1 & -2\\
3 & 3 & 5\\
\mu & -19 & -39
\end{array}\right|
=0\\
&\Rightarrow -5(-117+95)-1(-117-5\mu)-2(-57-3\mu)=0\\
&\Rightarrow -5(-22)+117+5\mu+114+6\mu=0\\
&\Rightarrow 11\mu=-110-231=-341\\
&\Rightarrow \mu=-31\\[4pt]
&\lambda^4-\mu=(-2)^4-(-31)=16+31=47
\end{aligned}
\)

Let \( \mathrm{M}=\left[\begin{array}{ccc}\mathrm{a}_1 & \mathrm{a}_2 & \mathrm{a}_3 \\ \mathrm{~b}_1 & \mathrm{~b}_2 & \mathrm{~b}_3 \\ \mathrm{c}_1 & \mathrm{c}_2 & \mathrm{c}_3\end{array}\right]\)

then \(\mathrm{M}\left(\begin{array}{l}1 \\ 0 \\ 0\end{array}\right)=\left(\begin{array}{l}1 \\ 2 \\ 3\end{array}\right)\)

\(\Rightarrow \mathrm{a}_1=1, \mathrm{~b}_1=2, \mathrm{c}_1=3\)

\(M\left(\begin{array}{l}0 \\ 1 \\ 0\end{array}\right)=\left(\begin{array}{l}0 \\ 1 \\ 2\end{array}\right)\)

\(\Rightarrow \mathrm{a}_2=0, \mathrm{~b}_2=1, \mathrm{c}_2=2\)

\(M\left(\begin{array}{l}0 \\ 0 \\ 1\end{array}\right)=\left(\begin{array}{c}-1 \\ 1 \\ 1\end{array}\right)\)

\(\Rightarrow \mathrm{a}_3=-1, \mathrm{~b}_3=1, \mathrm{c}_3=1\)

Now \(M\left(\begin{array}{l}x \\ y \\ z\end{array}\right)=\left(\begin{array}{c}1 \\ 7 \\ 11\end{array}\right)\)

\(\Rightarrow \mathrm{x}+0 \mathrm{y}-\mathrm{z}=1\)

\(2 x+y+z=7\)

\(3 x+2 y+z=11\)

On solving, we get:

\(\mathrm{x}=2, \mathrm{y}=2, \mathrm{z}=1\)

\(\therefore x+y+z=5\)

$\mathit{\text{A}}=\left[\begin{array}{ll}2& 3\\ 3& 5\end{array}\right]\Rightarrow {\mathit{\text{A}}}^2=\left[\begin{array}{ll}13& 21\\ 21& 34\end{array}\right]$

$\left|A\right|=1$, Now

$\left|{\mathit{\text{A}}}^{2025}-3{\mathit{\text{A}}}^{2024}+{\mathit{\text{A}}}^{2023}\right|$

$=\left|A^{2023}\mathit{(}{A}^2\mathit{-}3A\mathit{+}I\mathit{)}\right|$

$=|\mathit{\text{A}}{|}^{2023}\left|{\mathit{\text{A}}}^2-3\mathit{\text{A}}+\mathit{\text{I}}\right|$

$$\begin{gathered} =1·\left|\begin{array}{cc}13-6+1& 21-9+0\\ 21-9+0& 34-15+1\end{array}\right| \\ =1.\left|\begin{array}{cc}8& 12\\ 12& 20\end{array}\right|=160-144=16 \end{gathered}$$

$$\begin{gathered} A=\left[\begin{array}{ccc}2& 2\sqrt{2}& 4\\ 2\sqrt{2}& 4& 4\sqrt{2}\\ 4& 4\sqrt{2}& 8\end{array}\right] \\ {A}^2=\left[\begin{array}{ccc}2& 2\sqrt{2}& 4\\ 2\sqrt{2}& 4& 4\sqrt{2}\\ 4& 4\sqrt{2}& 8\end{array}\right]\left[\begin{array}{ccc}2& 2\sqrt{2}& 4\\ 2\sqrt{2}& 4& 4\sqrt{2}\\ 4& 4\sqrt{2}& 8\end{array}\right] \end{gathered}$$

Now, sum of elements in third row =$56+112+56\sqrt{2}$

=$168+56\sqrt{2}$

Now, $\alpha +\beta =168+56=224$

Here,

\(D=\left|\begin{array}{lll}1 & 1 & 1 \\ 1 & 2 & 5 \\ 2 & 3 & \lambda\end{array}\right|=\lambda-6\)

For infinitely many solutions, \(D=0\).

