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Exercise 1.2 · Q1

Q.Find the rank of the following matrices by minor method:

(i) (2−4−12)\begin{pmatrix} 2 & -4 \\ -1 & 2\end{pmatrix}
(ii) (−134−73−4)\begin{pmatrix} -1 & 3 \\ 4 & -7 \\ 3 & -4\end{pmatrix}
(iii) (1−2−103−6−31)\begin{pmatrix} 1 & -2 & -1 & 0 \\ 3 & -6 & -3 & 1\end{pmatrix}
(iv) (1−2324−651−1)\begin{pmatrix} 1 & -2 & 3 \\ 2 & 4 & -6 \\ 5 & 1 & -1\end{pmatrix}
(v) (012102438102)\begin{pmatrix} 0 & 1 & 2 & 1 \\ 0 & 2 & 4 & 3 \\ 8 & 1 & 0 & 2\end{pmatrix}
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For each matrix we test the highest-order minor its shape permits (min⁡(m,n)\min(m,n) for an m×nm\times n matrix). If that minor is 00, we drop to the next lower order and keep testing until a nonzero minor appears; that order is the rank ρ(A)\rho(A).

Step 1. Part (i): test the only 2×22\times2 minor. A=(2−4−12)A=\begin{pmatrix} 2 & -4 \\ -1 & 2\end{pmatrix} is itself 2×22\times2, so the highest possible order is 22.

∣A∣=2(2)−(−4)(−1)=4−4=0.|A| = 2(2)-(-4)(-1) = 4-4 = 0.

The order-22 minor vanishes, so we drop to order 11.

Step 2. Part (i): test an order-11 minor. The entry a11=2≠0a_{11}=2\neq0, so a nonzero 1×11\times1 minor exists. Hence ρ(A)=1\rho(A)=1.

Step 3. Part (ii): test an order-22 minor. A=(−134−73−4)A=\begin{pmatrix} -1 & 3 \\ 4 & -7 \\ 3 & -4\end{pmatrix} is 3×23\times2, so the highest possible order is min⁡(3,2)=2\min(3,2)=2. Take the minor formed by rows 1,21,2:

∣−134−7∣=(−1)(−7)−(3)(4)=7−12=−5≠0.\begin{vmatrix} -1 & 3 \\ 4 & -7\end{vmatrix} = (-1)(-7)-(3)(4) = 7-12 = -5 \neq 0.

A nonzero order-22 minor exists, so ρ(A)=2\rho(A)=2 (no need to check the other row pairs).

Step 4. Part (iii): test an order-22 minor. A=(1−2−103−6−31)A=\begin{pmatrix} 1 & -2 & -1 & 0 \\ 3 & -6 & -3 & 1\end{pmatrix} is 2×42\times4, so the highest possible order is 22. Take the minor formed by columns 1,41,4:

∣1031∣=1(1)−0(3)=1≠0.\begin{vmatrix} 1 & 0 \\ 3 & 1\end{vmatrix} = 1(1)-0(3) = 1 \neq 0.

A nonzero order-22 minor exists, so ρ(A)=2\rho(A)=2.

Step 5. Part (iv): test the order-33 minor (the whole determinant). A=(1−2324−651−1)A=\begin{pmatrix} 1 & -2 & 3 \\ 2 & 4 & -6 \\ 5 & 1 & -1\end{pmatrix} is 3×33\times3, so the highest possible order is 33. Expanding along row 1:

∣A∣=1∣4−61−1∣−(−2)∣2−65−1∣+3∣2451∣|A| = 1\begin{vmatrix}4&-6\\1&-1\end{vmatrix} -(-2)\begin{vmatrix}2&-6\\5&-1\end{vmatrix} +3\begin{vmatrix}2&4\\5&1\end{vmatrix}

=1(−4+6)+2(−2+30)+3(2−20)=1(2)+2(28)+3(−18)=2+56−54=4.= 1(-4+6) +2(-2+30) +3(2-20) = 1(2)+2(28)+3(-18) = 2+56-54 = 4.

Since ∣A∣=4≠0|A|=4\neq0, ρ(A)=3\rho(A)=3.

Step 6. Part (v): test an order-33 minor. A=(012102438102)A=\begin{pmatrix} 0 & 1 & 2 & 1 \\ 0 & 2 & 4 & 3 \\ 8 & 1 & 0 & 2\end{pmatrix} is 3×43\times4, so the highest possible order is 33. Try the minor from columns 1,2,31,2,3:

∣012024810∣=0(⋅)−1(0⋅0−4⋅8)+2(0⋅1−2⋅8)=0−1(−32)+2(−16)=32−32=0.\begin{vmatrix} 0&1&2\\0&2&4\\8&1&0\end{vmatrix} = 0(\cdot)-1(0\cdot0-4\cdot8)+2(0\cdot1-2\cdot8) = 0-1(-32)+2(-16) = 32-32=0.

This one vanishes, so try columns 1,2,41,2,4 instead:

∣011023812∣=0(⋅)−1(0⋅2−3⋅8)+1(0⋅1−2⋅8)=−1(−24)+1(−16)=24−16=8≠0.\begin{vmatrix} 0&1&1\\0&2&3\\8&1&2\end{vmatrix} = 0(\cdot)-1(0\cdot2-3\cdot8)+1(0\cdot1-2\cdot8) = -1(-24)+1(-16) = 24-16=8 \neq 0.

A nonzero order-33 minor exists, so ρ(A)=3\rho(A)=3.

✓Final answer

(i) ρ(A)=1\rho(A)=\boxed{1}; (ii) ρ(A)=2\rho(A)=\boxed{2}; (iii) ρ(A)=2\rho(A)=\boxed{2}; (iv) ρ(A)=3\rho(A)=\boxed{3}; (v) ρ(A)=3\rho(A)=\boxed{3}.

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