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Miscellaneous Exercise 2(A) · Q36

Q.If A=[213101111]A = \begin{bmatrix} 2 & 1 & 3 \\ 1 & 0 & 1 \\ 1 & 1 & 1 \end{bmatrix} then reduce it to I3I_3 by using row transformations.

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Step 1: A=[213101111]A=\begin{bmatrix} 2 & 1 & 3 \\ 1 & 0 & 1 \\ 1 & 1 & 1 \end{bmatrix}. Use R1↔R2R_1\leftrightarrow R_2 to get a leading 11: A∼[101213111]A\sim\begin{bmatrix}1&0&1\\2&1&3\\1&1&1\end{bmatrix}.

Step 2: Clear column 1 below the pivot: R2→R2−2R1R_2\to R_2-2R_1 gives row 2 =(0,1,1)=(0,1,1); R3→R3−R1R_3\to R_3-R_1 gives row 3 =(0,1,0)=(0,1,0). Matrix: [101011010]\begin{bmatrix}1&0&1\\0&1&1\\0&1&0\end{bmatrix}.

Step 3: The (2,2) entry is already 11; clear column 2 elsewhere: R3→R3−R2R_3\to R_3-R_2 gives row 3 =(0,0,−1)=(0,0,-1). Matrix: [10101100−1]\begin{bmatrix}1&0&1\\0&1&1\\0&0&-1\end{bmatrix}.

Step 4: Scale row 3: R3→−R3R_3\to -R_3 gives row 3 =(0,0,1)=(0,0,1). Matrix: [101011001]\begin{bmatrix}1&0&1\\0&1&1\\0&0&1\end{bmatrix}.

Step 5: Clear column 3 above the new pivot: R1→R1−R3R_1\to R_1-R_3 gives row 1 =(1,0,0)=(1,0,0); R2→R2−R3R_2\to R_2-R_3 gives row 2 =(0,1,0)=(0,1,0).

Step 6: Final matrix: [100010001]=I3\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}=I_3.

✓Final answer

A∼I3A\sim I_3 via R1↔R2R_1\leftrightarrow R_2; R2→R2−2R1R_2\to R_2-2R_1, R3→R3−R1R_3\to R_3-R_1; R3→R3−R2R_3\to R_3-R_2; R3→−R3R_3\to-R_3; R1→R1−R3R_1\to R_1-R_3, R2→R2−R3R_2\to R_2-R_3.

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