So, \(\lambda-6=0 \Rightarrow \lambda=6\)

Now, \(D_1=\left|\begin{array}{ccc}6 & 1 & 1 \\ 10 & 2 & 5 \\ \mu & 3 & \lambda\end{array}\right|=2 \lambda+3 \mu-60\)

\(D_2=\left|\begin{array}{ccc}1 & 6 & 1 \\ 1 & 10 & 5 \\ 2 & \mu & \lambda\end{array}\right|=4(\lambda-\mu+10)\)

and \(D_3=\left|\begin{array}{ccc}1 & 1 & 6 \\ 1 & 2 & 10 \\ 2 & 3 & \mu\end{array}\right|=\mu-16\)

For the system to have infinitely many solutions,

\(D=D_1=D_2=D_3=0\)

From \(D_3=0, \mu-16=0 \Rightarrow \mu=16\)

Thus, \(\lambda+\mu=6+16=22\)

\(\Delta=\left|\begin{array}{lll}1 & 2 & 1 \\ 2 & 1 & \alpha \\ 8 & 4 & \beta\end{array}\right|=0\)

\(\Delta=1(\beta-4 \alpha)-2(2 \beta-8 \alpha)+1(8-8)=0\)

\(\beta-4 \alpha-4 \beta+16 \alpha=0\)

\(-3 \beta+12 \alpha=0 \Rightarrow \beta=4 \alpha\)

When \(\beta=4 \alpha\), third equation has coefficient row \(4(2,1, \alpha)\),

but RHS \(18 \neq 4 \times 5=20\), so the system has no solution

\(\frac{\beta}{\alpha}=\frac{4 \alpha}{\alpha}=4\)

Given $[4 a b]\left[\begin{array}{rrr}1& 16& 13\\ -1& -1& 2\\ -2& -14& -8\end{array}\right]=[0 0 0]$

$$\begin{array}{l} \Rightarrow a + 2 b = 4 \dots \dots (1) \\ a + 1 4 b = 6 4 \dots \dots (2) \\ 2 a - 3 b = - 5 2 \dots \dots (3) \\ \end{array}$$

Solving (1), (2), (3),

we get $a = -6, b = 5$

Given equation is

$\frac{4}{\alpha^4} + \frac{m}{\alpha^a} + \frac{n}{\alpha^b} = 3$ ( $\because \alpha$ is root of $x^2 + x + 1 = 0$ if $\alpha = w$ or $w^2$ )

$$\begin{array}{l} \Rightarrow \frac {4}{w} + m + \frac {n}{w ^ {2}} = 3 (\because a = - 6, b = 5) \\ \Rightarrow 4 w ^ {2} + m + n w = 3 \\ \Rightarrow 4 w ^ {2} + 4 w + m = 3 \quad \text {if} \quad n = 4 \\ \end{array}$$

$$\begin{array}{l}\Rightarrow 4(w^2+w)+m=3\\ \Rightarrow 4(-1)+m=3\\ \Rightarrow n=7,n=4\\ \therefore m+n=11\end{array}$$

\(\begin{align*}
A&=\begin{bmatrix}
a_1 & b_1 & c_1\\
a_2 & b_2 & c_2\\
a_3 & b_3 & c_3
\end{bmatrix}\\[4pt]
A\begin{bmatrix}
0\\
0\\
1
\end{bmatrix}
&=
\begin{bmatrix}
c_1\\
c_2\\
c_3
\end{bmatrix}
=
\begin{bmatrix}
1\\
1\\
2
\end{bmatrix}\\[4pt]
&\Rightarrow c_1=1,\; c_2=1,\; c_3=2\\[6pt]
A\begin{bmatrix}
1\\
0\\
1
\end{bmatrix}
&=
\begin{bmatrix}
c_1+a_1\\
c_2+a_2\\
c_3+a_3
\end{bmatrix}
=
\begin{bmatrix}
-1\\
0\\
1
\end{bmatrix}\\[4pt]
&\Rightarrow a_1=-2,\; a_2=-1,\; a_3=-1\\[6pt]
A\begin{bmatrix}
1\\
1\\
0
\end{bmatrix}
&=
\begin{bmatrix}
a_1+b_1\\
a_2+b_2\\
a_3+b_3
\end{bmatrix}
=
\begin{bmatrix}
1\\
1\\
0
\end{bmatrix}\\[4pt]
&\Rightarrow b_1=3,\; b_2=2,\; b_3=1\\[6pt]
&\Rightarrow
A=
\begin{bmatrix}
-2 & 3 & 1\\
-1 & 2 & 1\\
-1 & 1 & 2
\end{bmatrix}\\[6pt]
&\Rightarrow
A-2I=
\begin{bmatrix}
-4 & 3 & 1\\
-1 & 0 & 1\\
-1 & 1 & 0
\end{bmatrix}\\[6pt]
&|A-2I|=0\\[6pt]
\text{Now, }\;
\begin{bmatrix}
-4 & 3 & 1\\
-1 & 0 & 1\\
-1 & 1 & 0
\end{bmatrix}
\begin{bmatrix}
x_1\\
x_2\\
x_3
\end{bmatrix}
&=
\begin{bmatrix}
4\\
1\\
1
\end{bmatrix}\\[6pt]
-4x_1+3x_2+x_3&=4 \tag{1}\\
-x_1+x_3&=1 \tag{2}\\
-x_1+x_2&=1 \tag{3}
\end{align*}\)

