31 votes 31 votes Let $P = \begin{bmatrix}1 & 1 & -1 \\2 & -3 & 4 \\3 & -2 & 3\end{bmatrix}$ and $Q = \begin{bmatrix}-1 & -2 &-1 \\6 & 12 & 6 \\5 & 10 & 5\end{bmatrix}$ be two matrices. Then the rank of $ P+Q$ is ___________ . Linear Algebra gatecse-2017-set2 linear-algebra eigen-value numerical-answers + – Arjun asked Feb 14, 2017 • edited Feb 15, 2017 by mcjoshi Arjun 11.8k views answer comment Share Follow See all 2 Comments See all 2 2 Comments reply ayush.5 commented Oct 12, 2020 reply Follow Share 1st column of P+Q is linear combination of 2nd and 3rd column as col1=2*col2-col3. So rank cannot be 3 3 votes 3 votes Souvik33 commented Dec 7, 2022 reply Follow Share Anyone who tried to solve usingRank(A+B) = Rank(A)+Rank(B) and got answer as 3, the problem is it is not an equality equation.The equation is:Rank(A+B) ≤ Rank(A)+Rank(B) and 2 is clearly less than 3 2 votes 2 votes Please log in or register to add a comment.
Best answer 52 votes 52 votes $P +Q = \begin{bmatrix}0 & -1 &-2 \\8 & 9 & 10 \\8 & 8 & 8\end{bmatrix}$ $\det(P+Q) = 0$, So Rank cannot be $3$, but there exists a $2*2$ submatrix such that determinant of submatrix is not $0$. So, $\text{Rank}(P+Q) = 2$ mcjoshi answered Feb 14, 2017 • edited Jun 19, 2021 by Lakshman Bhaiya mcjoshi comment Share Follow See all 9 Comments See all 9 9 Comments reply Ayush Upadhyaya commented Jul 14, 2018 reply Follow Share $P+Q=$$\begin{bmatrix} 0 & -1 & -2\\ 8&9 &10 \\ 8& 8 &8 \end{bmatrix}$ Do $R_3=R_3-R_2$ $\begin{bmatrix} 0 & -1 & -2\\ 8&9 &10 \\ 0& -1 &-2 \end{bmatrix}$ Now number of linearly Independent rows=2. Hence rank=2 15 votes 15 votes jatin khachane 1 commented Nov 18, 2018 reply Follow Share @Ayush Upadhyaya Which of these are linearly independent ?? Does Linearly independent means ..one row can't be expressed as linear combination of any other ??? 0 votes 0 votes Lakshman Bhaiya commented Nov 18, 2018 i edited by Lakshman Bhaiya Nov 29, 2019 reply Follow Share @ jatin khachane 1 $P+Q=$$\begin{bmatrix} 0 & -1 & -2\\ 8&9 &10 \\ 8& 8 &8 \end{bmatrix}$ Do $R_3\rightarrow R_3-R_2$ $P+Q=\begin{bmatrix} 0 & -1 & -2\\ 8&9 &10 \\ 0& -1 &-2 \end{bmatrix}$ Again do $R_3\rightarrow R_3-R_1$ $P+Q=\begin{bmatrix} 0 & -1 & -2\\ 8&9 &10 \\ 0& 0 &0 \end{bmatrix}$ Now the number of linearly Independent rows=2. Hence rank=2 Which of these are linearly independent ?? $R_{1}$ and $R_{2}$ are linearly independent. Here clearly see $R_{3}$ is linearly dependent on $R_{1}$ (or) $R_{1}$ is linearly dependent on $R_{3}.$ $\implies$ One relation gives only one dependent. $(OR)$ Again do $R_1\rightarrow R_1-R_3$ $P+Q=\begin{bmatrix} 0 & 0 & 0\\ 8&9 &10 \\ 0& -1 &-2 \end{bmatrix}$ Now the number of linearly Independent rows=2. Hence rank=2 Does Linearly independent means ..one row can't be expressed as linear combination of any other ??? Yes, this is true. 6 votes 6 votes jatin khachane 1 commented Nov 18, 2018 reply Follow Share Can we say that...number of non-zero rows ..are equal to no of independent rows ?? 1 votes 1 votes Lakshman Bhaiya commented Nov 18, 2018 reply Follow Share see this $A=\begin{bmatrix} 1 &2 &3 \\ 4 &5 &6 \\ 3 &6 &9 \end{bmatrix}$ Here the number of Non-zero rows are $3$ Now, we can do some operation on matrix $A$ $R_{3}\rightarrow R_{3}-3R_{1}$ $A=\begin{bmatrix} 1 &2 &3 \\ 4 &5 &6 \\ 3-3 &6-6 &9-9 \end{bmatrix}$ $A=\begin{bmatrix} 1 &2 &3 \\ 4 &5 &6 \\ 0 &0 &0 \end{bmatrix}$ Here the number of independent rows are $2$ But after do some operation and if non-zero rows are independent,then we can say that non-zero rows are independent rows. 1 votes 1 votes Amal commented Jan 29, 2020 reply Follow Share jatin khachane 1 The rank of a matrix is defined as (a) the maximum number of linearly independent column vectors in the matrix or (b) the maximum number of linearly independent row vectors in the matrix. Now, What does "making a row zero means" ? Actual logic behind the making of a row zero is , We are checking whether that row can be expressed as linear combination of other rows. In the above comment, Lakshman made last row as zero by the operation $R_{3}\rightarrow R_{3}- 3R_{1}$. [ It means, $R_{3} = 3R_{1} + 0R_{2}$ that means we can represent $3^{rd}$ row as linear combination of $1^{st}$ row and $2^{nd}$ row ]. So $R_{3}$ is linearly dependent on $R_{1}$ and $R_{2}$. But we were not able to make $1^{st}$ and $2^{nd}$ row zero. Which means these rows can't be expressed as linear combination of other rows. ie These two rows are Linearly independent. So in the above comment example, number of linearly independent rows =2, Hence rank =2 9 votes 9 votes shashankrustagi commented Dec 5, 2020 reply Follow Share 1st and 2nd or 3rd and 2nd 0 votes 0 votes abir_banerjee commented Oct 9, 2022 reply Follow Share Rank of matrix tells about a lot of things :- Number of linear independent rows . Number of linear independent columns. Number of non zero rows One thing to note is that number of linear independent rows are equal to number of linear independent columns. 2 votes 2 votes Nandhakumar commented Oct 12, 2023 reply Follow Share Rank also tells the number of non zero eigen values right? 0 votes 0 votes Please log in or register to add a comment.
7 votes 7 votes $P +Q = \begin{bmatrix}0 & -1 &-2 \\8 & 9 & 10 \\8 & 8 & 8\end{bmatrix}$ Applying $R_2 \leftarrow R_2+R_1$, we get $P +Q = \begin{bmatrix}0 & -1 &-2 \\8 & 8 & 8 \\8 & 8 & 8\end{bmatrix}$ Since there are only $2$ independent rows $\implies Rank(P+Q) = 2$ Satbir answered Oct 23, 2019 Satbir comment Share Follow See all 0 reply Please log in or register to add a comment.