Q.If $A = \begin{pmatrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{pmatrix}$, find $A^{-1}$. Use it to solve the system of equations $2x - 3y + 5z = 11$, $3x + 2y - 4z = -5$, $x + y - 2z = -3$. OR Using elementary row transformations, find the inverse of the matrix $A = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 5 & 7 \\ -2 & -4 & -5 \end{pmatrix}$.

Concept & Intuition

We are asked for $\left|\frac{A^{-1}}{-2}\right|$, which means the determinant of the matrix $A^{-1}$ multiplied by the scalar $-\frac12$. Two key facts do all the work:

1. For any $n \times n$ matrix $M$ and scalar $c$, $|cM| = c^n |M|$.
2. $|A^{-1}| = 1/|A|$ (since $A A^{-1}=I$ and $|A||A^{-1}|=1$).

So we just combine these: factor out the scalar, then replace $|A^{-1}|$ with $1/|A|$, and plug in $|A|=-4$.

Step-by-step

1. Interpret the expression

   $\frac{A^{-1}}{-2}$ means $-\frac12 \cdot A^{-1}$. So we want $|\,(-\frac12) A^{-1}\,|$.

2. Factor out the scalar

   For a $3\times 3$ matrix, $|cM| = c^3 |M|$. Here $c = -\frac12$, so

3. Use the inverse determinant property Since $|A^{-1}| = \frac{1}{|A|}$ and $|A| = -4$, we have $$|A^{-1}| = \frac{1}{-4} = -\frac14. …$$

We have three commodities: onion (let’s call the price per kg $x$), wheat ($y$), and rice ($z$). The given costs translate to:

1. $4x + 3y + 2z = 500$
2. $1x + 2y + 3z = 300$
3. $6x + 2y + 3z = 575$

The matrix form $AX = B$ means $A$ is the $3 \times 3$ coefficient matrix, $X = [x, y, z]^T$, and $B = [500, 300, 575]^T$.

We are asked for $\left|5A^{-1}\right|$. The key idea: for any square matrix $A$, $\left|kA^{-1}\right| = k^n \cdot |A^{-1}|$, where $n$ is the order (here $n=3$). And $|A^{-1}| = 1/|A|$. So the problem reduces to finding $|A|$.

Step-by-step solution:

1. Write the coefficient matrix $A$ From the equations:

2. Compute the determinant $|A|$ Expand along the first row:

• First minor: $(2)(3) - (3)(2) = 6 - 6 = 0$
• Second minor: $(1)(3) - (3)(6) = 3 - 18 = -15$
• Third minor: $(1)(2) - (2)(6) = 2 - 12 = -10$

So:

$|A| = 4(0) - 3(-15) + 2(-10) = 0 + 45 - 20 = 25$ …

For $A^{-1}$ to exist, $A$ must be non-singular: $\det A \neq 0$.

Expand along the first row:

$\det A = 4\,(2\cdot 3 - 5\cdot 1)\;-\;\lambda\,(0\cdot 3 - 5\cdot 1)\;+\;(-3)\,(0\cdot 1 - 2\cdot 1)$ …

We are given

Notice that $A^T = -A$; that is, $A$ is skew-symmetric. For any skew-symmetric matrix of odd order, the determinant is always zero. Here the order is 3 (odd), so $\det A = 0$. A matrix with zero determinant is singular and has no inverse.

Let’s verify quickly:

1. Compute the determinant directly:

2. Since $\det A = 0$, $A$ is singular, so $A^{-1}$ does not exist.

3. Among the options:

   • (A) and (C) involve $\operatorname{adj} A$ scaled by a nonzero constant — but the inverse doesn’t exist, so these are meaningless. …

1. The core relationship. For any square matrix $A$, the product $A \cdot \text{adj}(A) = |A| I$. Taking determinants of both sides gives $|A| \cdot |\text{adj}(A)| = |A|^n$, where $n$ is the order of the matrix. For a $3 \times 3$ matrix ($n=3$), this simplifies to $|\text{adj}(A)| = |A|^{2}$.

