Chapter 4: Determinants

Expanding the $2 \times 2$ determinant using the standard formula:

$\left|\begin{array}{rr}2 & 4 \\ -5 & -1\end{array}\right| = 2(-1) – 4(-5) = -2 + 20 = 18$

(i) Applying the $2 \times 2$ determinant formula directly:
$\left|\begin{array}{rr}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta\end{array}\right| = \cos\theta(\cos\theta) – (-\sin\theta)(\sin\theta) = \cos^2\theta + \sin^2\theta = 1$

(ii) Expanding using the standard formula:
$\left|\begin{array}{cc}x^2-x+1 & x-1 \\ x+1 & x+1\end{array}\right| = (x^2-x+1)(x+1) – (x-1)(x+1)$
$= (x^3+1) – (x^2-1) = x^3 – x^2 + 2$

Start by computing $2A$ by multiplying every element of $A$ by 2:
$2A = 2\left[\begin{array}{ll}1 & 2 \\ 4 & 2\end{array}\right] = \left[\begin{array}{ll}2 & 4 \\ 8 & 4\end{array}\right]$

L.H.S: $|2A| = \left|\begin{array}{ll}2 & 4 \\ 8 & 4\end{array}\right| = 2(4) – 4(8) = 8 – 32 = -24$

R.H.S: $4|A| = 4\left|\begin{array}{ll}1 & 2 \\ 4 & 2\end{array}\right| = 4(2-8) = 4(-6) = -24$

Since L.H.S. $=$ R.H.S. $= -24$, the result is verified.

First scale every entry of $A$ by 3 to get $3A$:
$3A = \left[\begin{array}{rrr}3 & 0 & 3 \\ 0 & 3 & 6 \\ 0 & 0 & 12\end{array}\right]$

L.H.S: Expanding $|3A|$ along the first column (which has two zeros):
$|3A| = 3[3(12) – 6(0)] = 3 \times 36 = 108$

R.H.S: Expanding $|A|$ along the first column:
$27|A| = 27 \times 1(4-0) = 27 \times 4 = 108$

Since L.H.S. $=$ R.H.S. $= 108$, the result is verified.

(i) Expanding along the first row:
$= 3\left|\begin{array}{rr}0 & -1 \\ -5 & 0\end{array}\right| -(-1)\left|\begin{array}{rr}0 & -1 \\ 3 & 0\end{array}\right| + (-2)\left|\begin{array}{rr}0 & 0 \\ 3 & -5\end{array}\right|$
$= 3(0-5) + 1(0+3) – 2(0-0) = -15 + 3 – 0 = -12$

(ii) Expanding along the first row:
$= 3\left|\begin{array}{rr}1 & -2 \\ 3 & 1\end{array}\right| -(-4)\left|\begin{array}{rr}1 & -2 \\ 2 & 1\end{array}\right| + 5\left|\begin{array}{rr}1 & 1 \\ 2 & 3\end{array}\right|$
$= 3(1+6) + 4(1+4) + 5(3-2) = 21 + 20 + 5 = 46$

(iii) Expanding along the first row:
$= 0\left|\begin{array}{rr}0 & -3 \\ 3 & 0\end{array}\right| – 1\left|\begin{array}{rr}-1 & -3 \\ -2 & 0\end{array}\right| + 2\left|\begin{array}{rr}-1 & 0 \\ -2 & 3\end{array}\right|$
$= 0 – (0-6) + 2(-3-0) = 0 + 6 – 6 = 0$

(iv) Expanding along the first row:
$= 2\left|\begin{array}{rr}2 & -1 \\ -5 & 0\end{array}\right| -(-1)\left|\begin{array}{rr}0 & -1 \\ 3 & 0\end{array}\right| + (-2)\left|\begin{array}{rr}0 & 2 \\ 3 & -5\end{array}\right|$
$= 2(0-5) + 1(0+3) – 2(0-6) = -10 + 3 + 12 = 5$

Expanding $|A|$ along the first row:
$|A| = 1\left|\begin{array}{rr}1 & -3 \\ 4 & -9\end{array}\right| – 1\left|\begin{array}{rr}2 & -3 \\ 5 & -9\end{array}\right| + (-2)\left|\begin{array}{rr}2 & 1 \\ 5 & 4\end{array}\right|$
$= (-9+12) – (-18+15) – 2(8-5)$
$= 3 – (-3) – 6 = 3 + 3 – 6 = 0$

Note: A matrix for which $|A| = 0$ is called a singular matrix.

(i) Evaluate both determinants and set them equal:
$2(1) – 4(5) = 2x(x) – 4(6)$
$\Rightarrow 2 – 20 = 2x^2 – 24$
$\Rightarrow -18 = 2x^2 – 24$
$\Rightarrow 2x^2 = 6$
$\Rightarrow x^2 = 3$
$\Rightarrow x = \pm\sqrt{3}$

(ii) Evaluate both determinants and set them equal:
$10 – 12 = 5x – 6x$
$\Rightarrow -2 = -x$
$\Rightarrow x = 2$

Evaluating both sides:
$x^2 – 36 = 36 – 36$
$\Rightarrow x^2 – 36 = 0$
$\Rightarrow x^2 = 36$
$\Rightarrow x = \pm 6$

Option (B) is correct.

(i) Using the determinant formula for area:
Area $= \left|\frac{1}{2}\left|\begin{array}{lll}1 & 0 & 1 \\ 6 & 0 & 1 \\ 4 & 3 & 1\end{array}\right|\right|$
Expanding along the first row: $= \frac{1}{2}[1(0-3) – 0 + 1(18-0)]$
$= \left|\frac{1}{2}(-3+18)\right| = \left|\frac{15}{2}\right| = \dfrac{15}{2}$ sq. units

(ii) Setting up the determinant:
Area $= \left|\frac{1}{2}\left|\begin{array}{rrr}2 & 7 & 1 \\ 1 & 1 & 1 \\ 10 & 8 & 1\end{array}\right|\right|$
$= \frac{1}{2}[2(1-8) – 7(1-10) + 1(8-10)]$
$= \frac{1}{2}(-14 + 63 – 2) = \left|\frac{47}{2}\right| = \dfrac{47}{2}$ sq. units

(iii) Setting up the determinant:
Area $= \left|\frac{1}{2}\left|\begin{array}{rrr}-2 & -3 & 1 \\ 3 & 2 & 1 \\ -1 & -8 & 1\end{array}\right|\right|$
$= \frac{1}{2}[-2(2+8) – (-3)(3+1) + 1(-24+2)]$
$= \frac{1}{2}(-20 + 12 – 22) = \frac{1}{2}(-30) = -15$
Area $= |-15| = 15$ sq. units

The three points will be collinear if the area of the triangle they form equals zero.

Area $= \frac{1}{2}\left|\begin{array}{lll}a & b+c & 1 \\ b & c+a & 1 \\ c & a+b & 1\end{array}\right|$

Expanding along the first row:
$= \frac{1}{2}[a(c+a-a-b) – (b+c)(b-c) + 1(b(a+b) – c(c+a))]$
$= \frac{1}{2}[a(c-b) – (b^2 – c^2) + (ab + b^2 – c^2 – ac)]$
$= \frac{1}{2}(ac – ab – b^2 + c^2 + ab + b^2 – c^2 – ac)$
$= \frac{1}{2}(0) = 0$

Since the area is zero, the points A, B, and C are collinear.

(i) Setting up the area condition:
$\left|\frac{1}{2}\left|\begin{array}{lll}k & 0 & 1 \\ 4 & 0 & 1 \\ 0 & 2 & 1\end{array}\right|\right| = 4$
Expanding along the first row: $\left|\frac{1}{2}(k(0-2) – 0 + 1(8-0))\right| = 4$
$\Rightarrow \left|\frac{1}{2}(-2k+8)\right| = 4 \Rightarrow |-k+4| = 4$
$\Rightarrow -k + 4 = \pm 4$
Taking the positive sign: $-k + 4 = 4 \Rightarrow k = 0$
Taking the negative sign: $-k + 4 = -4 \Rightarrow k = 8$
Therefore $k = 0$ or $k = 8$.

