Q.$(AB)^{-1} = A^{-1} \cdot B^{-1}$, where $A$ and $B$ are invertible matrices satisfying commutative property with respect to multiplication.

We are given matrices $A$ and $B^{-1}$, and asked for $(AB)^{-1}$. The fundamental property of inverses is that $(AB)^{-1} = B^{-1}A^{-1}$, provided both $A$ and $B$ are invertible. So we need $A^{-1}$ first, then multiply $B^{-1}$ by $A^{-1}$.

Why this approach works:

Instead of finding $B$ (which would require inverting $B^{-1}$) and then multiplying $A$ and $B$ and inverting the product, we use the reversal rule. This saves work because we already have $B^{-1}$ and only need to invert $A$, a $3 \times 3$ matrix.

1. Find $A^{-1}$ using the adjugate method. For $A = \begin{bmatrix} 5 & 0 & 4 \\ 2 & 3 & 2 \\ 1 & 2 & 1 \end{bmatrix}$, compute the determinant:

Since $\det(A) = -1 \neq 0$, $A$ is invertible.

2. Compute the cofactor matrix.

   For each entry $a_{ij}$, the cofactor $C_{ij} = (-1)^{i+j} M_{ij}$, where $M_{ij}$ is the minor (determinant of the submatrix after removing row $i$, column $j$).

   • $C_{11} = + \begin{vmatrix} 3 & 2 \\ 2 & 1 \end{vmatrix} = 3\cdot1 - 2\cdot2 = 3 - 4 = -1$
   • $C_{12} = - \begin{vmatrix} 2 & 2 \\ 1 & 1 \end{vmatrix} = -(2\cdot1 - 2\cdot1) = -(2-2) = 0$
   • $C_{13} = + \begin{vmatrix} 2 & 3 \\ 1 & 2 \end{vmatrix} = 2\cdot2 - 3\cdot1 = 4 - 3 = 1$
   • $C_{21} = - \begin{vmatrix} 0 & 4 \\ 2 & 1 \end{vmatrix} = -(0\cdot1 - 4\cdot2) = -(0-8) = 8$
   • $C_{22} = + \begin{vmatrix} 5 & 4 \\ 1 & 1 \end{vmatrix} = 5\cdot1 - 4\cdot1 = 5 - 4 = 1$
   • $C_{23} = - \begin{vmatrix} 5 & 0 \\ 1 & 2 \end{vmatrix} = -(5\cdot2 - 0\cdot1) = -(10-0) = -10$
   • $C_{31} = + \begin{vmatrix} 0 & 4 \\ 3 & 2 \end{vmatrix} = 0\cdot2 - 4\cdot3 = 0 - 12 = -12$
   • $C_{32} = - \begin{vmatrix} 5 & 4 \\ 2 & 2 \end{vmatrix} = -(5\cdot2 - 4\cdot2) = -(10-8) = -2$
   • $C_{33} = + \begin{vmatrix} 5 & 0 \\ 2 & 3 \end{vmatrix} = 5\cdot3 - 0\cdot2 = 15 - 0 = 15$

   So the cofactor matrix is:

3. Transpose to get the adjugate, then divide by determinant. The adjugate is the transpose of the cofactor matrix:

Since $\det(A) = -1$, we have $A^{-1} = \frac{1}{\det(A)} \text{adj}(A) = -1 \cdot \text{adj}(A)$:

4. Multiply $B^{-1}$ by $A^{-1}$ to get $(AB)^{-1}$.

   Given $B^{-1} = \begin{bmatrix} 1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \end{bmatrix}$, compute $B^{-1} A^{-1}$:

   Let $C = B^{-1} A^{-1}$. Then $C_{ij}$ is the dot product of row $i$ of $B^{-1}$ with column $j$ of $A^{-1}$.

   • Row 1 of $B^{-1}$: $[1, 3, 3]$ …

The concept here is a fundamental property of matrix inverses: if you multiply a matrix by a nonzero scalar $k$, its inverse gets multiplied by $1/k$. Why? Because the inverse must undo the original matrix. If $A$ sends a vector $\mathbf{v}$ to $A\mathbf{v}$, then $kA$ sends it to $k(A\mathbf{v})$. To get back to $\mathbf{v}$, you need to first divide by $k$ (i.e., multiply by $1/k$) and then apply $A^{-1}$. So $(kA)^{-1} = \frac{1}{k} A^{-1}$. This is the intuition that drives the solution.

Now, let’s work through it step by step.

1. Start with the given relationship.

   We know $A^{-1} = \frac{1}{8} B$. This tells us that the inverse of $A$ is a scalar multiple of $B$.

2. We need the inverse of $8A$.

   Using the property $(kA)^{-1} = \frac{1}{k} A^{-1}$ for any nonzero scalar $k$, set $k = 8$. Then:

3. Substitute the expression for $A^{-1}$. From step 1, $A^{-1} = \frac{1}{8} B$. So:

4. Match with the options. The result $\frac{1}{64} B$ corresponds exactly to option (C). …

For invertible matrices,

Substituting the data: …

Simplify the matrix. Using $(XY)^{-1}=Y^{-1}X^{-1}$,

Find $A^{-1}$. With $A=\begin{pmatrix}1&0&1\\0&1&0\\1&0&2\end{pmatrix}$, expanding gives $\det A=1(2-0)-0+1(0-1)=1$. Its inverse is

Compute $BA^{-1}$ with $B=\begin{pmatrix}2&1&0\\1&1&2\\0&2&1\end{pmatrix}$: …

1. The rule being tested

For real numbers, if $a < b$ and $c < 0$, then

The sense of the inequality reverses. The reason: $a<b \iff b-a>0$. Multiplying a positive number $b-a$ by a negative number gives a negative number, so $c(b-a) < 0$, i.e. $cb < ca$.

2. Apply it

Here the multiplier is $\dfrac{1}{x}$, and since $x<0$ we have $\dfrac{1}{x}<0$. Given $a<b$:

3. Why the inequality is strict …

Step 1 — Check (B): $|Z_1Z_2|=|Z_1||Z_2|$.

Write $Z_1=r_1e^{i\theta_1},\ Z_2=r_2e^{i\theta_2}$. Then $Z_1Z_2=r_1r_2e^{i(\theta_1+\theta_2)}$, whose modulus is $r_1r_2=|Z_1||Z_2|$. TRUE — multiplication multiplies moduli and adds arguments.

Step 2 — Check (A) and (C): the conjugation properties.

With $Z=x+iy$, conjugation is reflection in the real axis, and it is a ring homomorphism:

Proof of the first: if $Z_1=a+ib,\ Z_2=c+id$, then $\overline{Z_1+Z_2}=\overline{(a+c)+i(b+d)}=(a+c)-i(b+d)=(a-ib)+(c-id)$ ✓. These are the identities options (A) and (C) are quoting — both TRUE.

Step 3 — Check (D): the triangle inequality.

The genuine theorem is

$\boxed{|Z_1+Z_2|\ \le\ |Z_1|+|Z_2|}$ …