Q.Use the product of matrices $\begin{bmatrix} 1 & 1 & 2 \\ 0 & 2 & 3 \\ 3 & 2 & 4 \end{bmatrix} \begin{bmatrix} 2 & 0 & 1 \\ 9 & 2 & 3 \\ 6 & 1 & 2 \end{bmatrix}$ to solve the following system of equations: $x - y + 2z = 1$, $2y - 3z = 1$, $3x - 2y + 4z = 2$.

The two matrices given are meant to be inverses of each other (their product should be I); using the second as A⁻¹ and multiplying by the system's constant column solves the equations directly.

A transcription note: the printed first matrix $\begin{bmatrix}1&1&2\\0&2&3\\3&2&4\end{bmatrix}$ does not match the sign pattern of the system's own coefficient matrix — the equations $x-y+2z=1,\ 2y-3z=1,\ 3x-2y+4z=2$ have coefficient matrix $A=\begin{bmatrix}1&-1&2\\0&2&-3\\3&-2&4\end{bmatrix}$ (with minus signs that the scan appears to have dropped). Using A together with $B=\begin{bmatrix}-2&0&1\\9&2&-3\\6&1&-2\end{bmatrix}$ (the corresponding corrected second matrix) gives $AB=I$ exactly — confirming these are the intended matrices and that $B=A^{-1}$. I've solved using these corrected signs, since they are the only version consistent with both the multiplication trick and the system itself.

Verify AB = I:

Row 1 of A times each column of B: $[1,-1,2]\cdot[-2,9,6]=-2-9+12=1$; $[1,-1,2]\cdot[0,2,1]=0-2+2=0$; $[1,-1,2]\cdot[1,-3,-2]=1+3-4=0$ → row (1,0,0)

Row 2: $[0,2,-3]\cdot[-2,9,6]=0+18-18=0$; $[0,2,-3]\cdot[0,2,1]=0+4-3=1$; $[0,2,-3]\cdot[1,-3,-2]=0-6+6=0$ → row (0,1,0)

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