(ii) Setting up the area condition:
$\left|\frac{1}{2}\left|\begin{array}{rrr}-2 & 0 & 1 \\ 0 & 4 & 1 \\ 0 & k & 1\end{array}\right|\right| = 4$
Expanding: $\left|\frac{1}{2}(-2(4-k) – 0 + 0)\right| = 4$
$\Rightarrow \left|\frac{1}{2}(-8+2k)\right| = 4 \Rightarrow |-4+k| = 4$
$\Rightarrow -4 + k = \pm 4$
Taking the positive sign: $k = 8$
Taking the negative sign: $k = 0$
Therefore $k = 0$ or $k = 8$.

(i) Let $P(x,y)$ be any point on the line. Since three collinear points form a triangle of zero area:
$\frac{1}{2}\left|\begin{array}{lll}x & y & 1 \\ 1 & 2 & 1 \\ 3 & 6 & 1\end{array}\right| = 0$
Multiplying both sides by 2 and expanding along the first row:
$x(2-6) – y(1-3) + 1(6-6) = 0$
$\Rightarrow -4x + 2y = 0$
Dividing by $-2$: $\boxed{y = 2x}$ is the required equation of the line.

(ii) Let $P(x,y)$ be any point on the line. Setting the area to zero:
$\frac{1}{2}\left|\begin{array}{lll}x & y & 1 \\ 3 & 1 & 1 \\ 9 & 3 & 1\end{array}\right| = 0$
Expanding along the first row:
$x(1-3) – y(3-9) + 1(9-9) = 0$
$\Rightarrow -2x + 6y = 0$
Dividing by $-2$: $\boxed{x – 3y = 0}$ is the required equation of the line.

Setting up the area condition:
$\left|\frac{1}{2}\left|\begin{array}{rrr}2 & -6 & 1 \\ 5 & 4 & 1 \\ k & 4 & 1\end{array}\right|\right| = 35$
Expanding along the first row:
$\left|\frac{1}{2}[2(4-4) – (-6)(5-k) + 1(20-4k)]\right| = 35$
$\Rightarrow \left|\frac{1}{2}(0 + 30 – 6k + 20 – 4k)\right| = 35$
$\Rightarrow \left|\frac{1}{2}(50-10k)\right| = 35 \Rightarrow |25-5k| = 35$
$\Rightarrow 25 – 5k = \pm 35$
Taking positive sign: $25 – 5k = 35 \Rightarrow k = -2$
Taking negative sign: $25 – 5k = -35 \Rightarrow k = 12$
Therefore $k = 12$ or $k = -2$. Option (D) is correct.

(i) Let $\Delta = \left|\begin{array}{rr}2 & -4 \\ 0 & 3\end{array}\right|$

$M_{11} = |3| = 3$;   $A_{11} = (-1)^{1+1} \cdot 3 = 3$
$M_{12} = |0| = 0$;   $A_{12} = (-1)^{1+2} \cdot 0 = 0$
$M_{21} = |-4| = -4$;   $A_{21} = (-1)^{2+1} \cdot (-4) = 4$
$M_{22} = |2| = 2$;   $A_{22} = (-1)^{2+2} \cdot 2 = 2$

(ii) Let $\Delta = \left|\begin{array}{ll}a & c \\ b & d\end{array}\right|$

$M_{11} = d$;   $A_{11} = (-1)^{1+1} d = d$
$M_{12} = b$;   $A_{12} = (-1)^{1+2} b = -b$
$M_{21} = c$;   $A_{21} = (-1)^{2+1} c = -c$
$M_{22} = a$;   $A_{22} = (-1)^{2+2} a = a$

(i) Let $\Delta = \left|\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right|$ (Identity matrix determinant)

$M_{11} = \left|\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right| = 1$;   $A_{11} = (-1)^2 \cdot 1 = 1$
$M_{12} = \left|\begin{array}{ll}0 & 0 \\ 0 & 1\end{array}\right| = 0$;   $A_{12} = (-1)^3 \cdot 0 = 0$
$M_{13} = \left|\begin{array}{ll}0 & 1 \\ 0 & 0\end{array}\right| = 0$;   $A_{13} = (-1)^4 \cdot 0 = 0$
$M_{21} = \left|\begin{array}{ll}0 & 0 \\ 0 & 1\end{array}\right| = 0$;   $A_{21} = (-1)^3 \cdot 0 = 0$
$M_{22} = \left|\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right| = 1$;   $A_{22} = (-1)^4 \cdot 1 = 1$
$M_{23} = \left|\begin{array}{ll}1 & 0 \\ 0 & 0\end{array}\right| = 0$;   $A_{23} = (-1)^5 \cdot 0 = 0$
$M_{31} = \left|\begin{array}{ll}0 & 0 \\ 1 & 0\end{array}\right| = 0$;   $A_{31} = (-1)^4 \cdot 0 = 0$
$M_{32} = \left|\begin{array}{ll}1 & 0 \\ 0 & 0\end{array}\right| = 0$;   $A_{32} = (-1)^5 \cdot 0 = 0$
$M_{33} = \left|\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right| = 1$;   $A_{33} = (-1)^6 \cdot 1 = 1$

(ii) Let $\Delta = \left|\begin{array}{rrr}1 & 0 & 4 \\ 3 & 5 & -1 \\ 0 & 1 & 2\end{array}\right|$

$M_{11} = \left|\begin{array}{rr}5 & -1 \\ 1 & 2\end{array}\right| = 11$;   $A_{11} = 11$
$M_{12} = \left|\begin{array}{rr}3 & -1 \\ 0 & 2\end{array}\right| = 6$;   $A_{12} = -6$
$M_{13} = \left|\begin{array}{ll}3 & 5 \\ 0 & 1\end{array}\right| = 3$;   $A_{13} = 3$
$M_{21} = \left|\begin{array}{ll}0 & 4 \\ 1 & 2\end{array}\right| = -4$;   $A_{21} = 4$
$M_{22} = \left|\begin{array}{ll}1 & 4 \\ 0 & 2\end{array}\right| = 2$;   $A_{22} = 2$
$M_{23} = \left|\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right| = 1$;   $A_{23} = -1$
$M_{31} = \left|\begin{array}{rr}0 & 4 \\ 5 & -1\end{array}\right| = -20$;   $A_{31} = -20$
$M_{32} = \left|\begin{array}{rr}1 & 4 \\ 3 & -1\end{array}\right| = -13$;   $A_{32} = 13$
$M_{33} = \left|\begin{array}{ll}1 & 0 \\ 3 & 5\end{array}\right| = 5$;   $A_{33} = 5$

The elements of the second row are $a_{21} = 2,\ a_{22} = 0,\ a_{23} = 1$.

Calculate each cofactor by omitting the corresponding row and column:
$A_{21} = (-1)^{2+1}\left|\begin{array}{ll}3 & 8 \\ 2 & 3\end{array}\right| = (-1)^3(9-16) = 7$
$A_{22} = (-1)^{2+2}\left|\begin{array}{ll}5 & 8 \\ 1 & 3\end{array}\right| = (-1)^4(15-8) = 7$
$A_{23} = (-1)^{2+3}\left|\begin{array}{ll}5 & 3 \\ 1 & 2\end{array}\right| = (-1)^5(10-3) = -7$

Expanding along the second row:
$\Delta = a_{21}A_{21} + a_{22}A_{22} + a_{23}A_{23} = 2(7) + 0(7) + 1(-7) = 14 – 7 = 7$

The elements of the third column are $a_{13} = yz,\ a_{23} = zx,\ a_{33} = xy$.

Finding each cofactor by omitting its corresponding row and column:
$A_{13} = (-1)^{1+3}\left|\begin{array}{ll}1 & y \\ 1 & z\end{array}\right| = (z-y)$
$A_{23} = (-1)^{2+3}\left|\begin{array}{ll}1 & x \\ 1 & z\end{array}\right| = -(z-x)$
$A_{33} = (-1)^{3+3}\left|\begin{array}{ll}1 & x \\ 1 & y\end{array}\right| = (y-x)$

Expanding along the third column:
$\Delta = a_{13}A_{13} + a_{23}A_{23} + a_{33}A_{33}$
$= yz(z-y) + zx[-(z-x)] + xy(y-x)$
$= yz^2 – y^2z – z^2x + zx^2 + xy^2 – x^2y$
$= (yz^2 – y^2z) + (xy^2 – xz^2) + (zx^2 – x^2y)$
$= yz(z-y) + x(y^2-z^2) – x^2(y-z)$
$= -yz(y-z) + x(y+z)(y-z) – x^2(y-z)$
$= (y-z)[-yz + xy + xz – x^2]$
$= (y-z)[x(z-x) + y(x-z)]$
$= (y-z)(z-x)(x-y) = (x-y)(y-z)(z-x)$

By the cofactor expansion theorem, the value of a determinant equals the sum of the products of elements of any one row (or column) with their corresponding cofactors.

