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Exercise 1.2 · Q3

Q.Find the inverse of each of the following by Gauss-Jordan method:

(i) (2−15−2)\begin{pmatrix} 2 & -1 \\ 5 & -2\end{pmatrix}
(ii) (1−1010−16−2−3)\begin{pmatrix} 1 & -1 & 0 \\ 1 & 0 & -1 \\ 6 & -2 & -3\end{pmatrix}
(iii) (123253108)\begin{pmatrix} 1 & 2 & 3 \\ 2 & 5 & 3 \\ 1 & 0 & 8\end{pmatrix}
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For each AA we form [A∣I][A \mid I] and use elementary row operations (applied to the whole augmented row) to drive the left block to II; the right block that results is A−1A^{-1}.

Step 1. Part (i): set up [A∣I][A\mid I]. A=(2−15−2)A=\begin{pmatrix} 2 & -1 \\ 5 & -2\end{pmatrix}.

[2−1105−201]\left[\begin{array}{cc|cc} 2&-1&1&0\\ 5&-2&0&1\end{array}\right]

Step 2. Part (i): clear below the pivot. R2→2R2−5R1=2(5,−2,0,1)−5(2,−1,1,0)=(10,−4,0,2)−(10,−5,5,0)=(0,1,−5,2)R_2 \to 2R_2-5R_1 = 2(5,-2,0,1)-5(2,-1,1,0) = (10,-4,0,2)-(10,-5,5,0) = (0,1,-5,2).

[2−11001−52]\left[\begin{array}{cc|cc} 2&-1&1&0\\ 0&1&-5&2\end{array}\right]

Step 3. Part (i): clear above the second pivot, then normalise row 1. R1→R1+R2=(2,0,−4,2)R_1\to R_1+R_2 = (2,0,-4,2), then R1→R1/2=(1,0,−2,1)R_1\to R_1/2 = (1,0,-2,1).

[10−2101−52]⇒A−1=(−21−52)\left[\begin{array}{cc|cc} 1&0&-2&1\\ 0&1&-5&2\end{array}\right] \Rightarrow A^{-1}=\begin{pmatrix}-2&1\\ -5&2\end{pmatrix}

(Check: ∣A∣=2(−2)−(−1)(5)=−4+5=1≠0|A|=2(-2)-(-1)(5)=-4+5=1\neq0, so AA is indeed invertible.)

Step 4. Part (ii): set up [A∣I][A\mid I]. A=(1−1010−16−2−3)A=\begin{pmatrix} 1 & -1 & 0 \\ 1 & 0 & -1 \\ 6 & -2 & -3\end{pmatrix}.

[1−1010010−10106−2−3001]\left[\begin{array}{ccc|ccc} 1&-1&0&1&0&0\\ 1&0&-1&0&1&0\\ 6&-2&-3&0&0&1\end{array}\right]

Step 5. Part (ii): clear below the first pivot. R2→R2−R1=(0,1,−1,−1,1,0)R_2\to R_2-R_1=(0,1,-1,-1,1,0), R3→R3−6R1=(0,4,−3,−6,0,1)R_3\to R_3-6R_1=(0,4,-3,-6,0,1).

[1−1010001−1−11004−3−601]\left[\begin{array}{ccc|ccc} 1&-1&0&1&0&0\\ 0&1&-1&-1&1&0\\ 0&4&-3&-6&0&1\end{array}\right]

Step 6. Part (ii): clear below the second pivot. R3→R3−4R2=(0,0,−3+4,−6+4,−4,1)=(0,0,1,−2,−4,1)R_3\to R_3-4R_2 = (0,0,-3+4,-6+4,-4,1) = (0,0,1,-2,-4,1).

[1−1010001−1−110001−2−41]\left[\begin{array}{ccc|ccc} 1&-1&0&1&0&0\\ 0&1&-1&-1&1&0\\ 0&0&1&-2&-4&1\end{array}\right]

Step 7. Part (ii): back-substitute to clear above the pivots. R2→R2+R3=(0,1,0,−3,−3,1)R_2\to R_2+R_3=(0,1,0,-3,-3,1); then R1→R1+R2 new=(1,0,0,1−3,0−3,0+1)=(1,0,0,−2,−3,1)R_1\to R_1+R_2^{\,\text{new}}=(1,0,0,1-3,0-3,0+1)=(1,0,0,-2,-3,1) (column 3 of R1R_1 was already 00).

[100−2−31010−3−31001−2−41]⇒A−1=(−2−31−3−31−2−41)\left[\begin{array}{ccc|ccc} 1&0&0&-2&-3&1\\ 0&1&0&-3&-3&1\\ 0&0&1&-2&-4&1\end{array}\right] \Rightarrow A^{-1}=\begin{pmatrix}-2&-3&1\\ -3&-3&1\\ -2&-4&1\end{pmatrix}

(∣A∣=1|A|=1, so A−1A^{-1} exists.)

Step 8. Part (iii): set up [A∣I][A\mid I]. A=(123253108)A=\begin{pmatrix} 1 & 2 & 3 \\ 2 & 5 & 3 \\ 1 & 0 & 8\end{pmatrix}.

[123100253010108001]\left[\begin{array}{ccc|ccc} 1&2&3&1&0&0\\ 2&5&3&0&1&0\\ 1&0&8&0&0&1\end{array}\right]

Step 9. Part (iii): clear below the first pivot. R2→R2−2R1=(0,1,−3,−2,1,0)R_2\to R_2-2R_1=(0,1,-3,-2,1,0), R3→R3−R1=(0,−2,5,−1,0,1)R_3\to R_3-R_1=(0,-2,5,-1,0,1). …

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