\(
(1)-\bigl[(2)+3(3)\bigr]
\)

\(
0=0
\)

\(
\Rightarrow \text{infinite solutions}
\)

A matrix obtained from \(I_3\) by elementary row transformations must be invertible.

So its determinant must be non-zero.

Option A has all rows equal, so determinant \(=0\). Hence A is false.

Option B:

\(\begin{vmatrix}1&1&1\\2&3&4\\1&2&1\end{vmatrix}=1(3-8)-1(2-4)+1(4-3)\)

\(=-5+2+1=-2\ne0\)

So B is possible.

Option C:

\(\begin{vmatrix}1&1&1\\2&3&4\\2&5&8\end{vmatrix}=1(24-20)-1(16-8)+1(10-6)\)

\(=4-8+4=0\)

So C is false.

Option D:

\(\begin{vmatrix}1&1&1\\-1&1&2\\0&2&3\end{vmatrix}=1(3-4)-1(-3-0)+1(-2-0)\)

\(=-1+3-2=0\)

So D is false.

$$\begin{gathered} \text{ Given matrix }A=\left[\begin{array}{cc}1& 0\\ 0& -1\end{array}\right] \\ \text{ is orthogonal matrix, because }A{A}^T=I\text{. } \end{gathered}$$

$$\begin{gathered} \text{ So, }{A}^T{X}^{50}A=A^T{X}^{49}\left(AP{A}^T\right)A \\ =A^T{X}^{49}AP\left(A^{T}A\right) \end{gathered}$$

$$\begin{gathered} =A^T{X}^{49}AP \\ =A^T{X}^{48}\left(AP{A}^T\right)AP \end{gathered}$$

$$\begin{gathered} =A^T{X}^{48}A{P}^2\dots \dots =A^{T}A{P}^{50} \\ =I{P}^{50}=P^{50} \end{gathered}$$

$$\begin{gathered} \because P=\left[\begin{array}{ll}1& 1\\ 0& 1\end{array}\right] \\ \Rightarrow P^2=\left[\begin{array}{ll}1& 2\\ 0& 1\end{array}\right] \end{gathered}$$

$$\begin{gathered} \Rightarrow P^3=\left[\begin{array}{ll}1& 3\\ 0& 1\end{array}\right]\dots \\ \Rightarrow P^{50}=\left[\begin{array}{cc}1& 50\\ 0& 1\end{array}\right] \\ \text{ So, }{A}^T{X}^{50}A=P^{50}=\left[\begin{array}{cc}1& 50\\ 0& 1\end{array}\right] \end{gathered}$$

$$\begin{gathered} \mathrm{Using} \mathrm{Cramer}\text{'}\mathrm{s} \mathrm{rule} \\ △=\left|\begin{array}{ccc}1& 1& 1\\ 1& 2& 5\\ 1& 5& \lambda \end{array}\right|=0 \\ \Rightarrow \lambda =17 \\ also, {△}_z=\left|\begin{array}{lll}1& 1& 6\\ 1& 2& 9\\ 1& 5& \mu \end{array}\right|\ne 0 \\ \mu \ne 18 \end{gathered}$$

$\mathit{\text{A}}=\left[\begin{array}{ll}2& 3\\ 3& 5\end{array}\right]\Rightarrow {\mathit{\text{A}}}^2=\left[\begin{array}{ll}13& 21\\ 21& 34\end{array}\right]$

$\left|A\right|=1$, Now

$\left|{\mathit{\text{A}}}^{2025}-3{\mathit{\text{A}}}^{2024}+{\mathit{\text{A}}}^{2023}\right|$