Option (D) represents the expansion along the first column: $a_{11}A_{11} + a_{21}A_{21} + a_{31}A_{31}$, which correctly equals $\Delta$.

Note: Options (A) and (C) each give zero, since they represent the sum of products of elements of one row (or column) with the cofactors of a different row (or column).

Option (D) is correct.

Let $A = \left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]$.

Finding each cofactor (for a $2\times2$ matrix, diagonal elements get a $+$ sign and non-diagonal elements get a $-$ sign):
$A_{11} = (-1)^2 \cdot 4 = 4$
$A_{12} = (-1)^3 \cdot 3 = -3$
$A_{21} = (-1)^3 \cdot 2 = -2$
$A_{22} = (-1)^4 \cdot 1 = 1$

The adjoint is the transpose of the cofactor matrix:
$\text{adj.}\ A = \left[\begin{array}{rr}4 & -3 \\ -2 & 1\end{array}\right]^{\prime} = \left[\begin{array}{rr}4 & -2 \\ -3 & 1\end{array}\right]$

Let $A = \left[\begin{array}{rrr}1 & -1 & 2 \\ 2 & 3 & 5 \\ -2 & 0 & 1\end{array}\right]$.

Computing all nine cofactors (alternating signs starting with $+$ at position $(1,1)$):
$A_{11} = +\left|\begin{array}{ll}3 & 5 \\ 0 & 1\end{array}\right| = 3$
$A_{12} = -\left|\begin{array}{rr}2 & 5 \\ -2 & 1\end{array}\right| = -(2+10) = -12$
$A_{13} = +\left|\begin{array}{rr}2 & 3 \\ -2 & 0\end{array}\right| = 0+6 = 6$
$A_{21} = -\left|\begin{array}{rr}-1 & 2 \\ 0 & 1\end{array}\right| = -(-1) = 1$
$A_{22} = +\left|\begin{array}{rr}1 & 2 \\ -2 & 1\end{array}\right| = 1+4 = 5$
$A_{23} = -\left|\begin{array}{rr}1 & -1 \\ -2 & 0\end{array}\right| = -(0-2) = 2$
$A_{31} = +\left|\begin{array}{rr}-1 & 2 \\ 3 & 5\end{array}\right| = -5-6 = -11$
$A_{32} = -\left|\begin{array}{rr}1 & 2 \\ 2 & 5\end{array}\right| = -(5-4) = -1$
$A_{33} = +\left|\begin{array}{rr}1 & -1 \\ 2 & 3\end{array}\right| = 3+2 = 5$

Taking the transpose of the cofactor matrix:
$\text{adj.}\ A = \left[\begin{array}{rrr}3 & -12 & 6 \\ 1 & 5 & 2 \\ -11 & -1 & 5\end{array}\right]^{\prime} = \left[\begin{array}{rrr}3 & 1 & -11 \\ -12 & 5 & -1 \\ 6 & 2 & 5\end{array}\right]$

Let $A = \left[\begin{array}{rr}2 & 3 \\ -4 & -6\end{array}\right]$.

Using the shortcut formula for a $2\times2$ matrix, swap the diagonal elements and change signs of off-diagonal elements:
$\text{adj.}\ A = \left[\begin{array}{rr}-6 & -3 \\ 4 & 2\end{array}\right]$

Computing $A(\text{adj.}\ A)$:
$A(\text{adj.}\ A) = \left[\begin{array}{rr}2 & 3 \\ -4 & -6\end{array}\right]\left[\begin{array}{rr}-6 & -3 \\ 4 & 2\end{array}\right] = \left[\begin{array}{ll}-12+12 & -6+6 \\ 24-24 & 12-12\end{array}\right] = \left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]$    …(i)

Computing $(\text{adj.}\ A)A$:
$(\text{adj.}\ A)A = \left[\begin{array}{rr}-6 & -3 \\ 4 & 2\end{array}\right]\left[\begin{array}{rr}2 & 3 \\ -4 & -6\end{array}\right] = \left[\begin{array}{rr}-12+12 & -18+18 \\ 8-8 & 12-12\end{array}\right] = \left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]$    …(ii)

Computing $|A|I$:
$|A| = 2(-6) – 3(-4) = -12 + 12 = 0$
$|A|I = 0 \cdot \left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] = \left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]$    …(iii)

From (i), (ii) and (iii): $A(\text{adj.}\ A) = (\text{adj.}\ A)A = |A|I$ ✓

Let $A = \left[\begin{array}{rrr}1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3\end{array}\right]$. First, compute all cofactors:

$A_{11} = +\left|\begin{array}{rr}0 & -2 \\ 0 & 3\end{array}\right| = 0,\quad A_{12} = -\left|\begin{array}{rr}3 & -2 \\ 1 & 3\end{array}\right| = -11,\quad A_{13} = +\left|\begin{array}{rr}3 & 0 \\ 1 & 0\end{array}\right| = 0$
$A_{21} = -\left|\begin{array}{rr}-1 & 2 \\ 0 & 3\end{array}\right| = 3,\quad A_{22} = +\left|\begin{array}{rr}1 & 2 \\ 1 & 3\end{array}\right| = 1,\quad A_{23} = -\left|\begin{array}{rr}1 & -1 \\ 1 & 0\end{array}\right| = -1$
$A_{31} = +\left|\begin{array}{rr}-1 & 2 \\ 0 & -2\end{array}\right| = 2,\quad A_{32} = -\left|\begin{array}{rr}1 & 2 \\ 3 & -2\end{array}\right| = 8,\quad A_{33} = +\left|\begin{array}{rr}1 & -1 \\ 3 & 0\end{array}\right| = 3$

Taking the transpose of the cofactor matrix:
$\text{adj.}\ A = \left[\begin{array}{rrr}0 & -11 & 0 \\ 3 & 1 & -1 \\ 2 & 8 & 3\end{array}\right]^{\prime} = \left[\begin{array}{rrr}0 & 3 & 2 \\ -11 & 1 & 8 \\ 0 & -1 & 3\end{array}\right]$

Computing $A(\text{adj.}\ A)$:
$A(\text{adj.}\ A) = \left[\begin{array}{rrr}0+11+0 & 3-1-2 & 2-8+6 \\ 0-0-0 & 9+0+2 & 6+0-6 \\ 0+0+0 & 3+0-3 & 2+0+9\end{array}\right] = \left[\begin{array}{rrr}11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11\end{array}\right]$   …(i)

Computing $(\text{adj.}\ A)A$:
$(\text{adj.}\ A)A = \left[\begin{array}{rrr}0+9+2 & 0+0+0 & 0-6+6 \\ -11+3+8 & 11+0+0 & -22-2+24 \\ 0-3+3 & 0-0+0 & 0+2+9\end{array}\right] = \left[\begin{array}{rrr}11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11\end{array}\right]$   …(ii)

Computing $|A|I$:
Expanding $|A|$ along the first row: $|A| = 1(0-0) – (-1)(9+2) + 2(0-0) = 11$
$|A|I = 11I_3 = \left[\begin{array}{rrr}11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11\end{array}\right]$   …(iii)

From (i), (ii) and (iii): $A(\text{adj.}\ A) = (\text{adj.}\ A)A = |A|I$ ✓

Let $A = \left[\begin{array}{rr}2 & -2 \\ 4 & 3\end{array}\right]$.

$|A| = 6 – (-8) = 14 \neq 0$, so $A$ is non-singular and $A^{-1}$ exists.

