Question

If $A=\left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & -1 & 0 \\ 3 & 3 & -4\end{array}\right]$, $B=\left[\begin{array}{l}1 \\ 1 \\ 2\end{array}\right]$ and $X=\left[\begin{array}{l}x_1 \\ x_2 \\ x_3\end{array}\right]$ such that $AX = B$, then the value of $x_1 + x_2 + x_3 =$

• A. 4
• B. 5
• C. 6
• D. 3

Hint:

Always look for quick combinations of your structural rows! If you add 4 times equation (1) directly to equation (3), you get: $4(x_1 - x_2 + x_3) + (3x_1 + 3x_2 - 4x_3) = 4(1) + 2 \implies 7x_1 - x_2 = 6$. Subtracting 3 times equation (2) ($6x_1 - 3x_2 = 3$) from this helps isolate variables without executing a full formal row reduction matrix table!

Answer 1

The Correct Option is D

Step 1: Read the system from $AX = B$.
Multiplying row by row, the matrix equation gives the three equations $x_1 - x_2 + x_3 = 1$, $\;2x_1 - x_2 = 1$, and $3x_1 + 3x_2 - 4x_3 = 2$.
Step 2: A clever shortcut using the column sums.
Notice the answer asks only for $x_1 + x_2 + x_3$. Let us hunt for constants $p, q, r$ so that $p(\text{Eq}1) + q(\text{Eq}2) + r(\text{Eq}3)$ produces $x_1 + x_2 + x_3$ on the left.
Step 3: Solve directly instead (cleaner here).
From Eq2, $x_2 = 2x_1 - 1$. Put this into Eq1: $x_1 - (2x_1 - 1) + x_3 = 1$, which simplifies to $-x_1 + 1 + x_3 = 1$, giving $x_3 = x_1$.
Step 4: Use the third equation.
Substitute $x_2 = 2x_1 - 1$ and $x_3 = x_1$ into Eq3: $3x_1 + 3(2x_1 - 1) - 4x_1 = 2$, i.e. $3x_1 + 6x_1 - 3 - 4x_1 = 2$.
Step 5: Solve for $x_1$.
This gives $5x_1 - 3 = 2$, so $5x_1 = 5$ and $x_1 = 1$.
Step 6: Back-substitute and add.
Then $x_2 = 2(1) - 1 = 1$ and $x_3 = x_1 = 1$. Therefore $x_1 + x_2 + x_3 = 1 + 1 + 1 = 3$.
\[ \boxed{x_1 + x_2 + x_3 = 3} \]