$=\left|A^{2023}\mathit{(}{A}^2\mathit{-}3A\mathit{+}I\mathit{)}\right|$

$=|\mathit{\text{A}}{|}^{2023}\left|{\mathit{\text{A}}}^2-3\mathit{\text{A}}+\mathit{\text{I}}\right|$

$$\begin{gathered} =1·\left|\begin{array}{cc}13-6+1& 21-9+0\\ 21-9+0& 34-15+1\end{array}\right| \\ =1.\left|\begin{array}{cc}8& 12\\ 12& 20\end{array}\right|=160-144=16 \end{gathered}$$

Since, equations have no solution,

$$\begin{gathered} △=\left|\begin{array}{ccc}1& 1& 1\\ 1& 2& 5\\ 1& 5& \lambda \end{array}\right|=0 \\ \Rightarrow \lambda =17 \\ also,{△}_z=\left|\begin{array}{lll}1& 1& 6\\ 1& 2& 9\\ 1& 5& \mu \end{array}\right|\ne 0 \\ \mu \ne 18 \end{gathered}$$

$\Delta =\left|\begin{array}{ccc}3& 1& \varphi \\ 2& \alpha & -1\\ 1& 2& 1\end{array}\right|=0$

$3\mathrm{\alpha }+4\mathrm{\beta }-\mathrm{\alpha \beta }+3=0$

${\mathrm{\Delta }}_3=\left|\begin{array}{ccc}3& 1& 3\\ 2& \mathrm{\alpha }& -3\\ 1& 2& 4\end{array}\right|=0$

$9\mathrm{\alpha }+19=0$

$\mathrm{\alpha }=\frac{-19}{9},\mathrm{\beta }=\frac{6}{11}$

$\Rightarrow 22\mathrm{\beta }-9\mathrm{\alpha }=31$

\begin{aligned}
& \Delta=0 \Rightarrow\left|\begin{array}{ccc}
2 & -1 & 1 \\
5 & \lambda & 3 \\
100 & -47 & \mu
\end{array}\right|=0 \\
& 2(\lambda \mu+141)+(5 \mu-300)-235-100 \lambda=0 \\
& \Delta_3=0 \Rightarrow\left|\begin{array}{ccc}
2 & -1 & 4 \\
5 & \lambda & 12 \\
100 & -47 & 212
\end{array}\right|=0 \\
& 6 \lambda=-12 \Rightarrow \lambda=-2 \\
& \text { Put } \lambda=-2 \text { in (1) } \\
& 2(-2 \mu+141)+5 \mu-300-235+200=0 \\
& \mu=53 \\
& \therefore \mu-2\lambda= 57
\end{aligned}

Given

\(
2x+\lambda y+3z=5
\)

\(
3x+2y-z=7
\)

\(
4x+5y+\mu z=9
\)

For infinitely many solutions,

\(
\Delta=
\begin{vmatrix}
2&\lambda&3\\
3&2&-1\\
4&5&\mu
\end{vmatrix}
=0
\)

Also,

\(
\Delta_x=
\begin{vmatrix}
5&\lambda&3\\
7&2&-1\\
9&5&\mu
\end{vmatrix}
=0
\)

\(
\Delta_y=
\begin{vmatrix}
2&5&3\\
3&7&-1\\
4&9&\mu
\end{vmatrix}
=0
\)

\(
\Delta_z=
\begin{vmatrix}
2&\lambda&5\\
3&2&7\\
4&5&9
\end{vmatrix}
=0
\)

First use \(\Delta_y=0\):

\(
2(7\mu+9)-5(3\mu+4)+3(27-28)=0
\)

\(
14\mu+18-15\mu-20-3=0
\)

\(
\mu=-5
\)

Now use \(\Delta_z=0\):

\(
2(18-35)-\lambda(27-28)+5(15-8)=0
\)

\(
-34+\lambda+35=0
\)

\(
\lambda=-1
\)

Check with \(\Delta\):

\(
\Delta
=
2(2\mu+5)-\lambda(3\mu+4)+3(15-8)
\)

\(
=
2(-10+5)-(-1)(-15+4)+21
\)

\(
=-10-11+21=0
\)

\(
\Delta_x
=
5(2\mu+5)-\lambda(7\mu+9)+3(35-18)
\)

\(
=5(-5)-(-1)(-35+9)+51
\)

\(
=-25-26+51=0
\)

Hence the condition is satisfied.