Using the adjoint formula for a $2\times2$ matrix (swap diagonals, change signs of off-diagonals):
$\text{adj.}\ A = \left[\begin{array}{rr}3 & 2 \\ -4 & 2\end{array}\right]$

$A^{-1} = \dfrac{1}{|A|}\ \text{adj.}\ A = \dfrac{1}{14}\left[\begin{array}{rr}3 & 2 \\ -4 & 2\end{array}\right]$

Let $A = \left[\begin{array}{ll}-1 & 5 \\ -3 & 2\end{array}\right]$.

$|A| = -2 – (-15) = 13 \neq 0$, so $A^{-1}$ exists.

Using the $2\times2$ adjoint formula:
$\text{adj.}\ A = \left[\begin{array}{rr}2 & -5 \\ 3 & -1\end{array}\right]$

$A^{-1} = \dfrac{1}{13}\left[\begin{array}{rr}2 & -5 \\ 3 & -1\end{array}\right]$

Let $A = \left[\begin{array}{lll}1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5\end{array}\right]$.

Expanding along the first row: $|A| = 1(10-0) – 2(0-0) + 3(0-0) = 10 \neq 0$, so $A^{-1}$ exists.

Computing all cofactors:
$A_{11} = 10,\quad A_{12} = 0,\quad A_{13} = 0$
$A_{21} = -10,\quad A_{22} = 5,\quad A_{23} = 0$
$A_{31} = 2,\quad A_{32} = -4,\quad A_{33} = 2$

$\text{adj.}\ A = \left[\begin{array}{rrr}10 & 0 & 0 \\ -10 & 5 & 0 \\ 2 & -4 & 2\end{array}\right]^{\prime} = \left[\begin{array}{rrr}10 & -10 & 2 \\ 0 & 5 & -4 \\ 0 & 0 & 2\end{array}\right]$

$A^{-1} = \dfrac{1}{10}\left[\begin{array}{rrr}10 & -10 & 2 \\ 0 & 5 & -4 \\ 0 & 0 & 2\end{array}\right]$

Let $A = \left[\begin{array}{rrr}1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & -1\end{array}\right]$.

Expanding along the first row: $|A| = 1(-3-0) – 0 + 0 = -3 \neq 0$, so $A^{-1}$ exists.

Computing all cofactors:
$A_{11} = -3,\quad A_{12} = 3,\quad A_{13} = -9$
$A_{21} = 0,\quad A_{22} = -1,\quad A_{23} = -2$
$A_{31} = 0,\quad A_{32} = 0,\quad A_{33} = 3$

$\text{adj.}\ A = \left[\begin{array}{rrr}-3 & 0 & 0 \\ 3 & -1 & 0 \\ -9 & -2 & 3\end{array}\right]$

$A^{-1} = \dfrac{1}{-3}\left[\begin{array}{rrr}-3 & 0 & 0 \\ 3 & -1 & 0 \\ -9 & -2 & 3\end{array}\right]$

Let $A = \left[\begin{array}{rrr}2 & 1 & 3 \\ 4 & -1 & 0 \\ -7 & 2 & 1\end{array}\right]$.

Expanding along the first row: $|A| = 2(-1) – 1(4) + 3(8-7) = -2 – 4 + 3 = -3 \neq 0$, so $A^{-1}$ exists.

Computing all cofactors:
$A_{11} = -1,\quad A_{12} = -4,\quad A_{13} = 1$
$A_{21} = 5,\quad A_{22} = 23,\quad A_{23} = -11$
$A_{31} = 3,\quad A_{32} = 12,\quad A_{33} = -6$

$\text{adj.}\ A = \left[\begin{array}{rrr}-1 & 5 & 3 \\ -4 & 23 & 12 \\ 1 & -11 & -6\end{array}\right]$

$A^{-1} = -\dfrac{1}{3}\left[\begin{array}{rrr}-1 & 5 & 3 \\ -4 & 23 & 12 \\ 1 & -11 & -6\end{array}\right]$

Let $A = \left[\begin{array}{rrr}1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4\end{array}\right]$.

Expanding along the first row: $|A| = 1(8-6) – (-1)(0+9) + 2(0-6) = 2 + 9 – 12 = -1 \neq 0$, so $A^{-1}$ exists.

Computing all cofactors:
$A_{11} = 2,\quad A_{12} = -9,\quad A_{13} = -6$
$A_{21} = 0,\quad A_{22} = -2,\quad A_{23} = -1$
$A_{31} = -1,\quad A_{32} = 3,\quad A_{33} = 2$

$\text{adj.}\ A = \left[\begin{array}{rrr}2 & 0 & -1 \\ -9 & -2 & 3 \\ -6 & -1 & 2\end{array}\right]$

$A^{-1} = \dfrac{1}{-1}\left[\begin{array}{rrr}2 & 0 & -1 \\ -9 & -2 & 3 \\ -6 & -1 & 2\end{array}\right] = \left[\begin{array}{rrr}-2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2\end{array}\right]$

Let $A = \left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & \cos\alpha & \sin\alpha \\ 0 & \sin\alpha & -\cos\alpha\end{array}\right]$.

Expanding along the first row:
$|A| = 1(-\cos^2\alpha – \sin^2\alpha) = -(\cos^2\alpha + \sin^2\alpha) = -1 \neq 0$, so $A^{-1}$ exists.

Computing all cofactors:
$A_{11} = -1,\quad A_{12} = 0,\quad A_{13} = 0$
$A_{21} = 0,\quad A_{22} = -\cos\alpha,\quad A_{23} = -\sin\alpha$
$A_{31} = 0,\quad A_{32} = -\sin\alpha,\quad A_{33} = \cos\alpha$

$\text{adj.}\ A = \left[\begin{array}{ccc}-1 & 0 & 0 \\ 0 & -\cos\alpha & -\sin\alpha \\ 0 & -\sin\alpha & \cos\alpha\end{array}\right]$

$A^{-1} = \dfrac{1}{-1}\left[\begin{array}{ccc}-1 & 0 & 0 \\ 0 & -\cos\alpha & -\sin\alpha \\ 0 & -\sin\alpha & \cos\alpha\end{array}\right] = \left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & \cos\alpha & \sin\alpha \\ 0 & \sin\alpha & -\cos\alpha\end{array}\right]$

Finding $A^{-1}$: $|A| = 15-14 = 1 \neq 0$
$A^{-1} = \dfrac{1}{1}\left[\begin{array}{rr}5 & -7 \\ -2 & 3\end{array}\right] = \left[\begin{array}{rr}5 & -7 \\ -2 & 3\end{array}\right]$

Finding $B^{-1}$: $|B| = 54-56 = -2 \neq 0$
$B^{-1} = \dfrac{1}{-2}\left[\begin{array}{rr}9 & -8 \\ -7 & 6\end{array}\right]$

Computing $AB$:
$AB = \left[\begin{array}{ll}18+49 & 24+63 \\ 12+35 & 16+45\end{array}\right] = \left[\begin{array}{ll}67 & 87 \\ 47 & 61\end{array}\right]$
$|AB| = 67(61) – 87(47) = 4087 – 4089 = -2 \neq 0$

L.H.S. $= (AB)^{-1}$:
$(AB)^{-1} = \dfrac{1}{-2}\left[\begin{array}{rr}61 & -87 \\ -47 & 67\end{array}\right]$   …(i)

R.H.S. $= B^{-1}A^{-1}$:
$B^{-1}A^{-1} = \dfrac{-1}{2}\left[\begin{array}{rr}9 & -8 \\ -7 & 6\end{array}\right]\left[\begin{array}{rr}5 & -7 \\ -2 & 3\end{array}\right]$
$= \dfrac{-1}{2}\left[\begin{array}{rr}45+16 & -63-24 \\ -35-12 & 49+18\end{array}\right] = \dfrac{-1}{2}\left[\begin{array}{rr}61 & -87 \\ -47 & 67\end{array}\right]$   …(ii)

From (i) and (ii), L.H.S. $=$ R.H.S., confirming $(AB)^{-1} = B^{-1}A^{-1}$ ✓

Start by squaring $A$:
$A^2 = \left[\begin{array}{rr}3 & 1 \\ -1 & 2\end{array}\right]\left[\begin{array}{rr}3 & 1 \\ -1 & 2\end{array}\right] = \left[\begin{array}{rr}8 & 5 \\ -5 & 3\end{array}\right]$

