Adjoint and inverse of matrices Questions in English

(C) We know that for any invertible matrix $M$,$\operatorname{adj}(M) = |M| M^{-1}$.
Let $M = Q^{-1} B P^{-1}$.
Then $\operatorname{adj}(M) = |Q^{-1} B P^{-1}| (Q^{-1} B P^{-1})^{-1}$.
Using the properties of determinants $|XY| = |X||Y|$ and $|X^{-1}| = \frac{1}{|X|}$,we have $|Q^{-1} B P^{-1}| = |Q^{-1}| |B| |P^{-1}| = \frac{1}{|Q|} |B| \frac{1}{|P|}$.
Since $|P| = 1$ and $|Q| = 1$,we get $|Q^{-1} B P^{-1}| = |B|$.
Now,calculating the inverse: $(Q^{-1} B P^{-1})^{-1} = (P^{-1})^{-1} B^{-1} (Q^{-1})^{-1} = P B^{-1} Q$.
Substituting these into the adjoint formula:
$\operatorname{adj}(Q^{-1} B P^{-1}) = |B| P B^{-1} Q$.
Since $\operatorname{adj} B = |B| B^{-1} = A$,we substitute $A$ for $|B| B^{-1}$.
Therefore,$\operatorname{adj}(Q^{-1} B P^{-1}) = P A Q$.

(B) We know that for a $3 \times 3$ matrix $A$,the determinant of its adjoint matrix is given by $\det(\text{adj}(A)) = (\det(A))^{n-1}$,where $n$ is the order of the matrix.
Here,$n = 3$,so $\det(P) = (\det(A))^{3-1} = (\det(A))^2$.
Given $\det(A) = 4$,we have $\det(P) = 4^2 = 16$.
Now,calculate the determinant of matrix $P = \begin{bmatrix} 1 & \alpha & 3 \\ 1 & 3 & 3 \\ 2 & 4 & 4 \end{bmatrix}$:
$\det(P) = 1(3 \times 4 - 3 \times 4) - \alpha(1 \times 4 - 3 \times 2) + 3(1 \times 4 - 3 \times 2)$
$\det(P) = 1(12 - 12) - \alpha(4 - 6) + 3(4 - 6)$
$\det(P) = 0 - \alpha(-2) + 3(-2) = 2\alpha - 6$.
Equating the two values: $2\alpha - 6 = 16$.
$2\alpha = 22 \Rightarrow \alpha = 11$.

(C) We know that for any square matrix $M$,the adjoint of the transpose is equal to the transpose of the adjoint.
That is,$\text{adj}(M^{\prime}) = (\text{adj } M)^{\prime}$.
Therefore,$\text{adj}(M^{\prime}) - (\text{adj } M)^{\prime} = (\text{adj } M)^{\prime} - (\text{adj } M)^{\prime} = O$,where $O$ is the null matrix.

(C) We know that for a square matrix $A$ of order $n$,the determinant of its adjoint matrix is given by $|\operatorname{adj} A| = |A|^{n-1}$.
Given that $A$ is a $3 \times 3$ matrix,so $n = 3$.
Therefore,$|\operatorname{adj} A| = |A|^{3-1} = |A|^2$.
Given $|B| = |\operatorname{adj} A| = 64$.
So,$|A|^2 = 64$.
Taking the square root on both sides,we get $|A| = \pm \sqrt{64} = \pm 8$.

(B) Let $X$ be a column vector of order $5 \times 1$ where all elements are $1$,i.e.,$X = [1, 1, 1, 1, 1]^T$.
Given that the sum of the elements of each row of $P$ is $1$,we can write this as $PX = X$.
Since $P$ is a non-singular matrix,$P^{-1}$ exists.
Multiplying both sides by $P^{-1}$,we get $P^{-1}(PX) = P^{-1}X$.
This simplifies to $(P^{-1}P)X = P^{-1}X$,which is $IX = P^{-1}X$.
Thus,$P^{-1}X = X$.
This implies that the sum of the elements of each row of $P^{-1}$ is also $1$.

(C) Given $A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix}$ and $I_{3} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$.
Then $A-3I_{3} = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix} - \begin{bmatrix} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{bmatrix} = \begin{bmatrix} -2 & 1 & 1 \\ 1 & -2 & 1 \\ 1 & 1 & -2 \end{bmatrix}$.
To check if the matrix is invertible,we calculate its determinant:
$\det(A-3I_{3}) = -2((-2)(-2) - (1)(1)) - 1((1)(-2) - (1)(1)) + 1((1)(1) - (-2)(1))$
$\det(A-3I_{3}) = -2(4-1) - 1(-2-1) + 1(1+2)$
$\det(A-3I_{3}) = -2(3) - 1(-3) + 1(3) = -6 + 3 + 3 = 0$.
Since the determinant of the matrix is $0$,the matrix $A-3I_{3}$ is non-invertible.

