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Exercise D · Q2

Q.Find inverse of the given matrices, by using Adjugate (matrix) method:

(i) [4−132]\begin{bmatrix} 4 & -1 \\ 3 & 2 \end{bmatrix}
(ii) [2−144023−27]\begin{bmatrix} 2 & -1 & 4 \\ 4 & 0 & 2 \\ 3 & -2 & 7 \end{bmatrix}.
Sikkim CbseNCERTSubjective· 5mImportance★★★★★
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Using A−1=1det⁡A adj AA^{-1}=\dfrac{1}{\det A}\,\text{adj }A with adj A=[cofactor]T\text{adj }A=[\text{cofactor}]^T.

A−1=1det⁡A adj AA^{-1}=\dfrac{1}{\det A}\,\text{adj }A, valid when det⁡A≠0\det A\neq0.

(i) A=[4−132]A=\begin{bmatrix} 4 & -1 \\ 3 & 2 \end{bmatrix}

  1. det⁡A=(4)(2)−(−1)(3)=8+3=11.\det A=(4)(2)-(-1)(3)=8+3=11.
  2. adj A=[21−34].\text{adj }A=\begin{bmatrix}2&1\\-3&4\end{bmatrix}.
  3. A−1=111[21−34].A^{-1}=\dfrac{1}{11}\begin{bmatrix}2&1\\-3&4\end{bmatrix}.

(ii) A=[2−144023−27]A=\begin{bmatrix} 2 & -1 & 4 \\ 4 & 0 & 2 \\ 3 & -2 & 7 \end{bmatrix}

  1. det⁡A=2(0⋅7−2⋅(−2))−(−1)(4⋅7−2⋅3)+4(4⋅(−2)−0⋅3)\det A=2(0\cdot7-2\cdot(-2))-(-1)(4\cdot7-2\cdot3)+4(4\cdot(-2)-0\cdot3) =2(4)+1(22)+4(−8)=8+22−32=−2.=2(4)+1(22)+4(-8)=8+22-32=-2.
  2. Cofactors: C11=4,  C12=−22,  C13=−8C_{11}=4,\;C_{12}=-22,\;C_{13}=-8; C21=−1,  C22=2,  C23=1C_{21}=-1,\;C_{22}=2,\;C_{23}=1; C31=−2,  C32=12,  C33=4.C_{31}=-2,\;C_{32}=12,\;C_{33}=4.
  3. adj A=[4−1−2−22212−814]\text{adj }A=\begin{bmatrix}4&-1&-2\\-22&2&12\\-8&1&4\end{bmatrix} (transpose of the cofactor matrix).
  4. A−1=1−2[4−1−2−22212−814]=[−212111−1−64−12−2].A^{-1}=\dfrac{1}{-2}\begin{bmatrix}4&-1&-2\\-22&2&12\\-8&1&4\end{bmatrix}=\begin{bmatrix}-2&\tfrac12&1\\11&-1&-6\\4&-\tfrac12&-2\end{bmatrix}.
  5. Check (row 1 of AA times col 1 of A−1A^{-1}): 2(−2)+(−1)(11)+4(4)=−4−11+16=1.2(-2)+(-1)(11)+4(4)=-4-11+16=1. ✓
✓Final answer

A(i)−1=111[21−34]A^{-1}_{(i)}=\dfrac{1}{11}\begin{bmatrix}2&1\\-3&4\end{bmatrix},   A(ii)−1=[−21/2111−1−64−1/2−2].\;A^{-1}_{(ii)}=\begin{bmatrix}-2&1/2&1\\11&-1&-6\\4&-1/2&-2\end{bmatrix}.

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