Now substitute into the expression:
$A^2 – 5A + 7I = \left[\begin{array}{rr}8 & 5 \\ -5 & 3\end{array}\right] – \left[\begin{array}{rr}15 & 5 \\ -5 & 10\end{array}\right] + \left[\begin{array}{ll}7 & 0 \\ 0 & 7\end{array}\right]$
$= \left[\begin{array}{rr}-7 & 0 \\ 0 & -7\end{array}\right] + \left[\begin{array}{ll}7 & 0 \\ 0 & 7\end{array}\right] = \left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right] = O$    (proven)

Finding $A^{-1}$: Multiply both sides of $A^2 – 5A + 7I = O$ by $A^{-1}$:
$A – 5I + 7A^{-1} = O$
$\Rightarrow 7A^{-1} = 5I – A = \left[\begin{array}{ll}5 & 0 \\ 0 & 5\end{array}\right] – \left[\begin{array}{rr}3 & 1 \\ -1 & 2\end{array}\right] = \left[\begin{array}{rr}2 & -1 \\ 1 & 3\end{array}\right]$
$\therefore A^{-1} = \dfrac{1}{7}\left[\begin{array}{rr}2 & -1 \\ 1 & 3\end{array}\right]$

First compute $A^2$:
$A^2 = \left[\begin{array}{ll}3 & 2 \\ 1 & 1\end{array}\right]\left[\begin{array}{ll}3 & 2 \\ 1 & 1\end{array}\right] = \left[\begin{array}{rr}11 & 8 \\ 4 & 3\end{array}\right]$

Substituting into $A^2 + aA + bI = O$:
$\left[\begin{array}{rr}11 & 8 \\ 4 & 3\end{array}\right] + a\left[\begin{array}{ll}3 & 2 \\ 1 & 1\end{array}\right] + b\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] = \left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]$

Matching corresponding entries gives us the system:
From position $(1,2)$: $8 + 2a = 0 \Rightarrow a = -4$
From position $(2,1)$: $4 + a = 0 \Rightarrow a = -4$ ✓ (consistent)
Substituting $a = -4$ into position $(1,1)$: $11 – 12 + b = 0 \Rightarrow b = 1$
Check position $(2,2)$: $3 – 4 + 1 = 0$ ✓

Therefore $a = -4$ and $b = 1$.

Step 1 — Compute $A^2$:
$A^2 = A \cdot A = \left[\begin{array}{rrr}4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14\end{array}\right]$

Step 2 — Compute $A^3 = A^2 \cdot A$:
$A^3 = \left[\begin{array}{rrr}8 & 7 & 1 \\ -23 & 27 & -69 \\ 32 & -13 & 58\end{array}\right]$

Step 3 — Verify $A^3 – 6A^2 + 5A + 11I = O$:
Substituting all values and simplifying step by step:
$= \left[\begin{array}{rrr}-11 & 0 & 0 \\ 0 & -11 & 0 \\ 0 & 0 & -11\end{array}\right] + \left[\begin{array}{rrr}11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11\end{array}\right] = O$ ✓

Finding $A^{-1}$: Multiply $A^3 – 6A^2 + 5A + 11I = O$ by $A^{-1}$:
$A^2 – 6A + 5I + 11A^{-1} = O$
$\Rightarrow 11A^{-1} = 6A – 5I – A^2$

$11A^{-1} = 6\left[\begin{array}{rrr}1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3\end{array}\right] – 5\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right] – \left[\begin{array}{rrr}4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14\end{array}\right] = \left[\begin{array}{rrr}-3 & 4 & 5 \\ 9 & -1 & -4 \\ 5 & -3 & -1\end{array}\right]$

$\therefore A^{-1} = \dfrac{1}{11}\left[\begin{array}{rrr}-3 & 4 & 5 \\ 9 & -1 & -4 \\ 5 & -3 & -1\end{array}\right]$

Step 1 — Compute $A^2$:
$A^2 = \left[\begin{array}{rrr}6 & -5 & 5 \\ -5 & 6 & -5 \\ 5 & -5 & 6\end{array}\right]$

Step 2 — Compute $A^3 = A^2 \cdot A$:
$A^3 = \left[\begin{array}{rrr}22 & -21 & 21 \\ -21 & 22 & -21 \\ 21 & -21 & 22\end{array}\right]$

Step 3 — Verify $A^3 – 6A^2 + 9A – 4I = O$:
Substituting and simplifying each matrix operation leads to:
$= \left[\begin{array}{rrr}-14 & 9 & -9 \\ 9 & -14 & 9 \\ -9 & 9 & -14\end{array}\right] + \left[\begin{array}{rrr}14 & -9 & 9 \\ -9 & 14 & -9 \\ 9 & -9 & 14\end{array}\right] = O$ ✓

Finding $A^{-1}$: Multiply by $A^{-1}$:
$A^2 – 6A + 9I – 4A^{-1} = O$
$\Rightarrow -4A^{-1} = -A^2 + 6A – 9I$
$\Rightarrow A^{-1} = \dfrac{1}{4}(A^2 – 6A + 9I)$

$= \dfrac{1}{4}\left(\left[\begin{array}{rrr}6 & -5 & 5 \\ -5 & 6 & -5 \\ 5 & -5 & 6\end{array}\right] – \left[\begin{array}{rrr}12 & -6 & 6 \\ -6 & 12 & -6 \\ 6 & -6 & 12\end{array}\right] + \left[\begin{array}{rrr}9 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 9\end{array}\right]\right)$

$= \dfrac{1}{4}\left[\begin{array}{rrr}3 & 1 & -1 \\ 1 & 3 & 1 \\ -1 & 1 & 3\end{array}\right]$

For a non-singular matrix of order $n \times n$, the general result is $|\text{adj.}\ A| = |A|^{n-1}$.

Substituting $n = 3$: $|\text{adj.}\ A| = |A|^{3-1} = |A|^2$

Option (B) is correct.

Start from the defining property of an inverse: $AA^{-1} = I$.

Taking determinants on both sides: $|AA^{-1}| = |I| \Rightarrow |A||A^{-1}| = 1$

Dividing both sides by $|A|$ (which is non-zero since $A$ is invertible):
$|A^{-1}| = \dfrac{1}{|A|}$, i.e., $\det(A^{-1}) = \dfrac{1}{\det A}$

Option (B) is correct.

Writing in matrix form $AX = B$:
$\left[\begin{array}{ll}1 & 2 \\ 2 & 3\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right] = \left[\begin{array}{l}2 \\ 3\end{array}\right]$

$|A| = \left|\begin{array}{ll}1 & 2 \\ 2 & 3\end{array}\right| = 3 – 4 = -1 \neq 0$

Since $|A| \neq 0$, a unique solution exists. The system is consistent.

Writing in matrix form $AX = B$:
$\left[\begin{array}{rr}2 & -1 \\ 1 & 1\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right] = \left[\begin{array}{l}5 \\ 4\end{array}\right]$

$|A| = \left|\begin{array}{rr}2 & -1 \\ 1 & 1\end{array}\right| = 2 – (-1) = 3 \neq 0$

Since $|A| \neq 0$, a unique solution exists. The system is consistent.

Writing in matrix form $AX = B$:
$\left[\begin{array}{ll}1 & 3 \\ 2 & 6\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right] = \left[\begin{array}{l}5 \\ 8\end{array}\right]$

$|A| = \left|\begin{array}{ll}1 & 3 \\ 2 & 6\end{array}\right| = 6 – 6 = 0$

Since $|A| = 0$, we check $(\text{adj.}\ A)B$:
$\text{adj.}\ A = \left[\begin{array}{rr}6 & -3 \\ -2 & 1\end{array}\right]$
$(\text{adj.}\ A)B = \left[\begin{array}{rr}6 & -3 \\ -2 & 1\end{array}\right]\left[\begin{array}{l}5 \\ 8\end{array}\right] = \left[\begin{array}{r}30-24 \\ -10+8\end{array}\right] = \left[\begin{array}{r}6 \\ -2\end{array}\right] \neq O$

Since $(\text{adj.}\ A)B \neq O$, the system is inconsistent (no solution).