(A) Given matrix $A = \begin{bmatrix} 1 & 2 \\ -4 & -1 \end{bmatrix}$.
First,we calculate the determinant of $A$:
$|A| = (1)(-1) - (2)(-4) = -1 + 8 = 7$.
Since $|A| \neq 0$,$A^{-1}$ exists.
The adjoint of a $2 \times 2$ matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ is given by $\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$.
Therefore,$\text{adj}(A) = \begin{bmatrix} -1 & -2 \\ 4 & 1 \end{bmatrix}$.
The inverse is given by $A^{-1} = \frac{1}{|A|} \text{adj}(A)$.
$A^{-1} = \frac{1}{7} \begin{bmatrix} -1 & -2 \\ 4 & 1 \end{bmatrix}$.

(B) Given the equation $AB = 3I$,where $A$ and $B$ are square matrices of the same order and $I$ is the identity matrix.
To find $A^{-1}$,we multiply both sides of the equation by $A^{-1}$ from the left:
$A^{-1}(AB) = A^{-1}(3I)$
Using the associative property of matrix multiplication,we get:
$(A^{-1}A)B = 3(A^{-1}I)$
Since $A^{-1}A = I$ and $A^{-1}I = A^{-1}$,the equation simplifies to:
$IB = 3A^{-1}$
$B = 3A^{-1}$
Dividing both sides by $3$,we obtain:
$A^{-1} = \frac{1}{3}B$

(B) Given the equation $A^2-A+I=0$.
We can rewrite this as $A^2-A = -I$.
Multiplying both sides by $A^{-1}$ from the right,we get $(A^2-A)A^{-1} = -I \cdot A^{-1}$.
This simplifies to $A^2 A^{-1} - A A^{-1} = -A^{-1}$.
Since $A A^{-1} = I$,we have $A I - I = -A^{-1}$.
This gives $A - I = -A^{-1}$.
Multiplying by $-1$ on both sides,we get $A^{-1} = I - A$.

(B) Given $q_{ij} = 2^{(i+j-1)}p_{ij}$. The matrix $Q$ can be written as:
$Q = \begin{bmatrix} 2^1 p_{11} & 2^2 p_{12} & 2^3 p_{13} \\ 2^2 p_{21} & 2^3 p_{22} & 2^4 p_{23} \\ 2^3 p_{31} & 2^4 p_{32} & 2^5 p_{33} \end{bmatrix}$
Taking common factors from each row:
$\det(Q) = (2^1 \cdot 2^2 \cdot 2^3) \begin{vmatrix} p_{11} & p_{12} & p_{13} \\ 2 p_{21} & 2 p_{22} & 2 p_{23} \\ 2^2 p_{31} & 2^2 p_{32} & 2^2 p_{33} \end{vmatrix} = 2^6 \cdot (1 \cdot 2 \cdot 2^2) \det(P) = 2^6 \cdot 2^3 \det(P) = 2^9 \det(P)$
Given $\det(Q) = 2^{10}$,so $2^9 \det(P) = 2^{10} \implies \det(P) = 2$.
We know that $\det(\text{adj}(\text{adj } P)) = \det(P)^{(n-1)^2}$,where $n=3$.
$\det(\text{adj}(\text{adj } P)) = \det(P)^{(3-1)^2} = \det(P)^4 = 2^4 = 16$.

(D) First,simplify the integrand: $f(x) = \int \frac{x^{18}(7x^{-8} + 9x^{-10})}{(x^9(x^{-9} + x^{-7} + 2))^2} dx = \int \frac{7x^{-8} + 9x^{-10}}{(x^{-9} + x^{-7} + 2)^2} dx$.
Let $t = x^{-9} + x^{-7} + 2$,then $dt = (-9x^{-10} - 7x^{-8}) dx$,so $-(7x^{-8} + 9x^{-10}) dx = dt$.
Thus,$f(x) = \int -t^{-2} dt = t^{-1} + C = \frac{1}{x^{-9} + x^{-7} + 2} + C = \frac{x^9}{1 + x^2 + 2x^9} + C$.
Given $\lim_{x \to 0} f(x) = 0$,we find $C = 0$. Also $f(1) = \frac{1}{1+1+2} = \frac{1}{4}$,which is consistent.
Now,$f'(x) = \frac{9x^8(1+x^2+2x^9) - x^9(2x + 18x^8)}{(1+x^2+2x^9)^2}$.
At $x=1$,$f'(1) = \frac{9(4) - 1(20)}{4^2} = \frac{36-20}{16} = 1$.
Matrix $A = \begin{bmatrix} 0 & 0 & 1 \\ 1/4 & 1 & 1 \\ \alpha^2 & 4 & 1 \end{bmatrix}$.
$|A| = 1(\frac{1}{4} \times 4 - \alpha^2 \times 1) = 1 - \alpha^2$.
Given $|B| = |\text{adj}(\text{adj } A)| = |A|^{(n-1)^2} = |A|^4 = 81$,so $|A| = \pm 3$.
$1 - \alpha^2 = 3 \Rightarrow \alpha^2 = -2$ (not possible for real $\alpha^2$) or $1 - \alpha^2 = -3 \Rightarrow \alpha^2 = 4$.

