If $A = \begin{bmatrix} 3 & 5 \end{bmatrix}$, $B = \begin{bmatrix} 7 & 3 \end{bmatrix}$, then find a non-zero matrix $C$ such that $AC = BC$.

If A = 3 5 , B = 7 3 , then find a non-zero matrix C such that AC = BC.

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A = \begin{bmatrix} 3 & 5 \end{bmatrix}

Madhya Pradesh MpbseShort· 3mImportance★★★★★

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Concept understanding — Matrix Equation Solving

Solving a System of Equations by the Matrix Method

A system of linear equations can be written as a single matrix equation and solved in one clean step using the inverse of a matrix. This is the Class-12 "matrix method" for simultaneous equations.

Writing the system as $AX=B$

Take the system

a_1x+b_1y+c_1z=d_1,\quad a_2x+b_2y+c_2z=d_2,\quad a_3x+b_3y+c_3z=d_3.

Collect the coefficients, the unknowns, and the constants into matrices:

A=\begin{pmatrix}a_1&b_1&c_1\\a_2&b_2&c_2\\a_3&b_3&c_3\end{pmatrix},\quad X=\begin{pmatrix}x\\y\\z\end{pmatrix},\quad B=\begin{pmatrix}d_1\\d_2\\d_3\end{pmatrix}.

Then the whole system is just

AX=B.

Solving when $A$ is invertible

If $\det(A)\neq0$ , then $A^{-1}$ exists, and multiplying both sides on the left by $A^{-1}$ gives

X=A^{-1}B,\qquad\text{where }\;A^{-1}=\frac{1}{\det(A)}\operatorname{adj}(A).

So you compute $\det(A)$ , then $\operatorname{adj}(A)$ , form $A^{-1}$ , and multiply by $B$ . The single column $X=A^{-1}B$ hands you $x$ , $y$ , $z$ at once, and because $A^{-1}$ is unique, the solution is unique.

Watch out

Multiply in the correct order: $X=A^{-1}B$ , not $BA^{-1}$ . Matrix multiplication is not commutative, and $BA^{-1}$ is not even defined here.

When $\det(A)=0$

If $\det(A)=0$ , $A^{-1}$ does not exist and the inverse method fails. The system is then either inconsistent (no solution) or has infinitely many solutions. Decide which by computing $(\operatorname{adj}A)\,B$ :

$(\operatorname{adj}A)\,B\neq O$ → no solution (inconsistent).
$(\operatorname{adj}A)\,B=O$ → infinitely many solutions (consistent, dependent). …

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The key idea is that $AC = BC$ is equivalent to $(A-B)C = 0$ , so we need a non-zero matrix $C$ that lies in the nullspace of the row vector $A-B$ . For $A = \begin{bmatrix}3 & 5\end{bmatrix}$ and $B = \begin{bmatrix}7 & 3\end{bmatrix}$ , we get $A-B = \begin{bmatrix}-4 & 2\end{bmatrix}$ , and one such non-zero $C$ is $\begin{bmatrix}1 \\ 2\end{bmatrix}$ .

We are given two row vectors $A$ and $B$ , each of size $1 \times 2$ . The equation $AC = BC$ is a matrix equation where $C$ must be a $2 \times 1$ column vector (so that the multiplication is defined). The goal is to find a non-zero $C$ that satisfies this.

The natural first step is to bring everything to one side:

AC - BC = 0 \quad \Rightarrow \quad (A - B)C = 0

Here $A - B$ is a $1 \times 2$ row vector, and $C$ is a $2 \times 1$ column vector. So $(A-B)C$ is a scalar (a $1 \times 1$ matrix). The equation says: the dot product of the row vector $A-B$ with the column vector $C$ equals zero.

That is, we need a non-zero vector $C$ that is orthogonal (in the usual dot-product sense) to the vector $A-B$ . This is a classic linear algebra problem: find a non-zero vector in the nullspace of a $1 \times 2$ matrix.

Let's compute $A - B$ :

A - B = \begin{bmatrix}3 & 5\end{bmatrix} - \begin{bmatrix}7 & 3\end{bmatrix} = \begin{bmatrix}3-7 & 5-3\end{bmatrix} = \begin{bmatrix}-4 & 2\end{bmatrix}

So the condition becomes:

\begin{bmatrix}-4 & 2\end{bmatrix} \begin{bmatrix}x \\ y\end{bmatrix} = 0 \quad \Rightarrow \quad -4x + 2y = 0

This is a single linear equation in two unknowns. It has infinitely many solutions. We just need one non-zero solution.

From $-4x + 2y = 0$ , we can solve for $y$ in terms of $x$ : $2y = 4x \Rightarrow y = 2x$ .
Choose any non-zero value for $x$ . The simplest is $x = 1$ , which gives $y = 2$ . …

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Q.If $A = \begin{bmatrix} 3 & 5 \end{bmatrix}$ , $B = \begin{bmatrix} 7 & 3 \end{bmatrix}$ , then find a non-zero matrix $C$ such that $AC = BC$ .

Concept understanding — Matrix Equation Solving

Solving a System of Equations by the Matrix Method

Writing the system as $AX=B$

Solving when $A$ is invertible

When $\det(A)=0$

Method: Finding a non-zero $C$ with $AC=BC$

Steps

Common Mistakes

Q.If A=[35]A = \begin{bmatrix} 3 & 5 \end{bmatrix}A=[3​5​], B=[73]B = \begin{bmatrix} 7 & 3 \end{bmatrix}B=[7​3​], then find a non-zero matrix CCC such that AC=BCAC = BCAC=BC.

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Q.If $A = \begin{bmatrix} 3 & 5 \end{bmatrix}$ , $B = \begin{bmatrix} 7 & 3 \end{bmatrix}$ , then find a non-zero matrix $C$ such that $AC = BC$ .

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