Skip to main contentSkip to solution

If A=[111102311]A=\begin{bmatrix}1&1&1\\1&0&2\\3&1&1\end{bmatrix} and B=[−2025−2−112−1]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=12x+y+z=6, x+2z=7, 3x+y+z=12 is

Solution

✅ Correct Option: 3

The coefficient matrix is AA, so X=A−1C=14BCX=A^{-1}C=\frac14BC with C=(6,7,12)TC=(6,7,12)^T. Computing BC=(12,4,8)TBC=(12,4,8)^T gives X=(3,1,2)TX=(3,1,2)^T. Check: 3+1+2=63+1+2=6, 3+2(2)=73+2(2)=7, 9+1+2=129+1+2=12.

Keyboard Shortcuts

  • Left arrow: Previous question
  • Right arrow: Next question
  • S key: Jump to solution
  • Q key: Jump to question