(B) Given $A+A^T=O$,$A$ is a skew-symmetric matrix. Let $A = \begin{bmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{bmatrix}$.
From $A\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} 3 \\ 3 \\ 2 \end{bmatrix}$,we get:
$-a = 3 \Rightarrow a = -3$
$-b+c = 2$
$3a + 2b = -3 \Rightarrow 3(-3) + 2b = -3 \Rightarrow 2b = 6 \Rightarrow b = 3$.
Then $c = 2+b = 5$.
So,$A = \begin{bmatrix} 0 & -3 & 3 \\ 3 & 0 & 5 \\ -3 & -5 & 0 \end{bmatrix}$.
Then $A+I = \begin{bmatrix} 1 & -3 & 3 \\ 3 & 1 & 5 \\ -3 & -5 & 1 \end{bmatrix}$.
$|A+I| = 1(1+25) + 3(3+15) + 3(-15+3) = 26 + 54 - 36 = 44$.
We need $\det(\text{adj}(2\text{adj}(A+I)))$.
Since $A+I$ is a $3 \times 3$ matrix,$\text{adj}(A+I)$ is also $3 \times 3$.
$\det(2\text{adj}(A+I)) = 2^3 |\text{adj}(A+I)| = 8 |A+I|^2 = 8(44)^2$.
Then $\det(\text{adj}(2\text{adj}(A+I))) = (8 \cdot 44^2)^2 = (2^3 \cdot (2^2 \cdot 11)^2)^2 = (2^3 \cdot 2^4 \cdot 11^2)^2 = (2^7 \cdot 11^2)^2 = 2^{14} \cdot 11^4$.
Comparing with $(2)^\alpha \cdot (3)^\beta \cdot (11)^\gamma$,we have $\alpha=14, \beta=0, \gamma=4$.
Thus,$\alpha+\beta+\gamma = 14+0+4 = 18$.

(C) We know that the fundamental property of the adjoint of a matrix is $(\text{adj} A) \cdot A = |A|I$,where $I$ is the identity matrix of order $3 \times 3$.
Taking the determinant on both sides,we get $|(\text{adj} A) \cdot A| = ||A|I|$.
Since $|A|$ is a scalar,we use the property $|kA| = k^n|A|$,where $n$ is the order of the matrix.
Here,$n = 3$,so $|(\text{adj} A) \cdot A| = |A|^3 |I|$.
Since the determinant of an identity matrix $|I| = 1$,we have $|(\text{adj} A) \cdot A| = |A|^3 \times 1 = |A|^3$.

(B) The inverse of a matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ is given by $A^{-1} = \frac{1}{|A|} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$.
Given $A = \begin{bmatrix} 2 & 3 \\ 1 & -4 \end{bmatrix}$,the determinant $|A| = (2)(-4) - (3)(1) = -8 - 3 = -11$.
Thus,$A^{-1} = \frac{1}{-11} \begin{bmatrix} -4 & -3 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 4/11 & 3/11 \\ 1/11 & -2/11 \end{bmatrix}$.
Comparing this with the given $A^{-1} = \begin{bmatrix} a & 3/11 \\ 1/11 & b \end{bmatrix}$,we get $a = 4/11$ and $b = -2/11$.
Therefore,$a+b = 4/11 + (-2/11) = 2/11$.

(D) First,calculate the determinant of $A$: $\det(A) = 1(0-3) - 1(-10-1) + 2(-6-0) = -3 + 11 - 12 = -4$.
Using the property $\text{adj}(\text{adj}(M)) = \det(M)^{n-2} M$ for an $n \times n$ matrix,here $n=3$,so $\text{adj}(\text{adj}(M)) = \det(M) M$.
Let $M = 2(\text{adj} A)^{-1}$. Since $\text{adj} A = \det(A) A^{-1}$,we have $M = 2(\det(A) A^{-1})^{-1} = 2 \det(A)^{-1} A = 2(-4)^{-1} A = -\frac{1}{2} A$.
Then $\text{adj}(\text{adj}(M)) = \det(M) M = \det(-\frac{1}{2} A) (-\frac{1}{2} A) = (-\frac{1}{2})^3 \det(A) (-\frac{1}{2} A) = \frac{1}{16} \det(A) A = \frac{1}{16} (-4) A = -\frac{1}{4} A$.
The sum of all elements of $A$ is $1+1+2-2+0+1+1+3+5 = 12$.
Therefore,the sum of all elements of $-\frac{1}{4} A$ is $-\frac{1}{4} \times 12 = -3$.

