Q.The sum of three numbers is 6. If we multiply third number by 3 and add second number to it, we get 11. By adding first and third numbers, we get double of the second number. Represent it algebraically and find the numbers using matrix method.

Concept: Matrix Equation Solving — representing a system of linear equations as $AX = B$ and solving via $X = A^{-1}B$.

Let the three numbers be $x$, $y$, $z$.

From the problem:

1. $x + y + z = 6$
2. $3z + y = 11 \quad \Rightarrow \quad y + 3z = 11$
3. $x + z = 2y \quad \Rightarrow \quad x - 2y + z = 0$

In matrix form $AX = B$:

Find $A^{-1}$ (or use row reduction). Determinant of $A$:

Adjugate matrix gives:

Then $X = A^{-1}B$: …

We have three unknown numbers. Let them be $x$, $y$, and $z$ — first, second, and third respectively. The problem gives three conditions, each translating directly into an equation.

The first condition: "The sum of three numbers is 6" gives

The second: "If we multiply third number by 3 and add second number to it, we get 11" means

The third: "By adding first and third numbers, we get double of the second number" means

So the system is:

Why use the matrix method? Because once we write this as $AX = B$, solving $X = A^{-1}B$ is systematic — no back-substitution guesswork, and it works cleanly for any $3 \times 3$ system with a unique solution.

Step 1: Write in matrix form $AX = B$

Step 2: Find the determinant of $A$ to check if the inverse exists

Compute each minor:

• First: $(1)(1) - (3)(-2) = 1 + 6 = 7$
• Second: $(0)(1) - (3)(1) = 0 - 3 = -3$
• Third: $(0)(-2) - (1)(1) = 0 - 1 = -1$

So $\det(A) = 1(7) - 1(-3) + 1(-1) = 7 + 3 - 1 = 9$.

Since $\det(A) = 9 \neq 0$, the inverse exists and the solution is unique.

Step 3: Find the adjoint of $A$ (transpose of the cofactor matrix)

First, find all cofactors $C_{ij} = (-1)^{i+j} M_{ij}$.

• $C_{11} = + \begin{vmatrix} 1 & 3 \\ -2 & 1 \end{vmatrix} = 7$
• $C_{12} = - \begin{vmatrix} 0 & 3 \\ 1 & 1 \end{vmatrix} = -(-3) = 3$
• $C_{13} = + \begin{vmatrix} 0 & 1 \\ 1 & -2 \end{vmatrix} = -1$
• $C_{21} = - \begin{vmatrix} 1 & 1 \\ -2 & 1 \end{vmatrix} = -(1 + 2) = -3$
• $C_{22} = + \begin{vmatrix} 1 & 1 \\ 1 & 1 \end{vmatrix} = 0$
• $C_{23} = - \begin{vmatrix} 1 & 1 \\ 1 & -2 \end{vmatrix} = -(-2 - 1) = 3$
• $C_{31} = + \begin{vmatrix} 1 & 1 \\ 1 & 3 \end{vmatrix} = 3 - 1 = 2$
• $C_{32} = - \begin{vmatrix} 1 & 1 \\ 0 & 3 \end{vmatrix} = -(3 - 0) = -3$
• $C_{33} = + \begin{vmatrix} 1 & 1 \\ 0 & 1 \end{vmatrix} = 1 - 0 = 1$

So the cofactor matrix is:

$\begin{pmatrix} 7 & 3 & -1 \\ -3 & 0 & 3 \\ 2 & -3 & 1 \end{pmatrix}$ …

Method: Translating a Word Problem into a Linear System, Then Solving by the Matrix Method

This method turns a worded description of relationships between unknown numbers into a system of linear equations, then applies the standard matrix (adjoint) method to solve it.

Steps

Step 1: Name the unknowns

Assign a variable to each unknown quantity described in the problem (e.g. $x,y,z$ for "first number," "second number," "third number").

Step 2: Translate each sentence into one equation

Go through the problem sentence by sentence. Each stated relationship ("the sum is...", "multiply ... and add...", "adding ... gives double...") becomes exactly one linear equation — translate literally, phrase by phrase, rather than trying to guess the final system from memory.

Step 3: Collect the equations and write them in standard form

Rearrange each equation so all variable terms are on one side and the constant is on the other, matching the standard $a_1x+b_1y+c_1z=d_1$ pattern.

Step 4: Package as $AX=B$ and solve via the adjoint method

Follow the standard matrix method: compute $\det(A)$, find all cofactors, transpose to get $\operatorname{adj}(A)$, then $X=\dfrac{1}{\det(A)}\operatorname{adj}(A)\,B$.

Step 5: Translate the numeric solution back into the word problem's language …

Common Mistakes

Mistake 1: Mistranslating a sentence into the wrong equation

Why it's wrong: "if we multiply the third number by 3 and add the second number to it, we get 11" translates to $3z+y=11$ — but it's easy to instead write $3(y+z)=11$ or $y+3z=13$ if the sentence is read too quickly. Correct approach: translate phrase by phrase, writing down each operation ("multiply third by 3," "add second," "get 11") as it's read, rather than trying to write the whole equation from memory of the sentence.

Mistake 2: A sign error in one of the nine cofactors during the adjoint computation, given three unknowns …