Let A=[\beginarraylll2 & 0 & 1 1 & 1 & 0 1 & 0 & 1\endarray], B=[B_1, B_2, B_3] , where B_1, B_2, B_3 are column matrics, and AB_1=[\beginarrayl 1 0 0 \endarray], AB_2=[\beginarrayl 2 3 0 \endarray], \quad AB_3=[\beginarrayl 3 2 1 \endarray] If \alpha=|B| and \beta is the sum of all the diagonal elements of B , then \alpha^3+\beta^3 is equal to \underline\hspace2cm .

Matrices and Determinants PYQ — JEE Main 2024 · 27 Jan · Shift 1 · Q57 with Solution

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Correct answer: 28

Interpret the matrix equation

Given $A=\begin{bmatrix}2&0&1\\1&1&0\\1&0&1\end{bmatrix}, \quad B=[B_1,B_2,B_3]$ and $AB_1=\begin{bmatrix}1\\0\\0\end{bmatrix},\quad AB_2=\begin{bmatrix}2\\3\\0\end{bmatrix},\quad AB_3=\begin{bmatrix}3\\2\\1\end{bmatrix}.$

Since multiplication by $A$ acts columnwise, $AB=[AB_1,AB_2,AB_3].$ So, $AB=\begin{bmatrix}1&2&3\\0&3&2\\0&0&1\end{bmatrix}.$ Let this matrix be $C$ . Then $AB=C \implies B=A^{-1}C.$

Find $\alpha=|B|$

Using determinant property, $|AB|=|A||B|.$ Hence, $|B|=\frac{|AB|}{|A|}=\frac{|C|}{|A|}.$

Now, $C=\begin{bmatrix}1&2&3\\0&3&2\\0&0&1\end{bmatrix}$ is upper triangular, so $|C|=1\cdot 3\cdot 1=3.$

Next, $A=\begin{bmatrix}2&0&1\\1&1&0\\1&0&1\end{bmatrix}.$ Expand along the first row: $|A|=2\begin{vmatrix}1&0\\0&1\end{vmatrix}+1\begin{vmatrix}1&1\\1&0\end{vmatrix}$ (because the middle term is zero, and for position $(1,3)$ the sign is positive).

So, $|A|=2(1)-1=1.$ Therefore, $\alpha=|B|=\frac{3}{1}=3.$

Find $\beta$ , the trace of $B$

We need $B=A^{-1}C$ .

First find $A^{-1}$ .

Augment $A$ with identity:

2&0&1&1&0&0\\ 1&1&0&0&1&0\\ 1&0&1&0&0&1 \end{array}\right].$$ Perform row operations: - $R_1 \leftrightarrow R_2$ $$\left[\begin{array}{ccc|ccc} 1&1&0&0&1&0\\ 2&0&1&1&0&0\\ 1&0&1&0&0&1 \end{array}\right]$$ - $R_2\to R_2-2R_1$, $R_3\to R_3-R_1$ $$\left[\begin{array}{ccc|ccc} 1&1&0&0&1&0\\ 0&-2&1&1&-2&0\\ 0&-1&1&0&-1&1 \end{array}\right]$$ - $R_2\to -\frac12 R_2$ $$\left[\begin{array}{ccc|ccc} 1&1&0&0&1&0\\ 0&1&-\frac12&-\frac12&1&0\\ 0&-1&1&0&-1&1 \end{array}\right]$$ - $R_1\to R_1-R_2$, $R_3\to R_3+R_2$ $$\left[\begin{array}{ccc|ccc} 1&0&\frac12&\frac12&0&0\\ 0&1&-\frac12&-\frac12&1&0\\ 0&0&\frac12&-\frac12&0&1 \end{array}\right]$$ - $R_3\to 2R_3$ $$\left[\begin{array}{ccc|ccc} 1&0&\frac12&\frac12&0&0\\ 0&1&-\frac12&-\frac12&1&0\\ 0&0&1&-1&0&2 \end{array}\right]$$ - $R_1\to R_1-\frac12R_3$, $R_2\to R_2+\frac12R_3$ $$\left[\begin{array}{ccc|ccc} 1&0&0&1&0&-1\\ 0&1&0&-1&1&1\\ 0&0&1&-1&0&2 \end{array}\right].$$ Thus, $$A^{-1}=\begin{bmatrix}1&0&-1\\-1&1&1\\-1&0&2\end{bmatrix}.$$ Now, $$B=A^{-1}C= \begin{bmatrix}1&0&-1\\-1&1&1\\-1&0&2\end{bmatrix} \begin{bmatrix}1&2&3\\0&3&2\\0&0&1\end{bmatrix}.$$ Multiply: First row: $$[1,0,-1]\cdot C = [1,2,2]$$ Second row: $$[-1,1,1]\cdot C = [-1,1,0]$$ Third row: $$[-1,0,2]\cdot C = [-1,-2,-1]$$ So, $$B=\begin{bmatrix}1&2&2\\-1&1&0\\-1&-2&-1\end{bmatrix}.$$ Therefore, the sum of diagonal elements is $$\beta=1+1+(-1)=1.$$ --- 4. **Compute** $\alpha^3+\beta^3$ $$\alpha=3,\quad \beta=1$$ so $$\alpha^3+\beta^3=3^3+1^3=27+1=28.$$ --- 5. **Comparison with stored answer** Derived answer = $28$. Stored correct answer = $28$. They match.