\(
\lambda=-1,\qquad \mu=-5
\)

\(
\lambda^2+\mu^2
=
(-1)^2+(-5)^2
=
1+25
=
26
\)

$B={(I+A)}^{-1}, A+C=I$

$\Rightarrow B\left(I+A\right)=\left(I+A\right)B=I$

$\Rightarrow B+BA=B+AB$

$\Rightarrow B+B\left(I-C\right)=B+\left(I-C\right)B$

$\Rightarrow 2B-BC=2B-CB$

$\Rightarrow BC=CB$

$\therefore CB\left[\begin{array}{l}{x}_1\\ x_2\end{array}\right]=\left[\begin{array}{cc}1& -5\\ -1& 2\end{array}\right]\left[\begin{array}{l}{x}_1\\ x_2\end{array}\right]=\left[\begin{array}{c}12\\ -6\end{array}\right]$

$\Rightarrow \left[\begin{array}{l}{x}_1\\ x_2\end{array}\right]={\left[\begin{array}{cc}1& -5\\ -1& 2\end{array}\right]}^{-1}\left[\begin{array}{c}12\\ -6\end{array}\right]=-\frac{1}{3}\left[\begin{array}{ll}2& 5\\ 1& 1\end{array}\right]\left[\begin{array}{l}12\\ -6\end{array}\right]$

$\Rightarrow \left[\begin{array}{l}{x}_1\\ x_2\end{array}\right]=\left[\begin{array}{l}2\\ -2\end{array}\right]$

$\therefore x_1+x_2=0$

Let \(x \sin \theta=y \sin \left(\theta+\frac{2 \pi}{3}\right)=z \sin \left(\theta+\frac{4 \pi}{3}\right)=\lambda\)

\(\Rightarrow \frac{\lambda}{x}=\sin \theta ; \frac{\lambda}{y}=\sin \left(\theta+\frac{2 \pi}{3}\right) ; \frac{\lambda}{z}=\sin \left(\theta+\frac{4 \pi}{3}\right)\)

\(\Rightarrow \lambda\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=\sin \theta+\sin \left(\theta+\frac{2 \pi}{3}\right)+\sin \left(\theta+\frac{2 \pi}{3}\right)\)

\(=\sin \theta+2 \sin \left(\frac{2 \theta+2 \pi}{2}\right) \cos \left(\frac{\pi}{3}\right)\)

\(=\sin \theta-2 \sin \theta \cdot \frac{1}{2}\)

\(\Rightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)

\(\Rightarrow x y+y z+z x=0\)

\((x+y+z)^2=x^2+y^2+z^2+2(x y+y z+x z)=x^2+y^2+z^2 \neq 0\)

Hence trace of \(R=x+y+z\neq 0\)

\(\operatorname{adj}(\operatorname{adj} R)=|R|^{3-2} R=x y z(R)\)

\(\operatorname{tr}(\operatorname{adj} \operatorname{adj} R)=x y z \operatorname{tr}(R)=x y z(x+y+z) \neq 0\)

Hence both statements are false

$$\begin{gathered} \text{ Given matrix }A=\left[\begin{array}{cc}1& 0\\ 0& -1\end{array}\right] \\ \text{ is orthogonal matrix, because }A{A}^T=I\text{. } \end{gathered}$$

$$\begin{gathered} \text{ So, }{A}^T{X}^{50}A=A^T{X}^{49}\left(AP{A}^T\right)A \\ =A^T{X}^{49}AP\left(A^{T}A\right) \end{gathered}$$

$$\begin{gathered} =A^T{X}^{49}AP \\ =A^T{X}^{48}\left(AP{A}^T\right)AP \end{gathered}$$

$$\begin{gathered} =A^T{X}^{48}A{P}^2\dots \dots =A^{T}A{P}^{50} \\ =I{P}^{50}=P^{50} \end{gathered}$$

$$\begin{gathered} \because P=\left[\begin{array}{ll}1& 1\\ 0& 1\end{array}\right] \\ \Rightarrow P^2=\left[\begin{array}{ll}1& 2\\ 0& 1\end{array}\right] \end{gathered}$$

$$\begin{gathered} \Rightarrow P^3=\left[\begin{array}{ll}1& 3\\ 0& 1\end{array}\right]\dots \\ \Rightarrow P^{50}=\left[\begin{array}{cc}1& 50\\ 0& 1\end{array}\right] \\ \text{ So, }{A}^T{X}^{50}A=P^{50}=\left[\begin{array}{cc}1& 50\\ 0& 1\end{array}\right] \end{gathered}$$