Matrix form: $A = \left[\begin{array}{rrr}1 & 1 & 1 \\ 2 & 3 & 2 \\ a & a & 2a\end{array}\right]$

Expanding $|A|$ along the first row:
$|A| = 1(6a-2a) – 1(4a-2a) + 1(2a-3a) = 4a – 2a – a = a$

Case I — $a \neq 0$: $|A| = a \neq 0$, so a unique solution exists. The system is consistent.

Case II — $a = 0$: The third equation becomes $0 = 4$, which is impossible. The system is inconsistent.

Rewriting in standard form and setting up the matrix equation $AX = B$:
$A = \left[\begin{array}{rrr}3 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0\end{array}\right],\quad B = \left[\begin{array}{r}2 \\ -1 \\ 3\end{array}\right]$

Expanding $|A|$ along the first row:
$|A| = 3(0-5) – (-1)(0+3) + (-2)(0-6) = -15 + 3 + 12 = 0$

Since $|A| = 0$, compute $(\text{adj.}\ A)B$. After finding all cofactors:
$\text{adj.}\ A = \left[\begin{array}{rrr}-5 & 10 & 5 \\ -3 & 6 & 3 \\ -6 & 12 & 6\end{array}\right]$
$(\text{adj.}\ A)B = \left[\begin{array}{r}-10-10+15 \\ -6-6+9 \\ -12-12+18\end{array}\right] = \left[\begin{array}{r}-5 \\ -3 \\ -6\end{array}\right] \neq O$

The system is inconsistent.

Setting up the matrix equation $AX = B$:
$A = \left[\begin{array}{rrr}5 & -1 & 4 \\ 2 & 3 & 5 \\ 5 & -2 & 6\end{array}\right]$

Expanding $|A|$ along the first row:
$|A| = 5(18+10) – (-1)(12-25) + 4(-4-15)$
$= 5(28) + (-13) + 4(-19) = 140 – 13 – 76 = 51 \neq 0$

Since $|A| \neq 0$, a unique solution exists. The system is consistent.

Setting up $AX = B$: $A = \left[\begin{array}{ll}5 & 2 \\ 7 & 3\end{array}\right],\quad X = \left[\begin{array}{l}x \\ y\end{array}\right],\quad B = \left[\begin{array}{l}4 \\ 5\end{array}\right]$

$|A| = 15 – 14 = 1 \neq 0$, so a unique solution exists. Using $X = A^{-1}B = \frac{1}{|A|}(\text{adj.}\ A)B$:

$\left[\begin{array}{l}x \\ y\end{array}\right] = \frac{1}{1}\left[\begin{array}{rr}3 & -2 \\ -7 & 5\end{array}\right]\left[\begin{array}{l}4 \\ 5\end{array}\right] = \left[\begin{array}{r}12-10 \\ -28+25\end{array}\right] = \left[\begin{array}{r}2 \\ -3\end{array}\right]$

Therefore $x = 2$ and $y = -3$.

Setting up $AX = B$: $A = \left[\begin{array}{rr}2 & -1 \\ 3 & 4\end{array}\right],\quad B = \left[\begin{array}{r}-2 \\ 3\end{array}\right]$

$|A| = 8 – (-3) = 11 \neq 0$. Using $X = \frac{1}{|A|}(\text{adj.}\ A)B$:

$\left[\begin{array}{l}x \\ y\end{array}\right] = \frac{1}{11}\left[\begin{array}{rr}4 & 1 \\ -3 & 2\end{array}\right]\left[\begin{array}{r}-2 \\ 3\end{array}\right] = \frac{1}{11}\left[\begin{array}{r}-8+3 \\ 6+6\end{array}\right] = \frac{1}{11}\left[\begin{array}{r}-5 \\ 12\end{array}\right]$

Therefore $x = -\dfrac{5}{11}$ and $y = \dfrac{12}{11}$.

Setting up $AX = B$: $A = \left[\begin{array}{rr}4 & -3 \\ 3 & -5\end{array}\right],\quad B = \left[\begin{array}{l}3 \\ 7\end{array}\right]$

$|A| = -20 – (-9) = -11 \neq 0$. Using $X = \frac{1}{|A|}(\text{adj.}\ A)B$:

$\left[\begin{array}{l}x \\ y\end{array}\right] = \frac{1}{-11}\left[\begin{array}{ll}-5 & 3 \\ -3 & 4\end{array}\right]\left[\begin{array}{l}3 \\ 7\end{array}\right] = \frac{1}{-11}\left[\begin{array}{r}-15+21 \\ -9+28\end{array}\right] = \frac{1}{-11}\left[\begin{array}{r}6 \\ 19\end{array}\right]$

Therefore $x = -\dfrac{6}{11}$ and $y = -\dfrac{19}{11}$.

Setting up $AX = B$: $A = \left[\begin{array}{ll}5 & 2 \\ 3 & 2\end{array}\right],\quad B = \left[\begin{array}{l}3 \\ 5\end{array}\right]$

$|A| = 10 – 6 = 4 \neq 0$. Using $X = \frac{1}{|A|}(\text{adj.}\ A)B$:

$\left[\begin{array}{l}x \\ y\end{array}\right] = \frac{1}{4}\left[\begin{array}{rr}2 & -2 \\ -3 & 5\end{array}\right]\left[\begin{array}{l}3 \\ 5\end{array}\right] = \frac{1}{4}\left[\begin{array}{r}6-10 \\ -9+25\end{array}\right] = \frac{1}{4}\left[\begin{array}{r}-4 \\ 16\end{array}\right] = \left[\begin{array}{r}-1 \\ 4\end{array}\right]$

Therefore $x = -1$ and $y = 4$.

Setting up $AX = B$: $A = \left[\begin{array}{rrr}2 & 1 & 1 \\ 1 & -2 & -1 \\ 0 & 3 & -5\end{array}\right],\quad B = \left[\begin{array}{l}1 \\ \frac{3}{2} \\ 9\end{array}\right]$

Expanding $|A|$ along the first row:
$|A| = 2(10+3) – 1(-5-0) + 1(3-0) = 26 + 5 + 3 = 34 \neq 0$

Computing all cofactors and taking the transpose to get $\text{adj.}\ A$:
$\text{adj.}\ A = \left[\begin{array}{rrr}13 & 8 & 1 \\ 5 & -10 & 3 \\ 3 & -6 & -5\end{array}\right]$

$\left[\begin{array}{l}x \\ y \\ z\end{array}\right] = \frac{1}{34}\left[\begin{array}{rrr}13 & 8 & 1 \\ 5 & -10 & 3 \\ 3 & -6 & -5\end{array}\right]\left[\begin{array}{l}1 \\ \frac{3}{2} \\ 9\end{array}\right] = \frac{1}{34}\left[\begin{array}{r}34 \\ 17 \\ -51\end{array}\right] = \left[\begin{array}{r}1 \\ \frac{1}{2} \\ -\frac{3}{2}\end{array}\right]$

Therefore $x = 1,\quad y = \dfrac{1}{2},\quad z = -\dfrac{3}{2}$.

Setting up $AX = B$: $A = \left[\begin{array}{rrr}1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1\end{array}\right],\quad B = \left[\begin{array}{l}4 \\ 0 \\ 2\end{array}\right]$

Expanding $|A|$: $|A| = 1(1+3) – (-1)(2+3) + 1(2-1) = 4 + 5 + 1 = 10 \neq 0$

Computing all cofactors and forming $\text{adj.}\ A$:
$\text{adj.}\ A = \left[\begin{array}{rrr}4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3\end{array}\right]$

$\left[\begin{array}{l}x \\ y \\ z\end{array}\right] = \frac{1}{10}\left[\begin{array}{rrr}4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3\end{array}\right]\left[\begin{array}{l}4 \\ 0 \\ 2\end{array}\right] = \frac{1}{10}\left[\begin{array}{r}20 \\ -10 \\ 10\end{array}\right] = \left[\begin{array}{r}2 \\ -1 \\ 1\end{array}\right]$

Therefore $x = 2,\quad y = -1,\quad z = 1$.