(A) Given $A = \begin{bmatrix} 2 & -2 \\ 4 & -2 \end{bmatrix}$ and $B = \begin{bmatrix} 3 & 9 \\ 1 & 3 \end{bmatrix}$.
First,calculate the determinant of $A$: $|A| = (2)(-2) - (-2)(4) = -4 + 8 = 4$.
Since $|A| \neq 0$,$A^{-1}$ exists. $A^{-1} = \frac{1}{4} \begin{bmatrix} -2 & 2 \\ -4 & 2 \end{bmatrix} = \begin{bmatrix} -0.5 & 0.5 \\ -1 & 0.5 \end{bmatrix}$.
Given $PA = B \implies P = BA^{-1} = \begin{bmatrix} 3 & 9 \\ 1 & 3 \end{bmatrix} \begin{bmatrix} -0.5 & 0.5 \\ -1 & 0.5 \end{bmatrix} = \begin{bmatrix} -1.5-9 & 1.5+4.5 \\ -0.5-3 & 0.5+1.5 \end{bmatrix} = \begin{bmatrix} -10.5 & 6 \\ -3.5 & 2 \end{bmatrix}$.
Given $AQ = B \implies Q = A^{-1}B = \begin{bmatrix} -0.5 & 0.5 \\ -1 & 0.5 \end{bmatrix} \begin{bmatrix} 3 & 9 \\ 1 & 3 \end{bmatrix} = \begin{bmatrix} -1.5+0.5 & -4.5+1.5 \\ -3+0.5 & -9+1.5 \end{bmatrix} = \begin{bmatrix} -1 & -3 \\ -2.5 & -7.5 \end{bmatrix}$.
Now,$P+Q = \begin{bmatrix} -10.5-1 & 6-3 \\ -3.5-2.5 & 2-7.5 \end{bmatrix} = \begin{bmatrix} -11.5 & 3 \\ -6 & -5.5 \end{bmatrix}$.
Then $2(P+Q) = \begin{bmatrix} -23 & 6 \\ -12 & -11 \end{bmatrix}$.
The sum of the diagonal elements is $-23 + (-11) = -34$. The absolute value is $|-34| = 34$.

(B) First,calculate the determinant of $A$: $|A| = -1(0-0) - 1(1-0) - 1(0-0) = -1$.
Since $A$ is a $3 \times 3$ matrix,the property $adj(adj(A)) = |A|^{n-2} A$ holds,where $n=3$. Thus,$adj(adj(A)) = |A|^{3-2} A = (-1)A = -A$.
Next,we know $adj(A) = |A|A^{-1}$. Therefore,$adj(A)(adj(adj(A))) = (|A|A^{-1})(|A|A) = |A|^2 I = (-1)^2 I = I$.
The given equation becomes $A^2 - \alpha A + \beta I = M$,where $M = \begin{bmatrix} 2 & -2 & 2 \\ -2 & 0 & -1 \\ 0 & 0 & -1 \end{bmatrix}$.
Calculate $A^2$: $\begin{bmatrix} -1 & 1 & -1 \\ 1 & 0 & 1 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} -1 & 1 & -1 \\ 1 & 0 & 1 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 2 & -1 & 1 \\ -1 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$.
Substituting these into the equation: $\begin{bmatrix} 2 & -1 & 1 \\ -1 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} - \alpha \begin{bmatrix} -1 & 1 & -1 \\ 1 & 0 & 1 \\ 0 & 0 & 1 \end{bmatrix} + \beta \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 2 & -2 & 2 \\ -2 & 0 & -1 \\ 0 & 0 & -1 \end{bmatrix}$.
Comparing the elements,for the element at $(1, 2)$: $-1 - \alpha = -2 \implies \alpha = 1$.
For the element at $(1, 3)$: $1 + \alpha = 2 \implies \alpha = 1$.
For the element at $(3, 3)$: $1 - \alpha + \beta = -1 \implies 1 - 1 + \beta = -1 \implies \beta = -1$.
Thus,$(\alpha - \beta)^2 = (1 - (-1))^2 = 2^2 = 4$.