CBSE 2023 · Region 3 · Set 1 · Q34 · 5 marks
If $\displaystyle \mathrm{A}=\left[\begin{array}{lll}1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3\end{array}\right]$, then show that $\displaystyle \mathrm{A}^{3}-6 \mathrm{~A}^{2}+7 \mathrm{~A}+2 \mathrm{I}=\mathrm{O}$.If $\displaystyle \mathrm{A}=\left[\begin{array}{cc}3 & 2 \\ 5 & -7\end{array}\right]$, then find $\displaystyle \mathrm{A}^{-1}$ and use it to solve the following system of equations : $\displaystyle 3 \mathrm{x}+5 \mathrm{y}=11,2 \mathrm{x}-7 \mathrm{y}=-3$.
If $\displaystyle \mathrm{A}=\left[\begin{array}{lll}1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3\end{array}\right]$, then show that $\displaystyle \mathrm{A}^{3}-6 \mathrm{~A}^{2}+7 \mathrm{~A}+2 \mathrm{I}=\mathrm{O}$.
If $\displaystyle \mathrm{A}=\left[\begin{array}{cc}3 & 2 \\ 5 & -7\end{array}\right]$, then find $\displaystyle \mathrm{A}^{-1}$ and use it to solve the following system of equations : $\displaystyle 3 \mathrm{x}+5 \mathrm{y}=11,2 \mathrm{x}-7 \mathrm{y}=-3$.
Marking-scheme solution
(a)
getting, $\displaystyle \mathrm{A}^{2}=\left[\begin{array}{ccc} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{array}\right]$
$$\text { getting, } \mathrm{A}^{$\displaystyle 3$}=\left[\begin{array}{lll}
21 & 0 & 34
12 & 8 & 23
34 & 0 & 55
\end{array}\right]
$$$$
\left[\begin{array}{lll}
21 & 0 & 34
12 & 8 & 23
34 & 0 & 55
\end{array}\right]-\left[\begin{array}{ccc}
30 & 0 & 48
12 & 24 & 30
48 & 0 & 78
\end{array}\right]+\left[\begin{array}{ccc}
7 & 0 & 14
0 & 14 & 7
14 & 0 & 21
\end{array}\right]+\left[\begin{array}{lll}
2 & 0 & 0
0 & 2 & 0
0 & 0 & 2
\end{array}\right]
$$(b)adj $\displaystyle \mathrm{A}=\left[\begin{array}{rr} -7 & -2 \\ -5 & 3 \end{array}\right]$
$$\begin{aligned}
& |\mathrm{A}|=-$\displaystyle 31$
& \mathrm{A}^{-$\displaystyle 1$}=\frac{-1}{31}\left[\begin{array}{rr}
-7 & -2
-5 & 3
\end{array}\right]
\end{aligned}
$$Given system of equations is
$$\left\lfloor\begin{array}{rr}
$\displaystyle 3$ & $\displaystyle 5$
$\displaystyle 2$ & -$\displaystyle 7$
\end{array}\right\rfloor\left[\begin{array}{l}
\mathrm{x}
\mathrm{y}
\end{array}\right]=\left[\begin{array}{c}
11
-3
\end{array}\right]
$$which is $\displaystyle \mathrm{A}^{\prime} \mathrm{X}=\mathrm{B}$, where $\displaystyle \mathrm{X}=\left[\begin{array}{l}\mathrm{x} \\
\mathrm{y}\end{array}\right], \mathrm{B}=\left[\begin{array}{c}11 \\
-3\end{array}\right]$
MatricesInvertible MatricesApplylong_answermedium
More from Matrices
- Let both AB^′ and B^′ A be defined for matrices A and B. If order of A is n × m, then the order of B is:2025 · asked 3×
- Given A=[ ] and B=[ ], find AB. Hence, solve the system of linear equations: OR If A=[ ], then find A^-1.…2025 · asked 3×
- Four friends Abhay, Bina, Chhaya and Devesh were asked to simplify 4 AB+3(AB+BA)-4 BA, where A and B are both…2025 · asked 3×
- Three students, Neha, Rani and Sam go to a market to purchase stationery items. Neha buys 4 pens, 3 notepads…2025 · asked 3×
- If inverse of matrix [ ] is the matrix [ ], then value of λ is:2024 · asked 3×
- If A=[ ] and (3 I+4 A)(3 I-4 A)=x^2 I, then the value (s) x is/are:2023 · asked 3×
- If A=[ ], find A^-1 and use it to solve the following system of equations: -2 y+z=7,2 x-y-z=8, x-2 y=10 OR If…2026 · asked 3×
- What is the total number of possible matrices of order 3 × 3 with each entry as √2 or √3?2025 · asked 3×
CBSE Class 12 Mathematics past-paper question from the 2023board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.