Setting up $AX = B$: $A = \left[\begin{array}{rrr}2 & 3 & 3 \\ 1 & -2 & 1 \\ 3 & -1 & -2\end{array}\right],\quad B = \left[\begin{array}{r}5 \\ -4 \\ 3\end{array}\right]$

$|A| = 2(4+1) – 3(-2-3) + 3(-1+6) = 10 + 15 + 15 = 40 \neq 0$

Computing all cofactors and forming $\text{adj.}\ A$:
$\text{adj.}\ A = \left[\begin{array}{rrr}5 & 3 & 9 \\ 5 & -13 & 1 \\ 5 & 11 & -7\end{array}\right]$

$\left[\begin{array}{l}x \\ y \\ z\end{array}\right] = \frac{1}{40}\left[\begin{array}{rrr}5 & 3 & 9 \\ 5 & -13 & 1 \\ 5 & 11 & -7\end{array}\right]\left[\begin{array}{r}5 \\ -4 \\ 3\end{array}\right] = \frac{1}{40}\left[\begin{array}{r}40 \\ 80 \\ -40\end{array}\right] = \left[\begin{array}{r}1 \\ 2 \\ -1\end{array}\right]$

Therefore $x = 1,\quad y = 2,\quad z = -1$.

Setting up $AX = B$: $A = \left[\begin{array}{rrr}1 & -1 & 2 \\ 3 & 4 & -5 \\ 2 & -1 & 3\end{array}\right],\quad B = \left[\begin{array}{r}7 \\ -5 \\ 12\end{array}\right]$

$|A| = 1(12-5) – (-1)(9+10) + 2(-3-8) = 7 + 19 – 22 = 4 \neq 0$

Computing all cofactors and forming $\text{adj.}\ A$:
$\text{adj.}\ A = \left[\begin{array}{rrr}7 & 1 & -3 \\ -19 & -1 & 11 \\ -11 & -1 & 7\end{array}\right]$

$\left[\begin{array}{l}x \\ y \\ z\end{array}\right] = \frac{1}{4}\left[\begin{array}{rrr}7 & 1 & -3 \\ -19 & -1 & 11 \\ -11 & -1 & 7\end{array}\right]\left[\begin{array}{r}7 \\ -5 \\ 12\end{array}\right] = \frac{1}{4}\left[\begin{array}{r}8 \\ 4 \\ 12\end{array}\right] = \left[\begin{array}{l}2 \\ 1 \\ 3\end{array}\right]$

Therefore $x = 2,\quad y = 1,\quad z = 3$.

Finding $A^{-1}$:
$|A| = 2(-4+4) – (-3)(-6+4) + 5(3-2) = 0 – 6 + 5 = -1 \neq 0$

Computing all cofactors:
$A_{11} = 0,\quad A_{12} = 2,\quad A_{13} = 1$
$A_{21} = -1,\quad A_{22} = -9,\quad A_{23} = -5$
$A_{31} = 2,\quad A_{32} = 23,\quad A_{33} = 13$

$\text{adj.}\ A = \left[\begin{array}{rrr}0 & -1 & 2 \\ 2 & -9 & 23 \\ 1 & -5 & 13\end{array}\right]$

$A^{-1} = \frac{1}{-1}\left[\begin{array}{rrr}0 & -1 & 2 \\ 2 & -9 & 23 \\ 1 & -5 & 13\end{array}\right] = \left[\begin{array}{rrr}0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13\end{array}\right]$

Solving the system using $A^{-1}$:
$\left[\begin{array}{l}x \\ y \\ z\end{array}\right] = A^{-1}B = \left[\begin{array}{rrr}0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13\end{array}\right]\left[\begin{array}{r}11 \\ -5 \\ -3\end{array}\right] = \left[\begin{array}{r}0-5+6 \\ -22-45+69 \\ -11-25+39\end{array}\right] = \left[\begin{array}{l}1 \\ 2 \\ 3\end{array}\right]$

Therefore $x = 1,\quad y = 2,\quad z = 3$.

Let $₹x,\ ₹y,\ ₹z$ per kg be the prices of onion, wheat and rice respectively.

Setting up $AX = B$:
$A = \left[\begin{array}{lll}4 & 3 & 2 \\ 2 & 4 & 6 \\ 6 & 2 & 3\end{array}\right],\quad X = \left[\begin{array}{l}x \\ y \\ z\end{array}\right],\quad B = \left[\begin{array}{l}60 \\ 90 \\ 70\end{array}\right]$

$|A| = 4(12-12) – 3(6-36) + 2(4-24) = 0 + 90 – 40 = 50 \neq 0$

Computing all cofactors:
$A_{11} = 0,\quad A_{12} = 30,\quad A_{13} = -20$
$A_{21} = -5,\quad A_{22} = 0,\quad A_{23} = 10$
$A_{31} = 10,\quad A_{32} = -20,\quad A_{33} = 10$

$\text{adj.}\ A = \left[\begin{array}{rrr}0 & -5 & 10 \\ 30 & 0 & -20 \\ -20 & 10 & 10\end{array}\right]$

$\left[\begin{array}{l}x \\ y \\ z\end{array}\right] = \frac{1}{50}\left[\begin{array}{rrr}0 & -5 & 10 \\ 30 & 0 & -20 \\ -20 & 10 & 10\end{array}\right]\left[\begin{array}{l}60 \\ 90 \\ 70\end{array}\right] = \frac{1}{50}\left[\begin{array}{l}250 \\ 400 \\ 400\end{array}\right] = \left[\begin{array}{l}5 \\ 8 \\ 8\end{array}\right]$

Therefore the cost of onion is ₹5/kg, wheat is ₹8/kg, and rice is ₹8/kg.

Let $\Delta = \left|\begin{array}{ccc}x & \sin\theta & \cos\theta \\ -\sin\theta & -x & 1 \\ \cos\theta & 1 & x\end{array}\right|$

Expanding along the first row:
$\Delta = x\left|\begin{array}{rr}-x & 1 \\ 1 & x\end{array}\right| – \sin\theta\left|\begin{array}{rr}-\sin\theta & 1 \\ \cos\theta & x\end{array}\right| + \cos\theta\left|\begin{array}{rr}-\sin\theta & -x \\ \cos\theta & 1\end{array}\right|$

$= x(-x^2-1) – \sin\theta(-x\sin\theta – \cos\theta) + \cos\theta(-\sin\theta + x\cos\theta)$
$= -x^3 – x + x\sin^2\theta + \sin\theta\cos\theta – \sin\theta\cos\theta + x\cos^2\theta$
$= -x^3 – x + x(\sin^2\theta + \cos^2\theta)$
$= -x^3 – x + x = -x^3$

Since the result $-x^3$ contains only $x$ and no $\theta$, the determinant is independent of $\theta$.

Expanding along the first row:
$= \cos\alpha\cos\beta(\cos\alpha\cos\beta – 0) – \cos\alpha\sin\beta(-\cos\alpha\sin\beta – 0) – \sin\alpha(-\sin\alpha\sin^2\beta – \sin\alpha\cos^2\beta)$
$= \cos^2\alpha\cos^2\beta + \cos^2\alpha\sin^2\beta + \sin^2\alpha(\sin^2\beta + \cos^2\beta)$
$= \cos^2\alpha(\cos^2\beta + \sin^2\beta) + \sin^2\alpha(\sin^2\beta + \cos^2\beta)$
$= \cos^2\alpha + \sin^2\alpha = \mathbf{1}$

We use the Reversal Law: $(AB)^{-1} = B^{-1}A^{-1}$. Since $A^{-1}$ is already given, we only need to find $B^{-1}$.

