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Worked Examples · Example 3

Q.Find the minors and cofactors of the elements of the second column of A=(2−13140521)A=\begin{pmatrix} 2 & -1 & 3 \\ 1 & 4 & 0 \\ 5 & 2 & 1 \end{pmatrix}, and hence evaluate ∣A∣|A|.

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Minors of column 2. Deleting the relevant row and column 2 from AA: M12=∣1051∣=1−0=1,M22=∣2351∣=2−15=−13,M32=∣2310∣=0−3=−3.M_{12}=\begin{vmatrix} 1 & 0 \\ 5 & 1 \end{vmatrix}=1-0=1,\quad M_{22}=\begin{vmatrix} 2 & 3 \\ 5 & 1 \end{vmatrix}=2-15=-13,\quad M_{32}=\begin{vmatrix} 2 & 3 \\ 1 & 0 \end{vmatrix}=0-3=-3.

Cofactors of column 2. Using Cij=(−1)i+jMijC_{ij}=(-1)^{i+j}M_{ij}: C12=(−1)1+2(1)=−1,C22=(−1)2+2(−13)=−13,C32=(−1)3+2(−3)=3.C_{12}=(-1)^{1+2}(1)=-1,\qquad C_{22}=(-1)^{2+2}(-13)=-13,\qquad C_{32}=(-1)^{3+2}(-3)=3.

Expansion along column 2. The entries of column 2 are −1,4,2-1, 4, 2, so ∣A∣=(−1)(−1)+(4)(−13)+(2)(3)=1−52+6=−45.|A| = (-1)(-1) + (4)(-13) + (2)(3) = 1 - 52 + 6 = -45.

Verification by expanding along the first row instead. C11=+∣4021∣=4C_{11}=+\begin{vmatrix} 4 & 0 \\ 2 & 1 \end{vmatrix}=4; C12=−∣1051∣=−1C_{12}=-\begin{vmatrix} 1 & 0 \\ 5 & 1 \end{vmatrix}=-1; C13=+∣1452∣=2−20=−18C_{13}=+\begin{vmatrix} 1 & 4 \\ 5 & 2 \end{vmatrix}=2-20=-18. Then ∣A∣=(2)(4)+(−1)(−1)+(3)(−18)=8+1−54=−45|A| = (2)(4)+(-1)(-1)+(3)(-18) = 8+1-54 = -45 — the same value, confirming the column-2 expansion.

✓Final answer

M12=1, M22=−13, M32=−3M_{12}=1,\ M_{22}=-13,\ M_{32}=-3; C12=−1, C22=−13, C32=3C_{12}=-1,\ C_{22}=-13,\ C_{32}=3; and ∣A∣=−45|A| = -45.

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