Скачать или смотреть VTU 1st Sem | Math | Module 3 | Rank of Matrix using Row Operations | Find b if Rank = 3|1BMATS101

Welcome to Express VTU 4 All — your go-to channel for clear, step-by-step solutions of VTU questions.
In this video we solve a key problem from Module 3 — Rank of a Matrix (1BMATS101). We find the value(s) of the constant b for which the matrix has rank = 3, using elementary row transformations and determinant reasoning.
🔹 Question (given)

Find the constant b if the rank of

A=\begin{bmatrix}
1 & 1 & -1 & 0\\[4pt]
4 & 4 & -3 & 1\\[4pt]
b & 2 & 2 & 2\\[4pt]
9 & 3 & b & 3
\end{bmatrix}


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🔹 Strategy (short)

For a 4×4 matrix to have rank 3, its determinant must be zero (so rank ≤ 3) and at least one 3×3 minor must be nonzero (so rank = 3, not less). We simplify the determinant by using elementary row operations to reduce the first column to [1,0,0,0] and then compute the determinant of the remaining 3×3 block.


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🔹 Step-by-step solution (exam style)

1. Use row operations that do not change determinant up to sign (here we use row replacement, which preserves determinant):

R2 → R2 − 4 R1
R3 → R3 − b R1
R4 → R4 − 9 R1

After these operations the matrix becomes



\begin{bmatrix}
1 & 1 & -1 & 0\\[4pt]
0 & 0 & 1 & 1\\[4pt]
0 & 2-b & b+2 & 2\\[4pt]
0 & -6 & b+9 & 3
\end{bmatrix}.

Because the first column is now [1,0,0,0], the determinant of A equals the determinant of the lower 3×3 block formed by columns 2–4 and rows 2–4.

2. Compute determinant of the 3×3 block



M=\begin{bmatrix}
0 & 1 & 1\\
2-b & b+2 & 2\\
-6 & b+9 & 3
\end{bmatrix}.

\det(A)=\det(M) = -b^{2} + 2b + 12.

3. For rank ≤ 3 we require det(A) = 0:



-b^{2} + 2b + 12 = 0 \quad\Longrightarrow\quad b^{2} - 2b - 12 = 0.

b = 1 \pm \sqrt{13}.

4. Finally, verify rank is exactly 3 (not less): substitute either root into A and compute rank (or check at least one 3×3 minor ≠ 0). Substitution shows the matrix reduces to an echelon form with exactly three nonzero rows, so rank = 3 for both values.




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✅ Final Answer

\boxed{\,b = 1 - \sqrt{13}\quad\text{or}\quad b = 1 + \sqrt{13}\,}


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🔍 Why this matters

Rank questions are heavily tested in VTU Module 3 (Systems of Linear Equations).

This method (zeroing a column to reduce determinant computation to a 3×3) is a neat exam trick — fast and less error-prone.

Understanding rank is essential for solving linear systems, consistency checks, and applications in linear algebra.



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💡 Exam tips

Always try simple row/column operations to simplify determinant calculation.

After det = 0, check at least one 3×3 minor (or compute rank) to confirm rank exactly 3.

Write your row operations clearly in the answer — marks are awarded for method.



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VTU 1st sem maths rank of matrix
1BMATS101 rank using row operations
VTU Module 3 rank determinant PYQ
Rank of matrix find b VTU
Elementary row transformations VTU solutions



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“Welcome back to Express VTU 4 All. In this video we find the values of b that make a given 4 by 4 matrix have rank 3. Instead of expanding a 4×4 determinant directly, we use quick row operations to zero the first column below the pivot — so the determinant reduces to a 3×3 block. Solving the resulting quadratic gives b = 1 ± sqrt(13). Finally we verify the rank is exactly 3 by checking the reduced form. Watch the full steps for the detailed row operations and the verification step — and don’t forget to like and subscribe for more VTU PYQ solutions.”