Finding $B^{-1}$:
$|B| = 1(3-0) – 2(-1-0) + (-2)(2-0) = 3 + 2 – 4 = 1 \neq 0$

Computing all cofactors of $B$:
$B_{11} = 3,\quad B_{12} = 1,\quad B_{13} = 2$
$B_{21} = 2,\quad B_{22} = 1,\quad B_{23} = 2$
$B_{31} = 6,\quad B_{32} = 2,\quad B_{33} = 5$

$B^{-1} = \frac{1}{1}\left[\begin{array}{lll}3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5\end{array}\right]$

Computing $(AB)^{-1} = B^{-1}A^{-1}$:
$(AB)^{-1} = \left[\begin{array}{lll}3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5\end{array}\right]\left[\begin{array}{rrr}3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2\end{array}\right]$
$= \left[\begin{array}{rrr}9-30+30 & -3+12-12 & 3-10+12 \\ 3-15+10 & -1+6-4 & 1-5+4 \\ 6-30+25 & -2+12-10 & 2-10+10\end{array}\right] = \left[\begin{array}{rrr}9 & -3 & 5 \\ -2 & 1 & 0 \\ 1 & 0 & 2\end{array}\right]$

Setting up: $|A| = 1(15-1) – (-2)(-10-1) + 1(-2-3) = 14 – 22 – 5 = -13 \neq 0$

Finding all cofactors of $A$:
$A_{11} = 14,\quad A_{12} = 11,\quad A_{13} = -5$
$A_{21} = 11,\quad A_{22} = 4,\quad A_{23} = -3$
$A_{31} = -5,\quad A_{32} = -3,\quad A_{33} = -1$

$\text{adj.}\ A = B = \left[\begin{array}{rrr}14 & 11 & -5 \\ 11 & 4 & -3 \\ -5 & -3 & -1\end{array}\right]$

$A^{-1} = \dfrac{-1}{13}\left[\begin{array}{rrr}14 & 11 & -5 \\ 11 & 4 & -3 \\ -5 & -3 & -1\end{array}\right]$   …(i)

Verification of Part (i):
$|B| = 14(-13) – 11(-26) – 5(-13) = -182 + 286 + 65 = 169$

After computing adj.$B$ (using cofactors of $B$) and factoring out $-13$:
$\text{adj.}\ B = -13\left[\begin{array}{rrr}1 & -2 & 1 \\ -2 & 3 & 1 \\ 1 & 1 & 5\end{array}\right]$

$(\text{adj.}\ A)^{-1} = B^{-1} = \dfrac{1}{169}(-13)\left[\begin{array}{rrr}1 & -2 & 1 \\ -2 & 3 & 1 \\ 1 & 1 & 5\end{array}\right] = \dfrac{-1}{13}\left[\begin{array}{rrr}1 & -2 & 1 \\ -2 & 3 & 1 \\ 1 & 1 & 5\end{array}\right]$   …(ii)

Computing adj.$(A^{-1})$ similarly and factoring, one gets:
$\text{adj.}(A^{-1}) = -\dfrac{1}{13}\left[\begin{array}{rrr}1 & -2 & 1 \\ -2 & 3 & 1 \\ 1 & 1 & 5\end{array}\right]$   …(iii)

Since (ii) $=$ (iii), we have $(\text{adj.}\ A)^{-1} = \text{adj.}(A^{-1})$ ✓

Verification of Part (ii):
$(A^{-1})^{-1} = \dfrac{1}{|A^{-1}|}\text{adj.}(A^{-1}) = \dfrac{1}{(-\frac{1}{13})}\left(-\dfrac{1}{13}\right)\left[\begin{array}{rrr}1 & -2 & 1 \\ -2 & 3 & 1 \\ 1 & 1 & 5\end{array}\right] = \left[\begin{array}{rrr}1 & -2 & 1 \\ -2 & 3 & 1 \\ 1 & 1 & 5\end{array}\right] = A$ ✓

Notice that the sum of entries in each column equals $2(x+y)$. Apply $R_1 \rightarrow R_1 + R_2 + R_3$:

$\Delta = \left|\begin{array}{ccc}2(x+y) & 2(x+y) & 2(x+y) \\ y & x+y & x \\ x+y & x & y\end{array}\right|$

Take $2(x+y)$ common from $R_1$:
$= 2(x+y)\left|\begin{array}{ccc}1 & 1 & 1 \\ y & x+y & x \\ x+y & x & y\end{array}\right|$

Now apply $C_2 \rightarrow C_2 – C_1$ and $C_3 \rightarrow C_3 – C_1$ to create zeros in the first row:
$= 2(x+y)\left|\begin{array}{ccc}1 & 0 & 0 \\ y & x & x-y \\ x+y & -y & -x\end{array}\right|$

Expanding along $R_1$:
$= 2(x+y) \cdot 1 \left|\begin{array}{cc}x & x-y \\ -y & -x\end{array}\right|$
$= 2(x+y)(-x^2 + y(x-y))$
$= 2(x+y)(-x^2 + xy – y^2)$
$= -2(x+y)(x^2 – xy + y^2)$
$= -2(x^3 + y^3)$

Since the first column consists entirely of 1s, apply $R_2 \rightarrow R_2 – R_1$ and $R_3 \rightarrow R_3 – R_1$ to create zeros:

$\Delta = \left|\begin{array}{ccc}1 & x & y \\ 0 & y & 0 \\ 0 & 0 & x\end{array}\right|$

Expanding along the first column:
$\Delta = 1 \cdot \left|\begin{array}{cc}y & 0 \\ 0 & x\end{array}\right| = xy$

Introduce substitutions $u = \frac{1}{x},\ v = \frac{1}{y},\ w = \frac{1}{z}$ to linearise the system:
$2u + 3v + 10w = 4,\quad 4u – 6v + 5w = 1,\quad 6u + 9v – 20w = 2$

Setting up $AX = B$: $A = \left[\begin{array}{rrr}2 & 3 & 10 \\ 4 & -6 & 5 \\ 6 & 9 & -20\end{array}\right]$

$|A| = 2(120-45) – 3(-80-30) + 10(36+36) = 150 + 330 + 720 = 1200 \neq 0$

Computing all cofactors and forming $\text{adj.}\ A$:
$\text{adj.}\ A = \left[\begin{array}{rrr}75 & 150 & 75 \\ 110 & -100 & 30 \\ 72 & 0 & -24\end{array}\right]$

$A^{-1} = \dfrac{1}{1200}\left[\begin{array}{rrr}75 & 150 & 75 \\ 110 & -100 & 30 \\ 72 & 0 & -24\end{array}\right]$

$\left[\begin{array}{l}u \\ v \\ w\end{array}\right] = \frac{1}{1200}\left[\begin{array}{rrr}75 & 150 & 75 \\ 110 & -100 & 30 \\ 72 & 0 & -24\end{array}\right]\left[\begin{array}{l}4 \\ 1 \\ 2\end{array}\right] = \frac{1}{1200}\left[\begin{array}{l}600 \\ 400 \\ 240\end{array}\right] = \left[\begin{array}{l}\frac{1}{2} \\ \frac{1}{3} \\ \frac{1}{5}\end{array}\right]$

Reverting to the original variables: $x = \frac{1}{u} = 2,\quad y = \frac{1}{v} = 3,\quad z = \frac{1}{w} = 5$.

$|A| = xyz \neq 0$ (since $x,y,z$ are all non-zero).

Computing the cofactors of the diagonal matrix $A$:
$A_{11} = yz,\quad A_{22} = xz,\quad A_{33} = xy$; all off-diagonal cofactors $= 0$

$\text{adj.}\ A = \left[\begin{array}{rrr}yz & 0 & 0 \\ 0 & xz & 0 \\ 0 & 0 & xy\end{array}\right]$

$A^{-1} = \dfrac{1}{xyz}\left[\begin{array}{rrr}yz & 0 & 0 \\ 0 & xz & 0 \\ 0 & 0 & xy\end{array}\right] = \left[\begin{array}{ccc}\frac{1}{x} & 0 & 0 \\ 0 & \frac{1}{y} & 0 \\ 0 & 0 & \frac{1}{z}\end{array}\right] = \left[\begin{array}{ccc}x^{-1} & 0 & 0 \\ 0 & y^{-1} & 0 \\ 0 & 0 & z^{-1}\end{array}\right]$

Option (A) is correct.

Expanding $|A|$ along the first row:
$|A| = 1(1+\sin^2\theta) – \sin\theta(-\sin\theta + \sin\theta) + 1(\sin^2\theta + 1)$
$= 1 + \sin^2\theta + 0 + 1 + \sin^2\theta = 2 + 2\sin^2\theta$   …(i)

Now analyse the range. Since $0 \leq \sin^2\theta \leq 1$ for all $\theta$:
$0 \leq 2\sin^2\theta \leq 2$
$\Rightarrow 2 \leq 2 + 2\sin^2\theta \leq 4$
$\Rightarrow 2 \leq \det(A) \leq 4$

The value of $\det(A)$ lies in the closed interval $[2,4]$.

Option (D) is correct.