If \( \mathrm{A}=\begin{bmatrix}1 & 1 & 1\\1 & 0 & 2\\3 & 1 & 1\end{bmatrix} \) and \( \mathrm{B}=\begin{bmatrix}-2 & 0 & 2\\5 & -2 & -1\\1 & 2 & -1\end{bmatrix} \) such that AB=4I (where I is identity matrix) then solution of system of equations x+y+z=6, x+2z=7, 3x+y+z=12 is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → x=3,y=1,z=2
The system is AX = C with the given A:
\( \displaystyle \begin{bmatrix}1 & 1 & 1\\1 & 0 & 2\\3 & 1 & 1\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}6\\7\\12\end{bmatrix} \)
From \( AB=4I \), \( A\left(\frac{1}{4}B\right)=I \), so the inverse comes for free: \( A^{-1}=\frac{1}{4}B \).
\( \displaystyle X=A^{-1}C \)
\( \displaystyle =\frac{1}{4}\begin{bmatrix}-2 & 0 & 2\\5 & -2 & -1\\1 & 2 & -1\end{bmatrix}\begin{bmatrix}6\\7\\12\end{bmatrix} \)
\( \displaystyle =\frac{1}{4}\begin{bmatrix}-12+0+24\\30-14-12\\6+14-12\end{bmatrix} \)
\( \displaystyle =\frac{1}{4}\begin{bmatrix}12\\4\\8\end{bmatrix} \)
\( \displaystyle =\begin{bmatrix}3\\1\\2\end{bmatrix} \)
So x = 3, y = 1, z = 2.
Shortcut: Substitute the options: x = 3, y = 1, z = 2 gives 3 + 1 + 2 = 6, 3 + 4 = 7 and 9 + 1 + 2 = 12. Options (2) and (4) fail x + 2z = 7 (they give 4 and 5), and Option (1) fails 3x + y + z = 12 (it gives 8).