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Exercise A · Q2

Q.(i) A=[0−4310−7220]A = \begin{bmatrix} 0 & -4 & 3 \\ 1 & 0 & -7 \\ 2 & 2 & 0 \end{bmatrix}, write the element a12a_{12}.

(ii) B=[−94−3−104220]B = \begin{bmatrix} -9 & 4 & -3 \\ -1 & 0 & 4 \\ 2 & 2 & 0 \end{bmatrix}, find the sum of elements at b22b_{22} and b32b_{32}.
(iii) C=[−94−3−104220]C = \begin{bmatrix} -9 & 4 & -3 \\ -1 & 0 & 4 \\ 2 & 2 & 0 \end{bmatrix}, find c21+c32−c13c_{21}+c_{32}-c_{13}.
Chandigarh CbseNCERTSubjective· 2mImportance★★★★★
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✓ Free question

Reading the required (row,column)(\text{row},\text{column}) entries gives a12=−4a_{12}=-4, sum =2=2, and the combination =4=4.

The entry xijx_{ij} of a matrix lies in row ii, column jj. Identify each entry then combine as asked.

  1. (i) A=[0−4310−7220]A=\begin{bmatrix}0&-4&3\\1&0&-7\\2&2&0\end{bmatrix}. Element a12a_{12} = row 11, column 22 =−4=-4.
  2. (ii) B=[−94−3−104220]B=\begin{bmatrix}-9&4&-3\\-1&0&4\\2&2&0\end{bmatrix}. b22b_{22} (row 22, col 22) =0=0; b32b_{32} (row 33, col 22) =2=2. Sum =0+2=2=0+2=2.
  3. (iii) C=[−94−3−104220]C=\begin{bmatrix}-9&4&-3\\-1&0&4\\2&2&0\end{bmatrix}. c21=−1c_{21}=-1 (row 22, col 11), c32=2c_{32}=2 (row 33, col 22), c13=−3c_{13}=-3 (row 11, col 33). Then

c21+c32−c13=−1+2−(−3)=−1+2+3=4.c_{21}+c_{32}-c_{13}=-1+2-(-3)=-1+2+3=4.

✓Final answer

  1. a12=−4a_{12}=-4;
  2. b22+b32=2b_{22}+b_{32}=2;
  3. c21+c32−c13=4c_{21}+c_{32}-c_{13}=4.

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