Skip to main contentSkip to solution

If [−20x4]+3[y02−1]=I\begin{bmatrix}-2 & 0\\ x & 4\end{bmatrix}+3\begin{bmatrix}y & 0\\ 2 & -1\end{bmatrix}=I, where II is 2×22\times2 unit matrix, then the value of xx and yy are:

Solution

✅ Correct Option: 2

Adding gives [−2+3y0x+61]\begin{bmatrix}-2+3y&0\\x+6&1\end{bmatrix}. Comparing with II: −2+3y=1-2+3y=1 so y=1y=1, and x+6=0x+6=0 so x=−6x=-6. The lower-right entry 4−3=14-3=1 is consistent.

Keyboard Shortcuts

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