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View written solutionFree

Correct answer: 28

Interpret the matrix equation

Since multiplication by $A$ acts columnwise, $AB=[AB_1,AB_2,AB_3].$ So, $AB=\begin{bmatrix}1&2&3\\0&3&2\\0&0&1\end{bmatrix}.$ Let this matrix be $C$ . Then $AB=C \implies B=A^{-1}C.$

Find $\alpha=|B|$

Using determinant property, $|AB|=|A||B|.$ Hence, $|B|=\frac{|AB|}{|A|}=\frac{|C|}{|A|}.$

Now, $C=\begin{bmatrix}1&2&3\\0&3&2\\0&0&1\end{bmatrix}$ is upper triangular, so $|C|=1\cdot 3\cdot 1=3.$

So, $|A|=2(1)-1=1.$ Therefore, $\alpha=|B|=\frac{3}{1}=3.$

Find $\beta$ , the trace of $B$

We need $B=A^{-1}C$ .

First find $A^{-1}$ .

Augment $A$ with identity:

2&0&1&1&0&0\\ 1&1&0&0&1&0\\ 1&0&1&0&0&1 \end{array}\right].$$ Perform row operations: - $R_1 \leftrightarrow R_2$ $$\left[\begin{array}{ccc|ccc} 1&1&0&0&1&0\\ 2&0&1&1&0&0\\ 1&0&1&0&0&1 \end{array}\right]$$ - $R_2\to R_2-2R_1$, $R_3\to R_3-R_1$ $$\left[\begin{array}{ccc|ccc} 1&1&0&0&1&0\\ 0&-2&1&1&-2&0\\ 0&-1&1&0&-1&1 \end{array}\right]$$ - $R_2\to -\frac12 R_2$ $$\left[\begin{array}{ccc|ccc} 1&1&0&0&1&0\\ 0&1&-\frac12&-\frac12&1&0\\ 0&-1&1&0&-1&1 \end{array}\right]$$ - $R_1\to R_1-R_2$, $R_3\to R_3+R_2$ $$\left[\begin{array}{ccc|ccc} 1&0&\frac12&\frac12&0&0\\ 0&1&-\frac12&-\frac12&1&0\\ 0&0&\frac12&-\frac12&0&1 \end{array}\right]$$ - $R_3\to 2R_3$ $$\left[\begin{array}{ccc|ccc} 1&0&\frac12&\frac12&0&0\\ 0&1&-\frac12&-\frac12&1&0\\ 0&0&1&-1&0&2 \end{array}\right]$$ - $R_1\to R_1-\frac12R_3$, $R_2\to R_2+\frac12R_3$ $$\left[\begin{array}{ccc|ccc} 1&0&0&1&0&-1\\ 0&1&0&-1&1&1\\ 0&0&1&-1&0&2 \end{array}\right].$$ Thus, $$A^{-1}=\begin{bmatrix}1&0&-1\\-1&1&1\\-1&0&2\end{bmatrix}.$$ Now, $$B=A^{-1}C= \begin{bmatrix}1&0&-1\\-1&1&1\\-1&0&2\end{bmatrix} \begin{bmatrix}1&2&3\\0&3&2\\0&0&1\end{bmatrix}.$$ Multiply: First row: $$[1,0,-1]\cdot C = [1,2,2]$$ Second row: $$[-1,1,1]\cdot C = [-1,1,0]$$ Third row: $$[-1,0,2]\cdot C = [-1,-2,-1]$$ So, $$B=\begin{bmatrix}1&2&2\\-1&1&0\\-1&-2&-1\end{bmatrix}.$$ Therefore, the sum of diagonal elements is $$\beta=1+1+(-1)=1.$$ --- 4. **Compute** $\alpha^3+\beta^3$ $$\alpha=3,\quad \beta=1$$ so $$\alpha^3+\beta^3=3^3+1^3=27+1=28.$$ --- 5. **Comparison with stored answer** Derived answer = $28$. Stored correct answer = $28$. They match.

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