If A = [a 1 0; 1 b d; 1 b c], B = [a 1 1; 0 d c; f g h], U = [f; g; h], V = [a²; 0; 0], X = [x; y; z

Q: Wait, why can't we just use the inverse matrix $A^{-1}$ to solve $AX = U$?

That is a classic trap! We can only use the inverse if the determinant is non-zero. Here, we are given that the system has infinitely many solutions, which forces $\det(A) = 0$. A matrix with a zero determinant is singular and does not have an inverse. We must rely on consistency analysis instead.

Q: How do I know which column to replace in Cramer's Rule when calculating $\Delta_x, \Delta_y,$ or $\Delta_z$?

Great question! Cramer's Rule is very systematic. To find $\Delta_x$, you replace the first column (the coefficients of $x$) with the constant vector $V$. To find $\Delta_y$, you replace the second column (the coefficients of $y$) with $V$, and so on. If any one of these determinants is non-zero while $\det(B) = 0$, the system is inconsistent and has no solution.

Q: Why does the condition $afd \neq 0$ matter so much in the second part?

It is the key that unlocks the proof! Without knowing $a, f,$ and $d$ are non-zero, we couldn't guarantee that our numerator determinants ($\Delta_x, \Delta_y$) are non-zero. By knowing they are non-zero, we can confidently say that the expressions we derived (like $a^2cf$) do not vanish, which is exactly what we need to prove the system is inconsistent.

Question

If $A = \begin{bmatrix} a & 1 & 0 \ 1 & b & d \ 1 & b & c \end{bmatrix}, B = \begin{bmatrix} a & 1 & 1 \ 0 & d & c \ f & g & h \end{bmatrix}, U = \begin{bmatrix} f \ g \ h \end{bmatrix}, V = \begin{bmatrix} a^2 \ 0 \ 0 \end{bmatrix}, X = \begin{bmatrix} x \ y \ z \end{bmatrix}$ and $AX = U$ has infinitely many solutions, prove that $BX = V$ has no unique solution. Also show that if $afd 
eq 0$, then $BX = V$ has no solution.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineers! Today, we are going to embark on a journey through the elegant world of linear algebra. We have a problem that might look like a daunting wall of variables, but it is actually a beautifully choreographed dance.

We are given two matrices, $A$ and $B$, and we need to prove something profound about their behavior. Let's break this down together.

### The Anatomy of Infinite Solutions

We start with the system $AX = U$. You are told it has infinitely many solutions. In the language of linear algebra, this is a massive clue.

It tells us two things immediately. First, the determinant of the coefficient matrix $A$ must be zero. If it were non-zero, the system would have a unique solution.

Second, the system must be consistent. This means the equations are not fighting each other; they are essentially saying the same thing in different ways. When we calculate $\det(A)$, we expand along the first row and find:

$$\det(A) = (ab - 1)(c - d) = 0$$

For this to be zero, either $ab = 1$ or $c = d$. This is our foundation.

### The Bridge Between Systems

Now, we need to connect this to matrix $B$. We write out the equations for $AX = U$ and notice something fascinating. Equations two and three are almost identical.

By subtracting them, we find a consistency condition:

$$(c - d)z = h - g$$

If $c = d$, then $g$ must equal $h$ for the system to exist. If $ab = 1$, we find another relationship:

$$a(cg - dh) = f(c - d)$$

These are not just random equations; they are the DNA of our system. They hold the secret to why $BX = V$ behaves the way it does.

### The Verdict on $\det(B)$

Next, we look at matrix $B$. We calculate its determinant, $\det(B)$, and expand along the first column. We get an expression involving $a, d, h, c, g,$ and $f$.

Now, watch the magic happen. When we substitute our consistency conditions from the previous phase into this expression, the terms cancel out perfectly.

Whether $c = d$ or $ab = 1$, the result is always zero. This proves that $\det(B) = 0$. Since the determinant is zero, the system $BX = V$ can never have a unique solution.

### The Final Blow

Finally, we tackle the condition $afd 
eq 0$. We need to prove that $BX = V$ has no solution. We use Cramer's Rule.

We know $\det(B) = 0$, so if we can find just one numerator determinant—let's pick $\Delta_y$—that is non-zero, we have proven the system is inconsistent. We replace the second column of $B$ with $V$ and calculate $\Delta_y$.

It simplifies to:

$$\Delta_y = a^2cf$$

Since $a, f,$ and $d$ are non-zero, we just need to ensure $c$ is not zero. Through careful case analysis, we find that in every scenario, $\Delta_y$ (or $\Delta_x$) is non-zero.

The system is officially inconsistent. You have just navigated a complex proof with logic and precision. Keep this mindset, and no matrix will ever be too intimidating for you!

Visualized Solution

System of Linear Equations

Condition for $A X = U$

Calculating $det (A)$

Simplifying $det (A) = 0$

Consistency of $A X = U$

Deriving Consistency Conditions

Calculating $det (B)$

Proving $det (B) = 0$

Analyzing $B X = V$ for $a f d \neq = 0$

Calculating $Δ_{y}$

Checking if $Δ_{y} \neq = 0$

Case 1: $c = d$

Case 2: $ab = 1$ and $c = 0$

Calculating $Δ_{x}$ for $c = 0$

Final Proof of No Solution

The Sigma Insight: Solution of System of Linear Equations (Matrix Method and Cramer's Rule)

Analyzing the Setup

The Anatomy of Infinite Solutions

The Bridge Between Systems

The Verdict on $det (B)$

The Final Blow

Similar Questions

Let $A$ be a $3 \times 3$ real matrix such that $A 101 = 2101, A - 1 01 = 4 - 1 01, A 010 = 2010$ . Then, the system $(A - 3 I) x y z = 123$ has

Let $A$ and $B$ be two $3 \times 3$ non-zero real matrices such that $A B$ is a zero matrix. Then

Let $A$ and $B$ be $3 \times 3$ real matrices such that $A$ is symmetric matrix and $B$ is skew-symmetric matrix. Then the system of linear equations $(A^{2} B^{2} - B^{2} A^{2}) X = O$ , where $X$ is a $3 \times 1$ column matrix of unknown variables and $O$ is a $3 \times 1$ null matrix, has :

For a real number $a$ , if the system $1 a a^{2} a 1 a a^{2} a 1 x y z = 1 - 1 1$ of linear equations, has infinitely many solutions, then $1 + a + a^{2} =$

For $α, β \in R$ , suppose the system of linear equations $x - y + z = 5$ , $2 x + 2 y + α z = 8$ , $3 x - y + 4 z = β$ has infinitely many solutions. Then $α$ and $β$ are the roots of

For the system of linear equations $2 x + 4 y + 2 a z = b$ , $x + 2 y + 3 z = 4$ , $2 x + 5 y + 2 z = 8$ which of the following is NOT correct?

The system of linear equations $x + y + z = 6$ , $2 x + 5 y + a z = 36$ , $x + 2 y + 3 z = b$ has

If the system of linear equations $2 x + 2 y + 3 z = a$ , $3 x - y + 5 z = b$ , $x - 3 y + 2 z = c$ where $a, b, c$ are non-zero real numbers, has more than one solution, then :

Visualized Solution

System of Linear Equations

Condition for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AX=U

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } det(A)

Simplifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } det(A)=0

Consistency of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AX=U

Deriving Consistency Conditions

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } det(B)

Proving .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } det(B)=0

Analyzing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } BX=V for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } afd=0

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Δy​

Checking if .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Δy​=0

Case 1: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c=d

Case 2: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ab=1 and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c=0

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Δx​ for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c=0