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Worked Examples · Example 11

Q.Find [1−1202−33−24][−20193−361−2]\begin{bmatrix} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{bmatrix}\begin{bmatrix} -2 & 0 & 1 \\ 9 & 3 & -3 \\ 6 & 1 & -2 \end{bmatrix}.

Arunachal CbseNCERTSubjective· 3mImportance★★★★★
14% · 11/80 Questions
✓ Free question

Multiply the two 3×33\times 3 matrices row-by-column to get a 3×33\times 3 product.

(PQ)ij=∑k=13pikqkj(PQ)_{ij}=\displaystyle\sum_{k=1}^{3}p_{ik}q_{kj} — entry (i,j)(i,j) = row ii of the first matrix dotted with column jj of the second.

Let P=[1−1202−33−24]P=\begin{bmatrix} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{bmatrix}, Q=[−20193−361−2]Q=\begin{bmatrix} -2 & 0 & 1 \\ 9 & 3 & -3 \\ 6 & 1 & -2 \end{bmatrix}.

  1. Row 1 of PP =(1,−1,2)=(1,-1,2):
    • Col 1: 1(−2)+(−1)(9)+2(6)=−2−9+12=11(-2)+(-1)(9)+2(6)=-2-9+12=1,
    • Col 2: 1(0)+(−1)(3)+2(1)=0−3+2=−11(0)+(-1)(3)+2(1)=0-3+2=-1,
    • Col 3: 1(1)+(−1)(−3)+2(−2)=1+3−4=01(1)+(-1)(-3)+2(-2)=1+3-4=0.
  2. Row 2 of PP =(0,2,−3)=(0,2,-3):
    • Col 1: 0(−2)+2(9)+(−3)(6)=0+18−18=00(-2)+2(9)+(-3)(6)=0+18-18=0,
    • Col 2: 0(0)+2(3)+(−3)(1)=0+6−3=30(0)+2(3)+(-3)(1)=0+6-3=3,
    • Col 3: 0(1)+2(−3)+(−3)(−2)=0−6+6=00(1)+2(-3)+(-3)(-2)=0-6+6=0.
  3. Row 3 of PP =(3,−2,4)=(3,-2,4):
    • Col 1: 3(−2)+(−2)(9)+4(6)=−6−18+24=03(-2)+(-2)(9)+4(6)=-6-18+24=0,
    • Col 2: 3(0)+(−2)(3)+4(1)=0−6+4=−23(0)+(-2)(3)+4(1)=0-6+4=-2,
    • Col 3: 3(1)+(−2)(−3)+4(−2)=3+6−8=13(1)+(-2)(-3)+4(-2)=3+6-8=1.
  4. Assemble:

PQ=[1−100300−21].PQ=\begin{bmatrix} 1 & -1 & 0 \\ 0 & 3 & 0 \\ 0 & -2 & 1 \end{bmatrix}.

✓Final answer

[1−100300−21]\begin{bmatrix} 1 & -1 & 0 \\ 0 & 3 & 0 \\ 0 & -2 & 1 \end{bmatrix}.

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