2. Apply it to the given data. We are told $P = \text{adj}(A)$ and $|A| = 4$. Therefore, the determinant of $P$ must be $|P| = |\text{adj}(A)| = |A|^2 = 4^2 = 16$.

3. Compute the determinant of $P$. We have $P = \begin{bmatrix} 1 & \alpha & 3 \\ 1 & 3 & 3 \\ 2 & 4 & 4 \end{bmatrix}$. Let's expand along the first row:

$$|P| = 1 \cdot (12 - 12) - \alpha \cdot (4 - 6) + 3 \cdot (4 - 6)$$ …

We are given

and asked for the inverse of $(A I)^t$.

Concept and intuition:

First, note that $A I = A$ (multiplying by the identity does nothing). So $(A I)^t = A^t$. The problem is really asking for $(A^t)^{-1}$. A key property: the inverse of a transpose is the transpose of the inverse, i.e. $(A^t)^{-1} = (A^{-1})^t$. So we can find $A^{-1}$ and then transpose it. This is often easier than inverting $A^t$ directly.

Let’s proceed step by step.

1. Find the determinant of $A$ to ensure it’s invertible.

Compute each:

• $\begin{vmatrix}2&3\\1&1\end{vmatrix} = 2\cdot1 - 3\cdot1 = -1$
• $\begin{vmatrix}1&3\\3&1\end{vmatrix} = 1\cdot1 - 3\cdot3 = 1 - 9 = -8$
• $\begin{vmatrix}1&2\\3&1\end{vmatrix} = 1\cdot1 - 2\cdot3 = 1 - 6 = -5$

So

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

2. Find the matrix of cofactors for $A$.

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

   • Row 1:

     $C_{11} = +\begin{vmatrix}2&3\\1&1\end{vmatrix} = -1$

     $C_{12} = -\begin{vmatrix}1&3\\3&1\end{vmatrix} = -(-8) = 8$

     $C_{13} = +\begin{vmatrix}1&2\\3&1\end{vmatrix} = -5$

   • Row 2:

     $C_{21} = -\begin{vmatrix}1&2\\1&1\end{vmatrix} = -(1\cdot1 - 2\cdot1) = -(1-2) = 1$

     $C_{22} = +\begin{vmatrix}-1&2\\3&1\end{vmatrix} = (-1\cdot1 - 2\cdot3) = -1 - 6 = -7$

     $C_{23} = -\begin{vmatrix}-1&1\\3&1\end{vmatrix} = -((-1)\cdot1 - 1\cdot3) = -(-1-3) = 4$

   • Row 3:

     $C_{31} = +\begin{vmatrix}1&2\\2&3\end{vmatrix} = 1\cdot3 - 2\cdot2 = 3 - 4 = -1$

     $C_{32} = -\begin{vmatrix}-1&2\\1&3\end{vmatrix} = -((-1)\cdot3 - 2\cdot1) = -(-3-2) = 5$

     $C_{33} = +\begin{vmatrix}-1&1\\1&2\end{vmatrix} = (-1)\cdot2 - 1\cdot1 = -2 - 1 = -3$

   So the cofactor matrix is

   $$$$

Concept: solving a 3x3 linear system (here fastest by elimination, then verified by substitution).

x - y + z = 4 ... (1)

x - 2y + 2z = 9 ... (2)

2x + y + 3z = 1 ... (3)

(2) - (1): -y + z = 5 => z = y + 5.

(3) - 2*(1): (2x + y + 3z) - (2x - 2y + 2z) = 1 - 8 => 3y + z = -7.

Substitute z = y + 5: 3y + y + 5 = -7 => 4y = -12 => y = -3, hence z = 2. …

$A=\begin{bmatrix}2&2\\3&4\end{bmatrix}$, $\det A=2\cdot4-2\cdot3=8-6=2$.

For a $2\times2$ matrix, $\text{adj }A=\begin{bmatrix}4&-2\\-3&2\end{bmatrix}$.

