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Exercise B · Q5

Q.Do as directed:

(i) For A=[6−5−74]A = \begin{bmatrix} 6 & -5 \\ -7 & 4 \end{bmatrix}, find A2−6AA^2 - 6A.
(ii) Evaluate [213][−10−1−110011][10−1]\begin{bmatrix} 2 & 1 & 3 \end{bmatrix}\begin{bmatrix} -1 & 0 & -1 \\ -1 & 1 & 0 \\ 0 & 1 & 1 \end{bmatrix}\begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix}.
(iii) Find a matrix AA such that [12−1049]=[9−14−213]−A\begin{bmatrix} 1 & 2 & -1 \\ 0 & 4 & 9 \end{bmatrix} = \begin{bmatrix} 9 & -1 & 4 \\ -2 & 1 & 3 \end{bmatrix} - A.
(iv) If the matrix X=[−15013]X = \begin{bmatrix} -1 & 5 \\ 0 & 13 \end{bmatrix} is equal to the matrix Y=[p−q2p+r2p−q3r+s]Y = \begin{bmatrix} p-q & 2p+r \\ 2p-q & 3r+s \end{bmatrix} then find value of pp, qq, rr and ss.
(v) Let P=[2−131]P = \begin{bmatrix} 2 & -1 \\ 3 & 1 \end{bmatrix} and Q=[1472]Q = \begin{bmatrix} 1 & 4 \\ 7 & 2 \end{bmatrix} then calculate 3P−2Q3P-2Q.
Andaman Nicobar CbseNCERTSubjective· 5mImportance★★★★★
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Five short matrix computations: a polynomial in AA, a row-matrix product, a matrix subtraction, an equality of matrices, and a linear combination.

Standard operations: A2=A⋅AA^2=A\cdot A; scalar multiple (kA)ij=k aij(kA)_{ij}=k\,a_{ij}; matrix equality means corresponding entries are equal.

(i) A2−6AA^2-6A for A=[6−5−74]A=\begin{bmatrix}6&-5\\-7&4\end{bmatrix}

  1. A2=[6(6)+(−5)(−7)6(−5)+(−5)(4)(−7)(6)+4(−7)(−7)(−5)+4(4)]=[71−50−7051]A^2=\begin{bmatrix}6(6)+(-5)(-7)&6(-5)+(-5)(4)\\(-7)(6)+4(-7)&(-7)(-5)+4(4)\end{bmatrix}=\begin{bmatrix}71&-50\\-70&51\end{bmatrix}.
  2. 6A=[36−30−4224]6A=\begin{bmatrix}36&-30\\-42&24\end{bmatrix}.
  3. A2−6A=[71−36−50+30−70+4251−24]=[35−20−2827]A^2-6A=\begin{bmatrix}71-36&-50+30\\-70+42&51-24\end{bmatrix}=\begin{bmatrix}35&-20\\-28&27\end{bmatrix}.

(ii) [213][−10−1−110011][10−1]\begin{bmatrix}2&1&3\end{bmatrix}\begin{bmatrix}-1&0&-1\\-1&1&0\\0&1&1\end{bmatrix}\begin{bmatrix}1\\0\\-1\end{bmatrix}

  1. First [213][−10−1−110011]=[−341]\begin{bmatrix}2&1&3\end{bmatrix}\begin{bmatrix}-1&0&-1\\-1&1&0\\0&1&1\end{bmatrix}=\begin{bmatrix}-3&4&1\end{bmatrix} (e.g. 2(−1)+1(−1)+3(0)=−32(-1)+1(-1)+3(0)=-3).
  2. Then [−341][10−1]=−3(1)+4(0)+1(−1)=−4\begin{bmatrix}-3&4&1\end{bmatrix}\begin{bmatrix}1\\0\\-1\end{bmatrix}=-3(1)+4(0)+1(-1)=-4.

(iii) Find AA from [12−1049]=[9−14−213]−A\begin{bmatrix}1&2&-1\\0&4&9\end{bmatrix}=\begin{bmatrix}9&-1&4\\-2&1&3\end{bmatrix}-A

  1. Rearrange: A=[9−14−213]−[12−1049]=[8−35−2−3−6]A=\begin{bmatrix}9&-1&4\\-2&1&3\end{bmatrix}-\begin{bmatrix}1&2&-1\\0&4&9\end{bmatrix}=\begin{bmatrix}8&-3&5\\-2&-3&-6\end{bmatrix}.

(iv) Solve [−15013]=[p−q2p+r2p−q3r+s]\begin{bmatrix}-1&5\\0&13\end{bmatrix}=\begin{bmatrix}p-q&2p+r\\2p-q&3r+s\end{bmatrix}

  1. Equate entries: p−q=−1, 2p+r=5, 2p−q=0, 3r+s=13p-q=-1,\ 2p+r=5,\ 2p-q=0,\ 3r+s=13. …

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