If $A = \begin{bmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 3 & 1 \end{bmatrix}$ then reduce it to $I_3$ by using column transformations.

If A = 1 0 0 \2 1 0 \3 3 1 then reduce it to I₃ by using column transformations.

Miscellaneous Exercise 2(A) · Q35

A = \begin{bmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 3 & 1 \end{bmatrix}

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Step 1: $A=\begin{bmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 3 & 1 \end{bmatrix}$ is already lower triangular with $1$ s on the diagonal, so only the off-diagonal entries $2,3,3$ need clearing, using COLUMN operations.

Step 2: Clear the (2,1) entry $=2$ using $C_1\to C_1-2C_2$ (row 1 is unaffected since $a_{12}=0$ , so this cannot disturb the (1,1) entry): column 1 becomes $(1-2(0),\ 2-2(1),\ 3-2(3))=(1,0,-3)$ . Matrix: $\begin{bmatrix}1&0&0\\0&1&0\\-3&3&1\end{bmatrix}$ .

Step 3: Clear the remaining (3,1) entry $=-3$ using $C_1\to C_1+3C_3$ (column 3's row-1 and row-2 entries are $0$ , so this cannot disturb the (1,1) or (2,1) entries just fixed): column 1 becomes $(1+3(0),\ 0+3(0),\ -3+3(1))=(1,0,0)$ . Matrix: $\begin{bmatrix}1&0&0\\0&1&0\\0&3&1\end{bmatrix}$ .

Step 4: Clear the last remaining off-diagonal entry (3,2) $=3$ using $C_2\to C_2-3C_3$ (column 3's row-1, row-2 entries are still $0$ , so column 1 is untouched): column 2 becomes $(0-3(0),\ 1-3(0),\ 3-3(1))=(0,1,0)$ .

Step 5: Final matrix is $\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}=I_3$ .

✓Final answer

A\sim I_3

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Q.If $A = \begin{bmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 3 & 1 \end{bmatrix}$ then reduce it to $I_3$ by using column transformations.

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Q.If $A = \begin{bmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 3 & 1 \end{bmatrix}$ then reduce it to $I_3$ by using column transformations.

Unlock everything free for 14 days