$A^{-1}=\dfrac12\begin{bmatrix}4&-2\\-3&2\end{bmatrix}=\begin{bmatrix}2&-1\\-3/2&1\end{bmatrix}.$ …

The Concept: Invertibility and the Determinant

For any square matrix, invertibility is equivalent to having a non-zero determinant. For a $2 \times 2$ matrix $M = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$, the determinant is $\det(M) = ad - bc$. The matrix is invertible if and only if $ad - bc \neq 0$.

The problem adds two constraints: $M$ is symmetric ($b = c$) and all entries are integers. So $M = \begin{pmatrix} a & b \\ b & d \end{pmatrix}$ with $a, b, d \in \mathbb{Z}$, and $\det(M) = ad - b^2$.

We need to check which condition guarantees that $ad - b^2 \neq 0$.

Step-by-Step Analysis

1. Understanding the conditions in (A) and (B)

Let $M = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$. Since $M$ is symmetric, $b = c$.

• Condition (A): "the first column of $M$ is the transpose of the second row of $M$"

  • First column: $\begin{pmatrix} a \\ b \end{pmatrix}$
  • Second row: $\begin{pmatrix} b & d \end{pmatrix}$, its transpose is $\begin{pmatrix} b \\ d \end{pmatrix}$
  • Equality gives: $\begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix} b \\ d \end{pmatrix}$, so $a = b$ and $b = d$, hence $a = b = d$.
  • Then $M = \begin{pmatrix} a & a \\ a & a \end{pmatrix}$, determinant $= a \cdot a - a \cdot a = 0$. So $M$ is not invertible.

• Condition (B): "the second row of $M$ is the transpose of the first column of $M$"

  • Second row: $\begin{pmatrix} b & d \end{pmatrix}$
  • Transpose of first column: $\begin{pmatrix} a \\ b \end{pmatrix}$
  • Equality gives: $\begin{pmatrix} b & d \end{pmatrix} = \begin{pmatrix} a & b \end{pmatrix}$, so $b = a$ and $d = b$, hence $a = b = d$ again.
  • Same matrix, determinant $= 0$. Not invertible.

2. Checking condition (C) …

Step 1 — The property that makes this a one-liner.

For square matrices of the same order, the determinant is multiplicative:

Also, transposing a matrix never changes its determinant: $|B'|=|B|$. Therefore

Actually multiplying the three $2\times2$ matrices out would work too, but it is far more error-prone.

Step 2 — Compute $|A|$.

Step 3 — Compute $|B|$. …

The core idea here is that a matrix equation of the form $A \mathbf{v} = \mathbf{b}$ is solved exactly like the scalar equation $a x = b$ — you multiply both sides by the inverse of $A$ (provided it exists). The only difference is that "division" becomes multiplication by $A^{-1}$, and order matters because matrix multiplication is not commutative.

Let the coefficient matrix be $A = \begin{bmatrix} 1 & 1 \\ -1 & 1 \end{bmatrix}$, the unknown vector be $\mathbf{v} = \begin{bmatrix} x \\ y \end{bmatrix}$, and the constant vector be $\mathbf{b} = \begin{bmatrix} 2 \\ 4 \end{bmatrix}$. So we have $A \mathbf{v} = \mathbf{b}$.

1. Check if $A$ is invertible.

   Compute the determinant:

   $\det(A) = (1)(1) - (1)(-1) = 1 + 1 = 2$.

   Since $\det(A) \neq 0$, the inverse exists.

2. Find $A^{-1}$.

   For a $2 \times 2$ matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$, the inverse is $\frac{1}{\det(A)} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$.

   So here:

   $A^{-1} = \frac{1}{2} \begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix}$.

3. Multiply both sides by $A^{-1}$ on the left.

   Since $A^{-1} A = I$, we get:

   $\mathbf{v} = A^{-1} \mathbf{b}$.

   Compute:

   $\begin{bmatrix} x \\ y \end{bmatrix} = \frac{1}{2} \begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} 2 \\ 4 \end{bmatrix}$.

4. Perform the multiplication.

   First row: $1 \cdot 2 + (-1) \cdot 4 = 2 - 4 = -2$.

   Second row: $1 \cdot 2 + 1 \cdot 4 = 2 + 4 = 6$. …