If [ \beginmatrix 1 & 1 0 & 1 \endmatrix ][ \beginmatrix 1 & 2 0 & 1 \endmatrix ][ \beginmatrix 1 & 3 0 & 1 \endmatrix ] .... [ \beginmatrix 1 & n - 1 0 & 1 \endmatrix ] = [ \beginmatrix 1 & 78 0 & 1 \endmatrix ] , then the inverse of [ \beginmatrix 1 & n 0 & 1 \endmatrix ] is

[ \beginmatrix 1 & - 13 0 & 1 \endmatrix ]

[ \beginmatrix 1 & 0 12 & 1 \endmatrix ]

[ \beginmatrix 1 & 0 13 & 1 \endmatrix ]

[ \beginmatrix 1 & - 12 0 & 1 \endmatrix ]

Matrices and Determinants PYQ — JEE Main 2019 · 9 Apr · Shift 1 · Q30 with Solution

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Correct answer: C

Let $A_k=\begin{bmatrix}1 & k\\ 0 & 1\end{bmatrix}.$ Then the given product is $A_1A_2A_3\cdots A_{n-1}=\begin{bmatrix}1&78\\0&1\end{bmatrix}.$
Use the multiplication rule for such upper triangular matrices:

=\begin{bmatrix}1&a+b\\0&1\end{bmatrix}.$$ So, $$A_1A_2\cdots A_{n-1}=\begin{bmatrix}1&1+2+3+\cdots +(n-1)\\0&1\end{bmatrix}.$$ Hence, $$1+2+3+\cdots +(n-1)=78.$$ 3. Now, $$\frac{(n-1)n}{2}=78.$$ So, $$n(n-1)=156.$$ Solve: $$13\cdot 12=156,$$ therefore $$n=13.$$ 4. We need the inverse of $$\begin{bmatrix}1&n\\0&1\end{bmatrix}=egin{bmatrix}1&13\\0&1\end{bmatrix}.$$ For matrices of the form $$\begin{bmatrix}1&a\\0&1\end{bmatrix},$$ the inverse is $$\begin{bmatrix}1&-a\\0&1\end{bmatrix},$$ because $$\begin{bmatrix}1&a\\0&1\end{bmatrix}\begin{bmatrix}1&-a\\0&1\end{bmatrix} =\begin{bmatrix}1&0\\0&1\end{bmatrix}.$$ Therefore, $$\left(\begin{bmatrix}1&13\\0&1\end{bmatrix}\right)^{-1} =\begin{bmatrix}1&-13\\0&1\end{bmatrix}.$$ 5. Checking options: - A: incorrect - B: incorrect - C: correct - D: incorrect So the correct answer is **C**.

View written solutionFree

Correct answer: C

Let $A_k=\begin{bmatrix}1 & k\\ 0 & 1\end{bmatrix}.$ Then the given product is $A_1A_2A_3\cdots A_{n-1}=\begin{bmatrix}1&78\\0&1\end{bmatrix}.$
Use the multiplication rule for such upper triangular matrices:

=\begin{bmatrix}1&a+b\\0&1\end{bmatrix}.$$ So, $$A_1A_2\cdots A_{n-1}=\begin{bmatrix}1&1+2+3+\cdots +(n-1)\\0&1\end{bmatrix}.$$ Hence, $$1+2+3+\cdots +(n-1)=78.$$ 3. Now, $$\frac{(n-1)n}{2}=78.$$ So, $$n(n-1)=156.$$ Solve: $$13\cdot 12=156,$$ therefore $$n=13.$$ 4. We need the inverse of $$\begin{bmatrix}1&n\\0&1\end{bmatrix}=egin{bmatrix}1&13\\0&1\end{bmatrix}.$$ For matrices of the form $$\begin{bmatrix}1&a\\0&1\end{bmatrix},$$ the inverse is $$\begin{bmatrix}1&-a\\0&1\end{bmatrix},$$ because $$\begin{bmatrix}1&a\\0&1\end{bmatrix}\begin{bmatrix}1&-a\\0&1\end{bmatrix} =\begin{bmatrix}1&0\\0&1\end{bmatrix}.$$ Therefore, $$\left(\begin{bmatrix}1&13\\0&1\end{bmatrix}\right)^{-1} =\begin{bmatrix}1&-13\\0&1\end{bmatrix}.$$ 5. Checking options: - A: incorrect - B: incorrect - C: correct - D: incorrect So the correct answer is **C**.